Class 12 Chemistry

Biomolecules

Question 1. Glucose or succrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:

• A glucose molecule contains five -OH groups while a sucrose molecule contains eight -OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water.
• Hence, they are soluble in water.
• But, cyclohexane and benzene do not contain -OH groups. Hence, they cannot undergo II- bonding with water and as a result, they are insoluble in water.

Question 2. What are the expected products of the hydrolysis of lactose?
Answer:

Lactose is composed of β-D galactose and β-D glucose. Tims, on hydrolysis, gives β-D galactose and β-D glucose.

Question 3. How do you explain the absence of an aldehyde group in the pentaacctate of D-glucose?
Answer:

D-glucose reacts with hydroxylamine (NH₂, OH) to form an oxime because of the presence of an aldehydic (-CHO) group of carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH₂, OH to give an oxime.

But pentaacctate of D-glucose does not react with NH₂, OH. This is because pentaacctate does not form an open chain structure.

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Question 4. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer:

Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept, thus giving rise to a dipolar ion known as a zwitter ion.

• Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. However, halo-acids do not exhibit such dipolar behaviour.
• For this reason, the melting points and the solubility of amino acids in water are higher than those of the corresponding halo acids.

Question 5. Where does the water present in the egg go after boiling the egg?
Answer:

When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg. the water present in it is absorbed by the coagulated protein through H-bonding.

Question 6. Why cannot vitamin C be stored in our bodies?
Answer: Vitamin C cannot be stored in our body because it is water-soluble. As a result, it is readily excreted in the urine.

Question 7. What products would be formed when a nucleotide from DMA containing thymine is hydrolysed?
Answer: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine. β-D-2- deoxyribose and phosphoric acid are obtained as products.

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:

• A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine.
• Therefore, on hydrolysis of DNA. the quantity of adenine produced is equal of that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
• But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.

Question 9. What are monosaccharides?
Answer:

• Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified based on a number of carbon atoms and the functional group present in them.
• Monosaccharides containing an aldehyde group are known as aldose and those containing a keto group are known as ketose.
• Monosaccharides are further classified as triose, tetrose, pentose, Itexose, and heptose according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3  carbon atoms is called aldotriose.

Question 10. What are reducing sugars?
Answer: Reducing sugars are carbohydrates that reduce l’ehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.

Question 11. Write two main functions of carbohydrates in plants.
Answer:

Two main functions of carbohydrates in plants are:

1. Polysaccharides such as starch serve as storage molecules.
2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 12. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose. maltose, galactose, fructose and lactose
Answer:

1. Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
2. Disaccharides: Maltose, lactose

Question 13. What do you understand by the term glycosidic linkage?
Answer:

• Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
• For example, in a sucrose molecule, two monosaccharide units, α- glucose and β- fructose, are joined together by a glycosidic linkage.

Question 14. What is glycogen? How is it different from starch?
Answer:

Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components – amylose (15-20%) and amylopectin (80-85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.

Question 15. What are the hydrolysis products of (1) sucrose and (2) lactose?
Answer: On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

The hydrolysis of lactose gives β-D-galactose and β-D-glucose.

Question 16. What is the basic structural difference between starch and cellulose?
Answer:

Starch consists of two components – amylose and atm lopectin. Amylose is a long linear chain of α-D-(+)-glucose units joined by C₁– C₄ glycosidic linkage (α-link).

Amylopectin is a branched-chain polymer of α—D—glucose units, in which the chain is formed by C₁ – C₄ glycosidic linkage and the branching occurs by C₁ – C₆, glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁ – C₄ glycosidic linkage (β-link).

Question 17. What happens when D-glucose is treated with the following reagents?

1. HI
2. Bromine water
3. HNO₃

Answer:

When D-glucose is heated with III for a long time, n-hexane is formed.

When D-glucose is treated with Br₂ water, D-gluonic acid is produced.

On being treated with HNO₃, D-glucose gets oxidised to give saeeliaric acid.

Question 18. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

1. Aldehydes give 2, 4-DNP test. Schiff’s test, and react with NaHSO₄ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free-CHO group is absent from glucose.
3. Glucose exists in two crystalline forms -α and β. The α-form (m.p. 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p.=423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 19. What are essential and non-essential amino acids? Give two examples of each type.
Answer:

Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food. For example: valine and leucine Non-essential amino acids are also required by the human body, but they can be synthesised in the body.

For example: Glycine, and Alanine

1. Peptide linkage
2. Primary structure
3. Denaturation

Question 20. Define the following as related to proteins

1. Peptide linkage
2. Primary structure
3. Denaturation

Answer:

Peptide linkage: The amide for pied between the -COOH group of one molecule of an amino acid and the -NH₂, group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.

Primary structure: The primary structure of a protein refers to the specific sequence in which various amino acids are present in it. i.e. the sequence of linkage between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

Denaturation: In a biological system, a protein is found to have a unique 3- dimensional structure and a unique biological activity. In such a situation, the protein is called native protein.

• However, when the native protein is subjected to physical changes such as changes in temperature or chemical changes such as changes in pH. its H-bonds are disturbed.
• This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation.
• During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.  One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 21. What are the common types of secondary structures of proteins?
Answer:

There are two common types of secondary .structure of proteins:

1. α- helix structure
2. β-pleated sheet structure

α- Helix Structure: In this structure, the Nil group of an amino acid residue forms H-bonds with the group of the adjacent turn of the right-handed screw (α-helix).

β-pleated Sheet Structure: This structure is called so because it looks like the pleated folds of draper). In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Question 22. What type of bonding helps in stabilising the a -helix structure of proteins?
Answer: The H-bonds formed between the -Ml group of each amino acid residue and the group of
the adjacent turns of the α-helix help in stabilising the helix.

Question 23. Differentiate between globular and fibrous proteins.
Answer:

Question 24. How do you explain the amphoteric behaviour of amino acids?
Answer: In an aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as a zwitter ion.

Therefore, in Zwitter ionic form, the amino acid can act both as an acid and as a base.

Thus, amino acids show amphoteric behaviour.

Question 25. What are enzymes?
Answer:

• Enzymes are proteins that catalyse biological reactions. They are very specific and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after a particular substrate or class of substrate and sometimes after a particular reaction.
• For example, enzymes used to catalyse the hydrolysis of maltose into glucose are named as maltase.

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named oxidoreductase enzymes, the name of an enzyme ends with ’-ase’.

Question 26. What is the effect of denaturation on the structure of proteins?
Answer:

• As a result of denaturation. globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation.
• secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 27. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:

Based on their solubility in water or fat. vitamins are classified into two groups.

1. Fat-soluble Vitamins: Vitamins that are soluble in fat and oils, but not in ‘water, belong to this group.
   • For example: Vitamins A, D, E, and K
2. Water-soluble Vitamins: Vitamins that are soluble in water belong to this group,
   • For example: B group vitamins (B₁, B₂, B₆, B₁₂, etc.) and vitamin C

However, biotin or vitamin H is neither soluble in water nor fat. Vitamin K is responsible for the coagulation of blood.

Question 28. Why are vitamin A and vitamin C essential to us? Give their important sources,
Answer:

• The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums).
• The source of vitamin A is fish liver oil, carrots, butter, and milk. The sources of vitamin C are citrus fruits. amla, and green leafy vegetables.

Question 29. What are nucleic acids? Mention their two important functions.
Answer:

Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids – deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.

Two main functions of nucleic acids are:

1. DNA is responsible for the transmission of inherent characteristics from one generation to the next. This process of transmission is called heredity.
2. Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell, even though the proteins are synthesised by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.

Question 30. What is the difference between a nucleoside and a nucleotide?
Answer:

A nucleoside is formed by the attachment of a base to the 1′ position of the sugar.

Nucleoside = Sugar + Base

On the other hand, all three basic components of nucleic acids (i.e. pentose sugar, phosphoric acid, and base) are present in a nucleotide.

Question 31. The two strands in DNA are not identical bill are complementary. Explain.
Answer:

In the helical structure of DNA. the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosin forms hydrogen bonds with guanine, while adenine forms hydrogen bonds with thymine. As a result, the two strands are complementary to each other.

Question 32. Write the important structure and functional difference between DNA and RNA.
Answer: The structural differences between DNA and RNA are as follows:

The functional differences between DNA and RNA are as follows:

Question 33. What are the different types of RNA found in the cell?
Answer:

1. Messenger RNA (m-RNA)
2. Ribosomal RNA (rRNA)
3. Transfer RNA (t-RNA)

Question 34. What are hormones and write to its functions?
Answer:

Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly into the bloodstream which transports them to the site of action.

In terms of chemical nature, some of these are steroids, for example., estrogens and androgens: some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine.

1. Insulin keeps the blood glucose level within a narrow limit. Hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.
2. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play a role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of the amino acid tyrosine.
3. Steroid hormones are produced by the adrenal cortex and gonads (testes in males and ovaries in females).
4. Glucocorticoids control carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.
5. Testosterone is the major sex hormone produced in males. It is responsible for the development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It’s responsible for the development of secondary female characteristics and participates in the control of the menstrual cycle.
6. Progesterone is responsible for preparing the uterus for the implantation of a fertilised egg.

Amine

Question 1. Classify the following amines as primary, secondary, or tertiary:

Answer:

1. Primary: (1) and (3)
2. Secondary: (4)
3. Tertiary: (2)

Question 2. Write structures of different isomeric amines corresponding to the molecular formula. C₄H₁₁N

1. Write the IUPAC names of all the isomers.
2. What type of isomerism is exhibited by different pairs of amines?

Answer: (1),(2) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula. C₄H₁₁N are given below:

(3) The pairs (1) and (2). (5) and (7) exhibit position isomerism. The pairs (1) and (3); (2) and (d) exhibit chain isomerism. The pairs (5) (6) and (7) exhibit metamerism.

All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa.

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Question 3. How will you convert

1. Benzene into aniline
2. Benzene into N, N-dimethylaniline
3. Cl—(CH₂)₄-Cl into hexan-1, 6-diamine?

Answer:

Question 4. Arrange the following in increasing order of their basic strength:

1. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 \text { and }\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)
3. \(\mathrm{CH}_3 \mathrm{NH}_2,\left(\mathrm{CH}_3\right)_2 \mathrm{NH},\left(\mathrm{CH}_3\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

Answer:

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_3 \mathrm{~N}<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\)

Question 5. Complete the following acid-base reactions and name the products:

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{HCl} \longrightarrow\)
2. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}+\mathrm{HCl} \longrightarrow\)

Answer:

Question 6. Write reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.

Answer: Aniline reacts with methyl iodide to produce N. N-dimethylaniline.

With excess methyl iodide, in the presence of Na₂CO₃ solution. N, N-dimethylaniline produces N, N, N-trimethylanilinum carbonate.

Question 7. Write the chemical reaction of aniline with benzyl chloride and write the name of the product obtained.
Answer:

Question 8. Write the structure of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers that will liberate nitrogen gas on treatment with nitrous acid.
Answer:

The structure of different isomers corresponding to the molecular formula, C₃H₉N are given below:

1 amine i.e., (1) propan-1-amine, (2) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

Question 9. Convert

1. 3-Methylaniline into 3-nitrotoluene.
2. Aniline into 1,3,5-tribromobenzene

Answer:

Question 10. Write chemical equations for the following reactions:

1. Reaction ofelhanolie NH₃ with C₂H₅CI.
2. Ammonolysis of benzyl chloride and the reaction of amine so formed with two moles of CH₃CI.

Answer:

Question 11. Write chemical equations for the following conversions:

1. \(\mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)

Answer:

Question 12. Write structures and IUPAC names of

1. The amide gives propanamide by Hoffmann bromamide reaction.
2. The amine produced by the Hoffmann degradation of ben/amide.

Answer:

Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. The structure and IUPAC name of the starling amide with four carbon atoms are given below:

Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is an aromatic primary amine-containing six carbon atoms.

Question 13. Arrange the following mg in decreasing order of their basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{NH}_3\)

Answer: The decreasing order of basic strength of the above amines and ammonia follows the following order:

⇒ \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

Question 14. How will you convert 4-nitrotoluene to 2-bromobenzoic acid?

Question 15. Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH} \mathrm{N} \mathrm{H}_2\)
2. \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_2 \mathrm{NH}_2\)
3. \(\mathrm{CH}_3 \mathrm{NHCH}\left(\mathrm{CH}_3\right)_2\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{C} \mathrm{NH}_2\)
5. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3\)
6. \(\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NCH}_3\)
7. \(\mathrm{m}-\mathrm{BrCC}_6 \mathrm{H}_4 \mathrm{NH}_2\)

Answer:

1. Propan – 2 – amine (p)
2. Propan – I – amine(p)
3. N – methyl – 2 – propanamide (s)
4. 2 – methyl propane – 2 – amine (p)
5. N – methyl aniline (s)
6. N – ethyl – N – methyl ethanolamine (t)
7. 3 – Bromo benzydamine or 3 – Bromoaniline (p)

Question 16. Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amines
3. Ethylamine and aniline
4. Aniline and benzylamine
5. Aniline and N – methylaniline

Answer: Methylamine gives an arylamine reaction on heating with CHCl₃ and alcoholic KOH. Gives foul smelling (Carbyl amine)

whereas Dimethylamine does not give this reaction.

Secondary amines were given Kinsberg’s test. It gives insoluble substance with Ginsberg’s reagent (C₆H₅SO₂Cl) which is not affected by base.

• Whereas tertiary amine does not react with Hinsberg’s reagent.
• Aniline gives azo dye test : Dissolve Aniline in cone. HCl and add ice – a cold solution of HNO₂ (NaNO₂ + dil HCl) at 273 K and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aromatic amine (ie Aniline).
• Whereas aliphatic amine (ie ethylamine) does not form a dye. It will give brisk effervescence due to the evolution of N₂ but the solution remains clear.

Nitrous acid test: Benzvlainine reacts with nitrous acid (HNO₂) to form a diazonium salt which being unstable even at low temperatures, decomposes with the evolution of N₂ gas.

Aniline reacts with HNO₂ to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N₂ gas.

Carbylamine test: Aniline being a primary amine gives a carbylamine test. i.e. when heated with an alcoholic solution of KOH and CHCl₃. it gives an offensive smell of phenyl isocyanide. In contrast, N-methyl aniline. being secondary amine does not give this test.

Question 17. Account for the following:

1. pK_b of aniline is more than that of methylamine.
2. Ethylamine is soluble in water whereas aniline is not.
3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
4. Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
5. Aniline does not undergo Friedel-Crafts reaction.
6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
7. Gabrial phthalimide synthesis is preferred for synthesizing primary amines,

Answer:

1. In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene ring. Resulting, the electron density of the nitrogen decreases. On the other hand, in CH₃NH₂ +1 effect of CH₃ increases the electron density of the N-atom. Thus, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.

2. Ethylamine dissolves in water because it forms hydrogen bonds with water molecules. In aniline due to the large, hydrocarbon part, the extent of H-bonding decreases considerably, and hence aniline is insoluble in water.

3. Methylaminc being more basic than water, accepts a proton from water liberating OH^– ions.

These OHQ ions combine with Fe³⁺ ions present in H₂O to form a brown precipitate of hydrated ferric oxide.

⇒ \(\mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}^{+3}+3 \mathrm{Cl}^{-}\)

4. Nitration is usually carried out with a mixture of cones. HNO₃, and cone. H₂SO₄. In the presence of these acids, most of the aniline gets protonated to form anilingus ion. Thus in the presence of acids, the reaction mixture consists of aniline and anilinium ion. Now -NH₂ groups in aniline are o, p-directing, and activating while the NH⁺₃ group in anilinium ion is m-directing.

5. Aniline being a Lewis base, reacts with Lewis acid AlCl₃ to form a salt.

As a result, N of aniline acquires a positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction, consequently, aniline does not undergo Friedel – Crafts reaction.

6. The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to the dispersal of the positive charge on the benzene ring as shown below.

7. Gabriel’s phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore It is preferred for synthesising (aliphatic) primary amines.

Question 18. Arrange the following:

1. In decreasing order of the pKi, values :

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

2. In increasing order of basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{CH}_3 \mathrm{NH}_2\)

3. In increasing order of basic strength:

1. Aniline, p-nitroaniline and p-toluidinc
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

4. In decreasing order of basic strength in the gas phase:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 .\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} .\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N} \text { and } \mathrm{NH}_3\)

5. In increasing order of boiling point:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} .\left(\mathrm{CH}_3\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

6. In increasing order of solubilit> in water:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Answer:

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}-\mathrm{CH}_3>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
3. \(p \text {-nitroaniline }<\text { aniline }<p \text { – toluidine }\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)
5. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3\)
6. \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
7. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Question 19. How will you convert:

1. Ethanoic acid into methenamine
2. Hexamentirile into 1-aminopentane
3. Methanol to ethanoic acid
4. Ethanamine into methamine
5. Ethanoic acid into propionic acid
6. Methenamine into ethanolamine
7. Nitromethane into dimethylamine
8. Propanic acid into ethanoic acid

Answer:

Question 20. Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:

Benzene sulphonyl chloride (C₆H₅SO₂Cl), Which is also known as llinsberg’s reagent, reacts with primary and secondary amine to form sulphonamide.

The hydrogen attached to nitrogen in sulphone amide is strongly acidic due to the presence of a strong electron-withdrawing sulphonyl group. Hence it is soluble in alkali.

Tertiary amines do not react with benzene sulphonyl chloride.

This property of amines reacting with benzene sulphonyl chloride differently is used for the distinction of primary, secondary, and tertiary amines.

Question 21. Write short notes on the following:

1. Carbylamine reaction
2. Diazotisation
3. Hafmann’s bromamide reaction
4. Coupling reaction
5. Ammonolysis
6. Acetylation
7. Gabriel phthalimide synthesis

Answer:

Carbylamine Reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanide or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction, this reaction is known as the carby lamine reaction or idiocy and test and is used as a least for primary amine.

Diazotization reaction: When a cold solution of a primary aromatic amine in a dilute mineral acid (MCI or IfS()4) is treated with a cold solution of nitrous acid at 273 – 278 K. arene diazonium salt is formed. This reaction is called the diazotization reaction.

For example: \(\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow \mathrm{HNO}_2+\mathrm{NaCl}\)

Hoffmann’s Bromamide Reactions: The conversions of a primary amide to a primary amine-containing one carbon atom less than the original amide on heating with a mixture of Br₂ in the presence of NaOH or KOH is called Haffmann’s bromamide reaction, for example.

This reaction is extremely useful for converting a higher homolog to the next lower ho mo log lie.

Coupling Reaction: The reaction of diazonium salts with phenols and aromatic amines to form a/o compounds of the general formula Ar — N = N — Ar is called a coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while with amines it occurs under faintly acidic conditions.

Animonolysis: The process of cleavage of the C – X bond by ammonia molecule is known as ammonolysis.

Acetylation: The process of introducing an acetyl group into a molecule is called acetylation. Common acetylating agents used arc acetyl chloride and acetic anhydride.

Gabriel Phthalimide Synthesis: In this reaction, phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N alkylphthalimide which is hydrolyzed to phthalic acid and primary amine by heating with HCl or KOH solution.

aromatic primary amines cannot be prepared by this method.

Question 22. Accomplish the following conversions:

1. Nirobenzene to benzoic acid
2. Benzene to m-bromophenol
3. Benzoic acid to aniline
4. Aniline to 2.4.6-tribromolluorobenxcne
5. Benzyl chloride to 2-phenylethanolamine
6. Chlorobenzene to p-chloroaniJine
7. Aniline to p-bromoaniline
8. Benzamide to toluene
9. Aniline to benzyl alcohol

Answer:

Question 23. Give the structure of A, B, and C in the following reaction:

Answer:

Question 24. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound C of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:

Question 25. Complete the following reactions:

Answer:

Question 26. Win aromatic primary amines can not be prepared by Gabriel phthalimide synthesis.
Answer:

The success of Gabriel’s phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.

As aryl halides do not undergo nucleophilic substitution reaction easily, aromatic primary amines cannot be prepared by Gabriel phthalimide reaction.

Question 27. Write the reactions of (1) aromatic and (2) aliphatic primary amines with nitrous acid.
Answer:

Question 28. Give a plausible explanation for each of the following:

1. Why are amines less acidic than alcohols of comparable molecular masses?
2. Why do primary amines have a higher boiling point than tertiary amines?
3. Why do aliphatic amines arc stronger bases than aromatic amines?

Answer:

Loss of a proton from an amine gives an amide ion (N^–H) while loss of a proton from alcohol gives an alkoxide ion (O^–R) as shown below.

As O is more electronegative than N, So negative charge on a more electronegative atom is more stable. As R—O is more stable than RNH^–. Therefore amines are less acidic than alcohols.

Due to the presence of two H-atoms on N-atoms of 1° amines, they undergo extensive intermolecular H-bonding while at 3° amines due to the absence of H-atoms on N-atom, there is no hydrogen bonding takes place. So primary amines have higher b.p. than tertiary amines of comparable molecular mass.

Inter Molecular Hydrogen Bonding In 1° Amines.

Aliphatic amines are stronger bases than aromatic amines because:

Due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less available for protonation.

The aryl amine ions have lower stability than the corresponding aliphatic amines i.e. protonation of aromatic amines is not favored.

Aldehyde Ketones And Carboxylic Acid

Question 1. Write the structure of the following reactions:

Answer:

Question 2. Arrange the following compounds in the increasing order of their boiling points. CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Answer:

Question 3. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Enthanal, Propanal, Propanone,, Butanone,
2. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Hint: Consider the steric effect and electronic effect.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Answer: Reactivity of carbonyl compound α (+)ve change

The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

If the +I effect increases then (+)ve change on sp² decreases.

So the reactivity order is: Butanone < Propanone < Propanal < Ethanal

-I group on >C = O in reuses (+)ve charge so reactivity order is Acetophenone < p-tolualdchyde < Benzaldehvde < p-Nitrohenzaldehyde

Question 4. Predict the product of the following reactions:

Answer:

Question 5. Show how each of the following compounds can be converted to benzoic acid.

1. Ethylbenzene
2. Acetophenone
3. Bromobenzene
4. Phenylethane (Styrene)

Answer:

Question 6. Which of each pair shown here would you expect to be stronger?

1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { or } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { or } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { or } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
4. 

Answer: Acidic Strength

1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { < } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { > } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { < } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
4. 

Question 7. Give names of the reagents to bring about the following transformations:

1. Hexan-1-ol to hexanal
2. Cyclohexanol to cyclohexanone
3. p-FluorotoIuene to p-fluorobenzaldchydc
4. Ethanenitrile to ethanal
5. Allyl alcohol to propenal
6. But-2-ene to ethanal

Answer:

1. C₅H₅NH+CrO₃Cl(PCC)
2. K₂Cr₂O₇ in an acidic medium
3. CrO in the presence of acetic anhydride 1. CrO₂Cl₂ 2. HOH
4. (Diisobutyl) aluminium hydride (DIBAL-H)
5. PCC
6. O₃/H₂O-Zn dust

Question 8. Arrange the following compounds in the increasing order of their boiling points:

⇒\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}, \mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_3, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

• The molecular masses of these compounds are in the range of 72 to 74. Since only butane-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest.
• Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former, n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows :

⇒\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3<\mathrm{H}_3 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5\)<\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)

Question 9. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Answer:

The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Question 10. An organic compound (A) with molecular formula CsHsO forms an orange-red precipitate with 2,4-DNP reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens’ or Foldings’ reagent nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.
Answer:

(A)forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ or Fehling reagent, (A) must be a ketone. (A) responds to the iodoform test. Therefore, it should be a methyl ketone.

• The molecular formula of (A) indicates a high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent.
• This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid.
• The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Question 11. Write chemical reactions to affect the following transformations:

1. Butan-1-ol to butanoic acid
2. Benzyl alcohol to phenylethanoid acid
3. 3-Nilrobromobenzene to 3-nitrobenzoic acid
4. 4-Methylacetophenone to benzene-1,4-dicarboxylic acid
5. Cyelohexene to hexane-1,6-dioic acid
6. Butanal to butanoic acid.

Answer:

Question 12. What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Inline
9. 2,4- DNP-derivative
10. Schiffs base

Answer: Cyanohydrin: The organic compounds have -CN and -OH groups in the structure.

Acetal: Acetals are gem-dial koxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

Semicarbazone: Semicarbazone,s are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones with aH in the presence of a base.

Hemiacetal: Hemiacetal are a-alkoxyalcohols Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas.

Oxime: Oximes are a class of organic compounds having a general formula.

On treatment with hydroxyl amine in a weakly acidic medium, aldehydes or ketones form oximes.

Ketal: Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

Imine:

Iminces Imines are produced when aldehydes and ketones react with R- NH₂.

2, 4-DNP-derivative: 2, 4-dinitrophenyIhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2. 4-dinitrophenylhydrazine in a weakly acidic medium.

To identify and characterize aldehydes and ketones, 2, 4-DNP derivatives are used.

Sehiffs Base: Is >c = NR (R is not Hydrogen) (R may be -Ph)

Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of a trace of an acid yield a Schiffs base.

Question 13. Draw structures of the following derivatives.

1. The 2, 4-dinitrophenylhydrazone of benzaldehyde
2. Cyclopropane oxime
3. Acctaldehydedimcthylacctal
4. The semicarbazone of cyclobutanone
5. The ethylene ketal of hexane-3-one
6. The methyl hemiacetal of formaldehyde

Answer:

Question 14. Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents.

1. PhMgBr and then H₃O⁺
2. Tollen’s reagent
3. Semicarbazide and weak acid
4. Excess ethanol and acid
5. Zinc amalgam and dilute hydrochloric acid

Answer:

Question 15. Which of the following compounds would undergo aldol condensation, which is the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

1. Methanal
2. 2-Methylpentanal
3. Benzaldehyde
4. Benzophenone
5. Cyclohexanone
6. 1-Phenylpropanone
7. Phenylacetaldehyde
8. Bulan-1-ol
9. 2,2-Dimethylbutanal

Answer: Aldehydes and ketones having at least one a-hydrogen undergo aldol condensation (2), (5), (6), and (7) give aldol condensation.

• Aldehydes (only) having no a-hydrogen undergo Cannizzaro reactions, (1), (3), (9) give Cannizzaro reaction.
• Compound (4) is a ketone having no a-hydrogen atom and compound (8) Bulan-l-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or Cannizzaro reactions.

Aldol condensation:

Cannizaro Reaction:

Question 16. How will you convert cthanal into the following compounds?

1. Butane-1,3-diol
2. But-2-enal
3. But-2-enoic acid

Answer:

When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.

Question 17. Write structural formulas and names of four possible aids from propanal and butanal. In each case, indicate which aldehyde acts as a nucleophile and which is an electrophile.
Answer:

Take two molecules of propanal. one which acts as a nucleophile and the other as an electrophile.

Taking two molecules of butanal. one which acts as a nucleophile and the other as an electrophile.

Take one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal act as an electrophile.

Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Question 18. An organic compound (A) (molecular formula (C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-cue. Write equations for the reactions involved.
Answer: Hydrolysis of ester gives acid and alcohol

Question 19. Arrange the following compounds in increasing order of their property as indicated:

Acetaldehyde, Acetone, Di-tert-butyl ketone. Methyl tert-butyl ketone

⇒ \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{Br})\mathrm{COOH}\), \(\mathrm{CH}_3\mathrm{CH}(\mathrm{Br})\mathrm{CH}_2\mathrm{COOH},\left(\mathrm{CH}_3\right)_2\mathrm{CHCOOH},\mathrm{CH}_3 \mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}\)

Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid

⇒ \(\mathrm{CH}_3 \mathrm{COCl}, \mathrm{CH}_3 \mathrm{CONH}_2, \mathrm{CH}_3 \mathrm{COOCH}_3,\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}\)

Answer: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

4-Methoxybezo

Question 20. Give simple chemical tests to distinguish between the following pairs of compounds.

1. Propanal and Propanonc
2. Acetophenone and Bcnzophenonc
3. Phenol and Benzoic acid
4. Benzoic acid and Ethyl benzoate
5. Pentan-2-one and Pentan-3-one
6. Benzaldehyde and Acetophenone
7. Ethanal And Propanal

Answer: 1. By Tollen’s test

Propanal reduces Tollen’s reagent, Fehling solution, and Benedict solution but propanonc (ketone) does not. reduces. Iodoform test: Propanone gives an iodoform test but propanal (CH₃-CH₂-CHO) does not.

2. Acetophenone and Benzophenone can be distinguished using the iodoform test. Acetophenone gives an iodoform test but benzophenone does not.

3. Phenol and benzoic acid can be distinguished by the ferric chloride test and NaHCO₃, test.

1. Phenol gives a violet colour with FeCl₃,
2. Benzoic acid gives CO₂ with NaHCO₃ while phenol does not.

4. Benzoic acid and Ethyl benzoate can be distinguished by the sodium bicarbonate test. Sodium bicarbonate test: Benzoic acid reacts with NaHCO₃ to produce brisk effervescence due to the evolution of CO₂ gas but ethylbenzoate does not.

5. Pentan-2-one and pentan-3-one can be distinguished by iodoform lest. Pentan-2-one gives an iodoform test but pentan-3-one does not.

6. Benzaldehyde and acetophenone can be distinguished by (1) Tollen’s test and (2) the Iodoform test. Benzaldehyde gives Tollen’s test whereas acetophenone gives an iodoform test.

7. Ethanal and propanal can be distinguished by the iodoform test. Ethanal gives an iodoform test but propanal does not.

Question 21. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic regent having not more than one carbon atom

1. Methoxybenzoic
2. M-Nitrobenzoic acid
3. P-Nitrobenzoic acid
4. Phenylacetic acid
5. P-Nitrobenzaldehyde

Answer:

Question 22. How will you about the following conversions in more than two steps?

1. Propanone to Propene
2. Benzoic acid to Benzaldehyde
3. Ethanol to 3-Hydroxybutyanal
4. Benzene to m- Nitroacetophenone
5. Benzaldehyde to Benzophenone
6. Bromobenze to 1-Phenylethanol
7. Benzaldehyde to 3-Phenylpropan-1-ol
8. Benzaldehyde to α-Hydroxyphenylacetic acid

Answer:

Question 23. Describe the following:

1. Acetylation
2. Carnnizzaro reaction
3. Cross aldol condensation
4. Decarboxylation

Answer: Acetylation: To introduce acetyl groupin an organic compound.

Carnnizzaro reaction: The self-oxidation-reduction (disproportionation) reaction of aldehydes having no ex-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

Cross-aldol condensation: The reaction between two different aldehydes, or two different ketones, or an aldehyde and a ketone (having α-H) in the presence of a base is called a cross-aldol condensation.

Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Question 24. Complete each synthesis by giving missing starting material, reagent or products.

Answer:

Question 25. Give a plausible explanation for each of the following:

1. Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-trimethyl cyclohexanone does not.
2. There are two, groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
3. During the preparation of esters, a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Answer: Cyclohexanones form cyanohydrins according to the following equation.

In 2, 2. 6 – trimethyl cyclohexanone, methyl groups at α-positions offer steric hindrance and as a result.

CN^– cannot attack effectively.

2, 2, 6 – trimethyl cyclohexanone,

For this reason, it does not form a cyanohydrin.

The electron density on the -NH₂, group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other -NH₂, the group is not involved in resonance: it can act as a nucleophile and can attack carbonyl – carbon atoms of aldehydes and ketones to produce semicarbazone.

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

Question 26. An organic compound contains 69.77% carbon. 11.63% hydrogen and the rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an additional compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:

% of carbon = 69.77%

% of hydrogen = 11.63%

% of oxygen = [100-(69.77 + 11.63)]% = 18.6%

Thus, the ratio of the number of carbon, hydrogen and oxygen atoms in the organic compound can be given as:

⇒ \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{69.77}{12}: \frac{11.63}{1}: \frac{18.6}{16}=5.81: 11.63: 1.16=5: 10: 1\)

Therefore, the empirical formula of the compound is C₅H₁₀O. Now, the empirical formula mass of the compound can be given as:

5 × 12 +10 × 1 + 1 × 16 = 86

Molecular mass of the compound = 86 (given)

Therefore, the molecular formula of the compound is given by C₅H₁₀O.

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.

Hence, the given compound is plan-2-one.

The given reactions can be explained by the following equations:

Question 27. Although phenoxide ion has more number of resonating structures than carboxylate ion. carboxylic acid is a stronger acid than phenol. Why?
Answer: Resonance structures of phenoxide ion are:

Unequivalent resonating structures

Equivalent resonating structures are more stable.

Alcohol Phenol And Ether

Question 1. Classify the following as primary, secondary, and tertiary alcohols:

Answer:

1. Primary alcohol > (1), (2), (3)
2. Secondary alcohol > (4), (5)
3. Tertiary alcohol > (6)

Question 2. Identify allylic alcohols in the above examples.
Answer:

• Allylic that means C = C – C – OH
• The alcohols given in (2) and (6) are allylic.

Question 3. Name the following compounds according to the IUPAC system.

Answer:

1. 3-Chloromethyl-2-isopropylpentan-1-ol
2. 2, 5 – Dimelhylhcxane – 1,3- cliol
3. 3-Bromocyclohexanol
4. Hex-1 -en-3-ol
5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Answer:

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. Write structures of the products of the following reactions:

Answer:

Question 6. Give structures of the products you would expect when each of the following alcohols reacts with

1. HCl-ZnCl₂
2. HBr and
3. SOCl₂
4. Bulan-1-ol
5. 2-Methylbutan-2-ol

Answer:

Tertiary alcohols react immediately with Lucas’ reagent.

Question 7. Predict the major product of acid-catalyzed dehydration of

1. 1-methylcyclohexane and
2. Butan-1-ol

Answer:

Question 8. Ortho para nitrophenols are more than phenol. DFrawn the resonance structure of the corresponding phenoxide ions.
Answer:

(Resonance structure of the phenoxide ion).

(Resonance structure of p-nitro phenoxide ion)

(Resonance structure of o-nitro phenoxide ion)

It can be observed that the presence of the nitro group increases the stability of phenoxide ions.

Question 9. Write the equations involved in the following reactions:

1. Reimer – Tiemann reaction
2. Kolbe – Schmidt reaction

Answer:

Question 10. Write the reactions of Williamson synthesis of 2-ethoxy-3-inethylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4- nitrobenzene and why?

Answer: Set (2) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzcne.

In set (1), the Bromine atom is stable by resonance.

Question 12. Predict the products of the following reactions:

Answer:

Question 13. Give IUPAC names of the following compounds:

Answer:

1. 4-ChIoro-2,3-dimethylpentan-1-oI
2. 2-Ethoxy propane
3. 2,6-Diinethylphenol
4. 1-Elhoxy-2-nilrocyclohexane

Question 14. You will notice that the reaction produces a primary alcohol with methanol, a secondary alcohol with other aldehydes, and a tertiary alcohol with ketones. Give the structures and IUPAC names of the products expected from the following reactions:

1. Catalytic reduction of butane.
2. Hydration of propene in the presence of dilute sulphuric acid.
3. Reaction of propanone with methyl magnesium bromide followed by hydrolysis.

Answer:

Question 15. Arrange the following sets of compounds in order of their increasing boiling points:

1. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
2. Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Answer:

1. Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
2. n-Butane, ethoxyethane, pentanal, and pentan-1-ol.

Question 16. Arrange the following compounds in increasing order of their acid strength: Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.
Answer: Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 17. Write the structures of the major products expected from the following reactions:

1. Mononitration of 3-methyl phenol
2. Denitration of 3-methyl phenol
3. Mononitration of phenyl methanoate.

Answer: The combined influence of -OH and -CH₃ groups determines the position of the incoming group.

Question 18. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

1. What would be the major product of this reaction?
2. Write a suitable reaction for the preparation of t-butyl ethyl ether.

Answer:

1. The major product of the given reaction is 2-methyl prop-1-one.

This is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

2. Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.

Question 19.

Answer:

Question 20. Write IUPAC names of the following compounds:

Answer:

1. 2, 2, 4-Trimethylpcntan-3-ol
2. 5-Ethylheplane-2, 4-diol
3. Butane-2, 3-diol
4. Propane-1, 2, 3-triol
5. 2-Methylphenol
6. 4-Methylphenol
7. 2. 5-Dimcthylphenol
8. 2, 6-Dimethylphenol
9. 1-Methoxy-2-methylpropane
10. Ethoxyben/ene
11. 1-Phenoxyheptane
12. 2-Elhoxvbutane

Question 21. Write structures of the compounds whose IUPAC names are as follows:
Answer:

1. 2-Methylbutan-2-ol
2. 1-Phenyl-2-ol
3. 3,5- Dimethylhexene-1,3,5-triol
4. 2,3-Diethylphenol
5. 1-Ethoxypropane
6. 2-Ethoxy-3-methyl pentane
7. Cyclohexylmethanol
8. 3-Cyclohexylpentan-3-ol
9. Cyclopent-3-en-1-ol
10. 3-Chloromethylpentan-1-ol

Answer:

Question 22. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names. Classify them as primary, secondary, and tertiary alcohols.
Answer:

1. Primary alcohols – (1), (2)(3), and (4)
2. Secondary alcohols – (5),(6) and (7)
3. Tertiary alcohol – (8)

Question 23. Explain why propanol has a higher boiling point than that of the hydrocarbon, butane.
Answer:

Propanol undergoes intermolecular H-bonding because of the presence of the -OH group. On the other hand, butane does not

Question 24. Explain why are alcohols comparatively more soluble in water than the corresponding hydrocarbons.
Answer:

Hydrogen bonding exists between alcohol and water molecules while hydrocarbons are nonpolar, they can not form an H-bond with water molecules.

Question 25. What is meant by hydroboration oxidation reaction? Illustrate it with an example.
Answer:

Question 26. Give the structures and IUPAC names of monohydric phenols of molecular formula C₇H₈O.
Answer:

Question 27. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile. Give reason.
Answer: O-nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Question 28. Give the equation of reaction of preparation of phenol from cumene.
Answer:

Question 29. Write a chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

Question 30. You are given a benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Question 31. Show how will you synthesize:

1. 1-phenylethanol from a suitable alkane.
2. CyelohexyI methanol using an alkyl halide by an S_N2 reaction
3. Pentan-1-ol using a suitable alkyl halide?

Answer:

Question 32. Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:

The acidic nature of phenol can be represented by the following two reactions.

1. Phenol reacts with sodium to give sodium phenoxide, liberating H₂

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Question 33. Explain why is ortho-nitro phenol more acidic than ortho-methoxy phenol.
Answer: Electron withdrawing effect of nitro group and electron releasing effect of methoxy group.

Question 34. Explain how the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
Answer: Electron density at the o- and p-position increases and thus the electrophile attacks easily.

Question 35. Give equations of the following reactions:

1. Oxidation of propane-1 -ol with KMnO₄ solution.
2. Bromine is CS₂ with phenol.
3. Dilute HNO₃ with phenol.
4. Treating phenol with chloroform in the presence of aqueous NaOH.

Answer:

Question 36. Explain the following with an example.

1. Williamson ether synthesis.
2. Unsymmetrical ether.

Answer:

1. Williamson’s ether synthesis:

Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxide.

This reaction involves an S_N2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

2. Unsymmetrieal ether:

An unsymmetrical ether is an ether where two alkyl groups on the two sides of an oxygen atom differ. For example: ethyl methyl ether (CH₃ -O-CH₂CH₃).

Question 37. How are the following conversions carried out?

1. Propene → Propan-2-ol
2. Benzyl chloride → Benzyl alcohol
3. Ethyl magnesium chloride → Propan-1-ol.
4. Methyl magnesium bromide → 2-Methylpropan-2-ol,

Answer:

Question 38. Give a reason for the higher boiling point of ethanol in comparison to methoxyrnethane.
Answer:

• Ethanol undergoes intermolecular H-bonding due to the presence of the -OH group, resulting in the association of molecules.
• On the other hand, methoxyrnethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxyrnethane.

Question 39. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer: The reaction of Williamson synthesis involves an S_N2 attack of an alkoxide ion on a primary alkyl
halide.

• But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced.
• This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.

Question 40. How is 1-propoxypropane synthesized from propan- 1-ol? Write the mechanism of this reaction.

Answer:

The mechanism of this reaction involves the following three steps:

Step 1: Protonation

Step 2: Nucleophilic attack

Step 3: Deprotonation

Question 41. The preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

• The formation of ethers by dehydration of alcohol is a bimolecular reaction (S_N2) involving the attack of an alcohol molecule on a protonated alcohol molecule.
• In this method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of others, alkenes are formed.

Question 42. Write the equation of the reaction of hydrogen iodide with:

1. 1-propoxypropane
2. Methoxybcnezcnc and
3. Benzyl ethyl ether

Answer:

Question 43. Explain the fact that in aryl alkyl ethers

1. The alkoxy group activates the benzene ring towards electrophilic substitution and
2. It directs the incoming substituents to ortho and para positions in the benzene ring.

Answer:

In arly alkyl ethers, due to the + R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

2. It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 44. Write the mechanism of the reaction of III with methoxymethyl.
Answer:

The mechanism of the reaction of III with methoxymethane involves the following steps:

Step 1: Protonation of methoxymelthane:

Step 2: Nucleophilic attack of I^–

Question 45. Write equations of the following reactions:

1. Friedel-Crafts reaction-alkylation of anisole.
2. Nitration of anisole.
3. Bromination of anisole in ethanoic acid medium.
4. Friedcl-Craft’s acetylation of anisole.

Answer:

Question 46. When 3-methyl butane-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.

Answer: The mechanism of the given reaction involves the following steps:

Step 1: Protonation

Step 2: Formation of 2° carbonation by the elimination of a water molecule

Step 3: Re-arrangement by the hydride-ion shift

Step 4: Nucleophilic attack

Haloalkanes And Haloarenes

Question 1. Write The structure of the following compounds:

1. 2-Cliloro-3-methylpentane
2. 1-Chloro-4-methylcyclohexane
3. 4-terl. ButyI-3-iodohcplane
4. 1, 4-DibmmobiU-2- one
5. 1-Bromo-4-sec, butyl-2-melhylbenzene

Answer:

Question 2. Draw the structure of major monohalo products in each of the following reactions:

Answer:

Question 3. Arrange each set of increasing boiling points.

1. Bromomethane, Bromoform, Chloromethane, Dibromethane.
2. 1-chloride, Isopropyl chloride, 1-chlorobutane.

Answer:

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Which alkyl halide from the following pairs would you accept to react more rapidly by an S_N2 mechanism? Explain your answer.

Answer:

Question 5. In the following pairs of halogen compounds, which compound undergoes a faster S_N1 reaction?

Answer:

Question 6. Identify A, B, C, D, E, R, and R¹ in the following:

Answer:

Question 7. Write the IUPAC name of the following:

Answer:

1. 4-Bromopent-2-ene
2. 3-Bromo-2-methylbut-1-ene
3. 4-Bromo-3-methyl pent-2-ene
4. 1-Bromo-2-methylbut-2-ene
5. 1-Bromobut-2-ene
6. 3-Bromo-2-methylpropene

Question 8. Write the product of the following reactions:

Answer:

Question 9. Haloalkanes react with KCN to form alkyl cyanides as the main product while AgCN forms isocyanides as the chief product. Explain.
Answer:

• KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond.
• However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

Question 10. In the following pairs of hydrogen compounds, which would undergo SN2 reaction faster?

Answer:

• It is a primary halide and therefore undergoes S_N2 reaction faster.
• As iodine is a better-leaving group because of its large size, it will be released at a faster rate in the presence of income nucleophiles.

Question 11. Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions:

1. The four isomeric bromobutan.es
2. C₆H5CH₂Br, C₆H₅CH(C₆H₅)BR, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}<\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{SN}_{\mathrm{N}} 1\right)\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{Br}_2 \mathrm{CH}_3>\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{S}_{\mathrm{N}} 2\right)\right.\)

Question 12. Although chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why?
Answer: Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Question 13. Name the following halides according to the IUPAC system and classify them as alkyl, alkyl, benzene(primary, secondary, tertiary), vinyl, or aryl halide:

1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{Cl}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{I}\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{C}_6 \mathrm{H}_5\)
5. \(\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CH}_2 \mathrm{Br}\)
7. \(\mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{CH}_2 \mathrm{CH}_3\)
8. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}(\mathrm{Cl}) \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
9. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2\)
10. \(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
11. \(\mathrm{m}-\mathrm{ClCH}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_3\)
12. \(\mathrm{O}-\mathrm{Br}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

Question 14. Give the IUPAC names of the following compounds

1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
2. \(\mathrm{CHF}_2 \mathrm{CBrClF}\)
3. \(\mathrm{ClCH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{Br}\)
4. \(\left(\mathrm{CCl}_3\right)_3 \mathrm{CCl}\)
5. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4\right)_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}=\mathrm{CClC}_6 \mathrm{H}_4 \mathrm{I}-\mathrm{p}\)

Answer:

Question 15. Write the structures of the following organic halogen compounds.

1. 2-Chloro-3-methyl pentane
2. p-Bromochlorobenzene
3. l-Chloro-4-ethylcyelohexane
4. 2-(2-Chlorophenyl)-1-iodooctane
5. 2-Bromobutane
6. 4-tert-butyl-3-iodoheptane
7. 1-Bromo-4-sec-butyl-2-methyl benzene
8. 1,4-Dibromobut-2-ene

Answer:

2-Chloro-3-methyl pentane

p-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane

2-(2-ChlorophenyI)-1-idodooctane

2-Bromobutane CH

4-Tert-butyl-3-iodoheptane

1-Bromo-4-sec-butyl-2-methyl benzene

1,4-Dibromobut-2-ene

Question 16. Which of the following has the highest dipole moment?

1. CH₂Cl₂
2. CHCl₃
3. CCl₄

Answer:

CCl₄ < CHCl₃ < CH₂Cl₂

Question 17. A hydrocarbon C₅H₁₀ does not react with chlorine in the dark but gives a single monochrome compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:

The reactions involved in the question are:

Question 18. Write the isomers of the compound having the formula C₄H₉Br.
Answer: There are four isomers of the compound having the formula Chi I- Hr. These isomers are given below.

7. Write the equations for the preparation of 1-iodobutane from

1. 1-butanol
2. 1-chlorobutane
3. but-1-ene

Answer:

Question 19. What are ambident nucleophiles? Explain with an example.
Answer: Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

Question 20. Which compound in each of the following pairs will react faster in S_N2 reaction with Oil?

1. CH₃Br or CH₃I
2. (CH₃)₃CCl or CH₃Cl

Answer: CH₃-I is more reactive because I⁺ is a better-leaving group than Br⁺

In the case of (CH₃)₃CCl. the attack of the nucleophile at the carbon atom is hindered by the presence of bulk)- substituents on that carbon atom bearing the leaving group.

Question 21. Predict all the alkanes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.

1. 1-Brorao-1-methylcyclohexane
2. 2-Chloro-2-methyl butane
3. 2,2, 3-Trirpethyl-3-bromopentane

Answer:

Question 22. How will you bring about the following conversion?

1. Ethanol to but-1-yne
2. Ethane to broiuoethene
3. Propene to I-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1-cue to but-2-ene
9. 1-Chlorobutane to n-octane
10. Benzene to biphenyl

Answer:

Question 23. Explain why

1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. Alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions.

Answer:

+M of -Cl and I of -Cl are in opposite directions so the dipole moment is decreased in chlorobenzene.

1. No H-bonding of halide with water.
2. Grignard reagents in the presence of moisture, react with H₂O to give alkanes.

Question 24. Give the uses of freon 12. DDT, carbon tetrachloride, and iodoform.
Answer:

• Uses of Freon-12 Freon-12 (dichlorodifluoromethane. CF₂Cl₂) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.
• The use of DDT (p-p-dichlorodiphenyltrichloroethane) is one of the best-known insecticides.

Uses of carbonletrachloride (CCI₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cans.
2. It is used as a solvent in the manufacture of pharmaceutical products.
3. It is used as a fire extinguisher.

Uses of iodoform (CHI₃) Iodoform was used earlier as an antiseptic. The antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 25. Write the structure of the major organic product in each of the following reactions.

1. 

Answer:

Question 26. Write the mechanism of the following reaction:

Answer: The given reaction is:

The given reaction is an S_N2 reaction. In this reaction. CM acts as the nucleophile and attacks the carbon atom to which Br is attached. CN ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 27. Arrange the compounds of each set in order of reactivity towards S_N2 displacement:

1. 2-Bromo-2-metln Ihutane. 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methyl butane.2-Bromo-2-methyl butane.3-Bromo-2-inethyIhutane
3. 1-Bromobutane. l-Bromo-2. 2-dimethyIpropane. 1-Bromo-2-methylbutane. 1-Bromo 3-methyIhutane

Answer:

(more branching at a nearer distance)

Hence, the increasing order of reactivity of the given compounds towards S_N2 is:

1-Bromo-2,2-dimethy Ipropane < 1-Bromo-2-methy Ihutane < 1-Bromo-3-methylbutane < 1-Bromobutane.

Question 28. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolyzed by aqueous KOH?
Answer: C₆H₅CHClC₆H₅ (more reactive)

(more stable intermediate)

Question 29. p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, molecules of p-dichlorobenzene are more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.

Question 30. How the following conversions can be carried out?

1. Propene to propane-1-ol
2. Ethanol to but-1-yne
3. 1-Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bronitobenzene
6. Benzyl alcohol to 2-phenylethanoid acid
7. Ethanol to propane nitrile
8. Aniline to chloro benzene
9. 2-Chlorobutane to 3, 4-dimethyl hexane
10. 2-Methyl-1-propane to 2-chloro-2-methyl propane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-chloropropane to 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphene
19. tert-Butyl bromide to isobutyl bromide
20. Aniline to phenyl isocyanide

Answer:

Question 31. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products, Explain.
Answer:

OH^– ion is highly solvated in an aqueous solution and as a result, the basic character of OH^– ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon. So

⇒ \(\mathrm{R}-\mathrm{Cl}+\mathrm{Aq} \cdot \mathrm{KOH} \longrightarrow \mathrm{R}-\mathrm{OH}+\mathrm{KCl}\)

On the other hand, an alcoholic solution of KOFI contains an alkoxide (RO^–) ion. which is a strong base. Thus, it can abstract hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

Question 32. Primary alkyl halide C₄H₉Br(a) reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with lIBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:

Two primary alkyl halides have the formula. C₄H₉Br. They are n-butyl bromide and isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (h) of molecular formula. C₈H₁₈. Which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence. compound (a) must be isobutyl bromide.

Thus, compound (d) is 2. 5-Dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). compound (b) is 2-methyl propene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence compound (e) is -2-Bromo-2-methy propane.

Question 33. What happens when

1. n-butyl chloride is treated with alcoholic KOH?
2. Broinohen/ene is treated with Mg in the presence of dry ether.
3. Chlorobenzene is subjected to hydrolysis.
4. Ethyl chloride is treated with aqueous KOH.
5. Methyl bromide is treated with sodium in the presence of dry ether.
6. Methyl chloride is treated with KCN.

Answer:

Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

 

Coordination Compounds

Question 1. Write the formulas for the following coordination compounds:

1. Tetraamminediaquacobalt(3) chloride
2. Potassium tetracyanonickelate (2)
3. Tris (ethane-1,2-diamine) chromium(3) chloride
4. Ammincbromideochloridonitrito-N-platinate(2)
5. Dichloridobis(ethane-1,2-diamine)platinum(4) nitrate
6. Iron(3) hexacyanoferrate(2)

Answer:

⇒ \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_3\)

⇒ \(\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]\)

⇒ \(\left[\mathrm{Cr}(\text { en })_3\right] \mathrm{Cl}_3\)

⇒ \(\left[\mathrm{Pl}(\mathrm{NH})_3 \mathrm{BrCl}\left(\mathrm{NO}_2\right)\right]^{-}\)

⇒ \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]\left(\mathrm{NO}_3\right)_2\)

⇒ \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)

Question 2. Write the IUPAC name of the following coordination compounds:

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)

⇒ \(\mathrm{K}_3\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)

⇒ \(\mathrm{K}_2\left[\mathrm{PdCl}_4\right]\)

⇒ \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}\)

Answer:

1. Hexarmninecobalt(3) chloride
2. Pentaamminechloridocobalt(3) chloride
3. Potassium hexacyanoferrate(3)
4. Potassium trioxalaloferrate(3)
5. Potassium tetrachloridopalladate(2)
6. Diamminechiorido(methylamine)platinum(2) chloride

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 3. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

⇒ \(\mathrm{K} \mid \mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\)

⇒ \(\left[\mathrm{Co}(\mathrm{en})_3\right] \mathrm{Cl}_3\)

⇒\(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]\left(\mathrm{NO}_3\right)_2\)

⇒ \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Cl}_2\right]\)

Answer: 1. Both geometrical (cis-, trans-) isomers for K[Cr(H₂O),(C₂O₄)₂] can exist. Also, optical isomers for cis-isomers exist.

Geometrical isomers

Truns-iso me r is optically inactive. On the other hand, cis-isomer is optically active.

2. Two-optical isomers for [CO(en)₃]Cl₃ exist.

Two optical isomers arc possible Tor this structure

3. [CO(NH₃)₅(NO₂)](NO₃)₂

A pair of optical isomers

It can also show linkage isomerism [CO( NH₃ )₅(NO₂)] ( NO₃)₂ and [CO(NH )₅ (ONO)]( NO₃)₂

It can also show ionization isomerism [Co(NH₃)₅ (NO₂)](NO₃)₂ and [CO(NH₃)₅ (NO₃)] (NO₃) (NO₂)

Geometrical ( cis-trails-) isomers of [Pt( NH₃)( H₂O)Cl₂] can exist.

Question 4. Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]CI are ionization isomers.
Answer:

When ionization isomers are dissolved in water, they ionize to give different ions, l liese ions then react differently

⇒ \(\left[\mathrm {CO} (\mathrm {NH}_{3})_{5}\mathrm {Cl} \left]\mathrm{SO}_4+\mathrm{Ba}^{2+} \longrightarrow \mathrm{BaSO}_4 \downarrow\right.\right.\) White precipitate

⇒ [CO(NH₃)₅Cl]SO₄ + Ag⁺ →  No reaction

⇒ [CO(NH₃)₅SO₄]Cl + Ba²⁺ →  No reaction

⇒ \(\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4 \mathrm]{Cl}+\mathrm{Ag}^{+} \longrightarrow \mathrm{AgCl} \downarrow\right.\)

Question 5. Explain based on valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [Ni(Cl₄]^2- ion with tetrahedral geometry Is paramagnetic.
Answer:

Ni is in the +2 oxidation state i.e., in the d⁸ configuration.

d⁸ configuration:

There are 4 CN^– ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN^– ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

It now undergoes dsp² hybridization. Since all electrons are paired, it is diamagnetic. in. the case of [NiCl₄]^2-, Cl ion is a weak field ligand. Therefore, it does not lead to the pairing of impaired 3d electrons. Therefore, it undergoes sp³ hybridization.

Since there are 2 unpaired electrons in this case, it is paramagnetic.

Question 6. [NiC₄]^2- is paramagnetic while [NitCO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:

Though both [NiCl₄]^2- and [Ni(CO)₄] are tetrahedral, their magnetic characters are different. This is due to a difference like ligands. Cl^– is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence. [NiCl₄]^2- is paramagnetic.

In Ni(CO)₄, Ni is in the zero oxidation state Le., it has a configuration of 3d⁸ 4s².

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also. it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)₄] is diamagnetic.

Question 7. [Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas ji’e(CN)6J’ is weakly paramagnetic. Explain.
Answer:

In both [Fe(H₂O)₆]³⁺ and [Fe(CN)₆]^3-, Fe exists in the +3 oxidation state i.e., in d⁵ configuration.

Since CN^– is a strong field ligand, it causes the pairing of impaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.

Therefore, \(\mu=\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732 B M\)

On the other hand. (H₂O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.

Therefore, \(\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35} \approx 6 B M\)

Thus, it is evident that [Fe(H₂O)₆]³⁺ is strongly paramagnetic, while [Fe(CN)₆]^3- is weakly paramagnetic.

Question 8. Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:

Question 9. Predict the number of unpaired electrons in the square planar [Pt(CN)₆]^2- ion.
Answer: [Pt(CN)₆]^2-

In this complex. Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp² hybridization. Now. the electronic configuration of Pd(+2) is 5d⁸.

CN^– being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in [Pt(CN)₄]^2-.

Question 10. The hexaquo manganese(ll) ion contains live unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Answer:

Question 11. Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)²⁺₄‘ ion. given that the β₄ for this complex is 2.1 × 10¹³,
Answer:

⇒ \(\beta^4=2.1 \times 10^{13}\)

The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant (f.

⇒ \(\frac{1}{\beta_4}=\frac{1}{2.1 \times 10^{13}}=4.7 \times 10^{-14}\)

Question 12. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:

1. Werner’s postulates explain the bonding in coordination compounds as follows: A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions.
2. A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound.
3. Primary valencies are usually ionizable. while secondary valencies are non-ionizable.

Question 13. FeSO₄ solution mixed with (NH₄)₂SO₄ solutions in a 1:1 molar ratio gives the least of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in a 1:4 molar ratio does not give the least of Cu²⁺ ion. Explain why?
Answer:

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+\mathrm{FeSO}_4+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Mohr’s Salt }}{\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}}\)

⇒ \(\mathrm{CuSO}_4+4 \mathrm{NH}_3+5 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Tetraam min copper(2)sulphate }}{\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 .5 \mathrm{H}_2 \mathrm{O}}\)

Both the compounds i.e., FeSO₄. (NH₄), SO₄. 6H₂O and [Cu(NH₃)₄] SO₄. 5H₂O falls under the category of addition compounds with only one major difference i.e., the former is an example of a double salt, while the latter is a coordination compound.

Question 14. Explain with two examples of each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:

Coordination entity: A coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion surrounded by a suitable number of neutra 1 molecules or negative ions (called ligands For example:

⇒\(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+},\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+} \text { etc. }\) ⇒ cationic complex

⇒ \(\left[\mathrm{PtCl}_4\right]^{2-},\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-} \quad \text { etc. }\) ⇒ anionic complex

⇒ \(\left[\mathrm{Ni}(\mathrm{CO})_4\right] \cdot\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \text { etc. }\) ⇒ neutral complex

Ligands: The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity or a coordinate complex are known as ligands. For example Cl, FLO. H, N CFLCFLNFL etc.

Coordination number: The total number of ligands (either neutral molecules or negative ions) that get attached to the central metal atom in the coordination sphere is called the coordination number of the central metal atom. It is also referred to as its legacy.

For example:

1. In the complex, K₂[PtCI]₆ there as six chloride ions attached to Pt in the coordinate sphere. Therefore, the coordination number of Pt is 6.
2. Similarly, in the complex [Ni(NH₃)₄ ]Cl₂, the coordination number of the central atom (Ni) is 4.

Coordination polyhedron: Coordination polyhedrons about the central atom can be
defined as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere. For example:

Homoleptic complexes: These are those complexes in which the metal ion is bound to only one kind of a donor group.

For example., [Co(NH₃)₆]³⁺ .[PtCl₄]^2- etc.

Heteroleptic complexes: Heteroleptic complexes are those complexes where the central metal ion is bound to more than one type of donor group.

For example., [Co(NH₃)₄ Cl₂]⁺ ,[Co(NH₃)₅Cl]²⁺

Question 16. What is meant by unidentate? dentate and ambidentate ligands? Give two examples for each.
Answer:

A ligand may contain one or more unshared pail’s of electrons which are called the donor sites of ligands. Now. depending on the number of these donor sites, ligands can be classified as follows:

Unidenatate ligands: Ligands with only one donor site arc are called unidentate ligands.

⇒ \(\ddot{\mathrm{N}} \mathrm{H}_3, \mathrm{Cl}^{-2} \text { etc. }\)

Didentate ligands: Ligands that have two donor sites arc called bidentate ligands. For example.,

Ethane- 1, 2-diamine

Ambidentate ligands: Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands. For example:

Question 17. Specify the oxidation numbers of the metals in the following coordination entities:

1. \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_2\right]^{2+}\)
2. \(\left[\mathrm{CoBr}_2(\mathrm{en})_2\right]^{+}\)
3. \(\left[\mathrm{PtCl}_4\right]^{2-}\)
4. \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)
5. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\)

Answer:

1.  \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_2\right]^{2+}\)

Let the oxidation number of Co be x.

The charge on the complex is +2

x- 1 = +2

x = +3

2. [PtCl₄]^–

Let the oxidation number of Pt be x.

The charge on the complex is -2,

x = +2

3.

x- 2= +1

x =+3

4. \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)

x – 2 = +1

x = +3

5. K₃[Fe(CN)₆]

x = +3

x – 3 = 0

x = +3

Question 18. Using IUPAC norms write the formulas for the following:

1. Tetrahydroxozincatc (2)
2. Potassium tetrachloridopalladate (2)
3. Diamminedichloridoplatinum (2)
4. Potassium tetracyanonickelate (2)
5. Pentaamminenitrito-O-cobalt (3)
6. Hexaamminecobalt (3) sulphate
7. Potassium tri(oxalate)chromate (3)
8. Hexaammineplatinum (4)
9. Tetrabromidocuprate (2)
10. Pentaamminenitrito-N-cobalt (3)

Answer:

1. \(\left[\mathrm{Zn}(\mathrm{OH})_4\right]^{2-}\)
2.  \(\mathrm{K}_2\left[\mathrm{PdCl}_4\right]\)
3. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]\)
4. \(\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]\)
5. \(\left[\mathrm{Co}(\mathrm{ONO})\left(\mathrm{NH}_3\right)_5\right]^{2+}\)
6. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]_2\left(\mathrm{SO}_4\right)_3\)
7. \(\mathrm{K}_3\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)
8. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_6\right]^{4+}\)
9. \(\left[\mathrm{Cu}(\mathrm{Br})_4\right]{^2-}\)
10. \(\left[\mathrm{Co}\left[\mathrm{NO}_2\right)\left(\mathrm{NH}_3\right)_5\right]^{2+}\)

Question 19. Using IUPAC norms write the systematic names of the following:

1. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)
2. \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NH}_2 \mathrm{CH}_3\right)\right] \mathrm{Cl}\)
3. \(\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\)
4. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}\)
5. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)
6. \(\left[\mathrm{NiCl}_4\right]^{2-}\)
7. \(\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2\)
8. \(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\)
9. \(\left[\mathrm{Ni}(\mathrm{CO})_4\right]\)

Answer:

1. Hexaamminecobalt (3) chloride
2. Diamminechlorido (methylamine) platinum (2) chloride
3. Hexaaquatitanium (3) ion
4. Tetraamminechloridonitrito-N-Cobalt (3) chloride
5. Hexaaquamangancse (2) ion
6. Tetrachloridonickelate (2) ion
7. Hexaamminenickcl (2) chloride
8. Tris (ethane- 1, 2-diamine) cobalt (3) ion
9. Tetracarbonylnickel (0)

Question 20. List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:

1. Geometrical Isomerism: This type of isomerism is common in heteroleptic complexes. It arises due to the different possible geometric arrangements of the ligands. For example:

2. Optical Isomerism: This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-superim possible.

3. Linkage Isomerism: This type of isomerism is found in complexes that contain ambidentate ligands.

4. Coordination Isomerism: This type of isomerism arises when the ligands are interchanged between cationic and anionic entities of different metal ions present in the complex.

For example \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right] \text { and }\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}(\mathrm{CN})_6\right]\)

5. Ionization Isomerism: This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. For example

⇒ \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Br} \text { and }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Br}\right] \mathrm{SO}_4\)

6. Solvate Isomerism: Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or merely present as a free solvent molecule in the crystal lattice.

Question 21. How many geometrical isomers are possible in the following coordination entities?

1. [Cr(C₂O₄)₃]³⁺
2. [Co(NH₃)₃Cl₃]

Answer:

For [Cr(C₂O₄)₃]³⁺ no geometrical isomer is possible as it is a bidentate ligand.

[Co(NH₃)₃Cl₃]  Two geometrical isomers are possible.

Question 22. Draw the structure of optical isomers of:

1. \(\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\)
2. \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}\)
3. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Answer: 1. \(\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\)

2. \(\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}\)

3. \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Question 23. Draw all the isomers (geometrical and optical) of:

1. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)
2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{en})_2\right]^{2+}\)
3. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2(\mathrm{en})\right]^{+}\)

Answer: 1. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)

In total, three isomers are possible.

2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{en})_2\right]^{2+}\)

Trans-isomers are optically inactive.

Cis-isomers are optically active.

3. \(\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}\)

Question 24. Write all the geometrical isomers of [Pt(NH₃)(Br)(G)(py)j and how many of these will exhibit optical isomers?
Answer: [Pt(NH₃)(Br)(Cl)(py)]

From the above prisoners, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerization. They do so only in the presence of unsymmetrical chelating agents.

Question 25. Aqueous copper sulphate solution (blue) gives:

1. A green precipitate with aqueous potassium fluoride, and
2. A bright green solution with aqueous potassium chloride Explain these experimental results.

Answer:

Aqueous CuSO₄ exists as [Cu(H₂O)₄]SO₄ It is blue due to the presence of [Cu(H₂O)₄]²⁺ ions.

When KF is added.

When KCI is added:

⇒ \(\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}_4\right)\right]^{2+}+4 \mathrm{Cl}^{-} \longrightarrow \underset{\text { (bright green) }}{\left[\mathrm{CuCl}_4\right]^{2-}}+4 \mathrm{H}_2 \mathrm{O}\)

In both these cases, the weak field ligand water is replaced by the F and Cl ions.

Question 26. What is the coordination entity formed when an excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S (g) is passed through this solution?
Answer:

⇒ \(\mathrm{CuSO}_{4(\mathrm{aq})}+4 \mathrm{KCN}_{(\mathrm{aq})} \longrightarrow \mathrm{K}_2\left[\mathrm{Cu}(\mathrm{CN})_4\right]_{(\mathrm{aq})}+\mathrm{K}_2 \mathrm{SO}_{4(\mathrm{aq})}\)

⇒ \(\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+}+4 \mathrm{CN}^{-} \longrightarrow\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}+4 \mathrm{H}_2 \mathrm{O}\)

Thus, the coordination entity formed in the process is K₂[Cu(CN)₄]. It is a very stable complex, which does not ionize to give Cu²⁺ ions when added to water. Hence, Cu²⁺ ions are not precipitated when H₂S, i is passed through the solution.

Question 27. Discuss the nature of bonding in the following coordination entities based on valence bond theory:

1. [Fe(CN)₆]^4-
2. [FeF₆]^3-
3. [Co(C₂CO₄)₃]^3-
4. [CoF₆]3-

Answer: [Fe(CN)₆]^4-

In the above coordination complex, iron exists in the +2 oxidation state.

Fe²⁺: Electronic configuration is 3d6

Orbitals of Fe²⁺ ion:

As CN^– ion is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since there are six ligands around the central metal ion, the most feasible hybridization is d²sp³.

d²sp³ hybridized orbitals of Fe²⁺ are:

6 electron pairs from CN^– ions occupy the six hybrid dfsp3 orbitals.

Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

2. [FeF₆]^3-

In this complex, the oxidation state of Fe is +3.

Orbitals of Fe^-3 ion:

There are 6 F^– ions. Thus, it will undergo d²sp³ or sp²d³ hybridization. As F^– is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbitals. Hence, the most feasible hybridization is sp³d².

sp³d² hybridized orbitals of Fe³⁺ it are:

Hence, the geometry of the complex is found to be octahedral.

3. [Co(C₂CO₄)₃]^3-

Cobalt exists in the +3 oxidation state in the given complex.

Orbitals of Co³⁺ ion:

Oxalate ion is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be sp³d².

sp³d² hybridized orbitals of Co³⁺

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

4. [CoF₂]^3-

Cobalt exists in the +3 oxidation state.

Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co!+ ion will undergo sp’d2 hybridization. sp³d² hybridized orbitals of Co³⁺ ion are:

Hence, the geometry of the complex is octahedral and paramagnetic.

Question 28. Draw a figure to show the splitting of d orbitals in an octahedral crystal field.
Answer:

Question 29. What is a spectrochemical series? Explain the difference between a weak field and a strong field ligand.
Answer:

A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

I-< Br- <S² <SCN <CI <N^3- <OH^– <C₂O ^2-₄ <~H₂O < NCS^– <edta^4- <NH₃<en^– <CN^– < CO

Question 30. What is crystal field splitting energy? How does the magnitude of Δ₀ decide the actual configuration of d-orbitals in a coordination entity?
Answer:

• The degenerate d-orbitals (in a spherical field environment) split into two levels i.e., e_g and t_2g in the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called crystal-field splitting while the energy difference between the two levels (e_g and t_2g) is called the crystal-field splitting energy.
• It is denoted by Δ₀. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has been filled in the three t_2g orbitals, the filling of the fourth electron takes place in two ways.
• It can enter the e orbitals (giving rise to t³_2g e¹_g like electron configuration) or the pairing of the electrons can take place in the t_2g orbitals (giving rise to t⁴_2g e⁰_g like electronic configuration).
• If the A₀ value of a ligand is less than the pairing energy (P), then the electrons enter the e orbital.  On the other hand, if the Δ₀ value of a ligand is more than the pairing energy (P), then the electrons enter the t_2g orbital.

Question 31. [Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:

In [Cr(NH₃)₆]³⁺ Cr is in the +3 oxidation state i.e., d³ configuration. Since there are three unpaired electrons in 3d orbitals.

Therefore, it undergoes d²sp³ hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic.

In [Ni(CN)₄]^2- Ni exists in the +2 oxidation state i.e., d⁸ configuration.

CN^– is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni²⁺ undergoes dsp² hybridization.

As there are no unpaired electrons, it is diamagnetic.

Question 32. A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:

• In [Ni(H₂O)₆]²⁺, H₂O is a weak field ligand. Therefore, there are unpaired electrons in Ni-‘. In this complex, the 3d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d-d transition is present.
• Hence,[Ni(H₂O)₆]²⁺  is coloured. In [Ni(CN)₄]^2- the electrons are all paired as CN is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]^2-. Hence, it is colourless.

Question 33. [Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:

• The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand.
• In the case of [Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺, the colour differs because there is a difference in the CFSE. Now CN^– is a strong field ligand having a higher CFSE value of compound to the CFSE value of water. This means that the absorption of energy for the infra d-d transition also differs. Hence the transmitted colour also differs.

Question 34. Discuss the nature of bonding in metal carbonyls.
Answer:

• The metal-carbon bonds in metal carbonyls have both σ and π characters. An σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal.
• A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the valent anti-bonding π* orbital (also known as back bonding of the carbonyl group).
• The σ bond strengthens the π bond and vice versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the metal.

Question 35. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
Answer:

1. \(\mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)
2. \(\text { cis- }\left[\mathrm{Cr}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)
3. \(\left(\mathrm{NH}_4\right)_2\left[\mathrm{CoF}_4\right]\)
4. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

Answer:

1. \(\mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]\)

The central metal ion is Co³⁺

Its coordination number is 6.

The oxidation state can be given as:

x – 6 = -3

x = +3

The d orbitals occupation for Co³⁺ is t⁶_2g e⁰_g.

2. \(\text { cis- }\left[\mathrm{Cr}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}\)

The central metal ion is Cr³⁺

The coordination number is 6.

The oxidation state can be given as:

x + 2(0) + 2(-1) = +1

x- 2 = +1

x = +3

The d orbitals occupation for Cr³⁺ is t³_2g.

3. \(\left(\mathrm{NH}_4\right)_2\left[\mathrm{CoF}_4\right]\)

The central metal ion is Co²⁺.

The coordination number is 4.

The oxidation state can be given as:

x- 4 = -2

x = +2

The d orbitals occupation for Co²⁺ is e ⁴_g t³_2g

4. \(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4\)

The central metal ion is Mn²⁺.

The coordination number is 6.

The oxidation state can be given as:

x + 0 = +2

x = +2

The d orbitals occupation for Mn²⁺ is t³_2g e²_g.

Question 36. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also, give the stereochemistry and magnetic moment of the complex:

1. \(\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right] \cdot 3 \mathrm{H}_2 \mathrm{O}\)
2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)
3. \(\mathrm{CrCl}_3(\mathrm{py})_3\)
4. \(\mathrm{Cs}\left[\mathrm{FeCl}_4\right]\)
5. \(\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right]\)

Answer:

1. \(\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right] \cdot 3 \mathrm{H}_2 \mathrm{O}\)

IUPAC Name: Potassium diaquadioxalatochromate (TIT) trihydrate.

The oxidation state of chromium = +3

Electronic configuration: 3d³: t³_2g

Coordination number = 6

Shape: octahedral

Stereochemistry:

Magnetic moment, \(\mu=\sqrt{n(n+2)}\)

\(\mu=\sqrt{3(3+2)}=\sqrt{5} \sim 4 \mathrm{BM}\)

2. \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)

TUPAC name: Pcntaamminechloridocoball (3) chloride

The oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration : d⁶: t⁶_2g.

Stereochemistry:

Magnetic Moment = 0

3. CrCl₃(py)₃

TUPAC name: Trichloridotripyridincchromiuixi (3)

The oxidation state of chromium = +3

Electronic configuration for d³: t³_2g

Coordination number = 6

Shape: octahedral.

Stereochemistry:

Both isomers are optically active. Therefore, a total of 4 isomers exist.

Magnetic moment, \(\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15} \sim 4 B M\)

4. Cs[FeCl₄]

IUPAC name: Caesium tetrachloroferrate (3)

The oxidation state of Fe = +3

Electronic configuration d⁵: e²_g t³_2g

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{5(5+2)}=\sqrt{35} \sim 6 \mathrm{BM}\)

5. K₄[Mn(CN)₆]

IUPAC Name: Potassium hexacyanomanganate (2)

The oxidation state of manganese = +2

Electronic configuration: d⁵: t⁵_2g

Coordination number = 6

Shape: octahedral

S(geochemistry:optically inactive

Magnetic moment,

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)

Question 37. What is meant by the stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:

The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.

⇒ \(\mathrm{M}+3 \mathrm{~L} \rightleftharpoons \mathrm{ML}_3\)

Stability constant,

⇒ \(\beta=\frac{\left[\mathrm{ML}_3\right]}{[\mathrm{M}][\mathrm{L}]^3}\)

For this reaction, the greater the value of the stability constant, the greater is the proportion of ML₃ in the solution.

Stability can be of two types:

1. Thermodynamic stability: The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.
2. Kinetic stability: This helps determine the speed with which the transformation will occur to attain the state of equilibrium.

Question 38. What is meant by the chelate effect? Give an example.
Answer:

When a ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.

For example: ⇒ \(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+6 \mathrm{NH}_{3 \mathrm({aq})} \longleftrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]_{(\mathrm{aq})}^{2+} \log \beta=7.99\)

⇒ \(\mathrm{Ni}_{(\mathrm{aq})}^{2+}+3 \mathrm{en}_{(\mathrm{aq})} \longleftrightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]_{(\mathrm{aq})}^{2+} \log \beta=18.1 \text { (more stable) }\)

Question 39. Discuss briefly giving an example in each case the role of coordination compounds in:

1. Biological system
2. Medicinal chemistry
3. Analytical Chemistry
4. Extraction/metallurgy of metals

Answer:

1. Role of coordination compounds in biological system: We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium.
2. Role of coordination compounds in medicinal chemistry: Certain coordination compounds of platinum (for example, cis-plutin) are used for inhibiting the growth of tumours.
3. Role of coordination compounds in analytical chemistry: During salt analysis, several basic radicals are detected with the help of the colour changes they exhibit with different reagents.
4. Role of coordination compounds in extraction or metallurgy of metals: From [Au(CN)₂]⁺ solution, gold is extracted by the addition of zinc metal.

Question 40. How many ions are produced from the complex Co(NH₃)₆Cl₂ in solution?

1. 6
2. 4
3. 3
4. 2

Answer: 3

The given complex can be written as Co(NH₃)₆Cl₂ Thus, [Co(NH₃)₆]²⁺ along with two Cl ions are produced.

Question 41. Amongst the following ions which one has the highest magnetic moment value?

1. [Cr(H₂O₆)]³⁺
2. [Fe(H₂O)₆]²⁺
3. [Zn(H₂O)₆]²⁺

Answer:

No. of unpaired electrons in [Cr(H₂O₆)]³⁺ = 3

⇒ \(\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt 15 {\sim 4 B M}\)

No. of unpaired electrons in [Fe(H₂O)₆]²⁺ = 4

⇒ \(\mu=\sqrt{4(4+2)}=\sqrt{24} \sim 5 \mathrm{BM}\)

No. of unpaired electrons in [Zn(H₂O)₆]²⁺ = 0

Hence, [Fe(H₂O)₆]²⁺ has the highest magnetic moment value.

Question 42. The oxidation number of cobalt in K[Co(CO)4] is

1. +1
2. +3
3. -1
4. -3

Answer:

We know that CO is a neutral ligand and K carries a charge of + 1. Therefore, the complex can be written as K⁺[Co(CO)₄]^–. Therefore, the oxidation number of Co in the given complex is -1. Hence, option (3) is correct.

Question 43. Amongst the following, the most stable complex is

1. [Fe(H₂O)₆]³⁺
2. [Fe(NH₃)₆]³⁺
3. [Fe(C₂O₄)₃]^3-
4. [FeCl₆]^3-

Answer:

We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C₂O₄)₃]^3-

Question 44. What will be the correct order for the wavelengths of absorption in the visible region for the following:

⇒\(\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-},\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+},\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Answer:

The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE value of the ligands increases in the spectrochemical series is as follows:

H₂O < NH₃ < NO₂^–

Thus, the amount of crystal-field splitting observed will be in the following order:

Δ_0[H2O] < Δ_0[NH3] < Δ_0[NO2-]

Hence, the wavelengths of absorption in the visible region will be in the following order:

⇒ \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}>\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}>\left[\mathrm{Ni}(\mathrm{NO})_6\right]^{4-}\)

D And F Block

Question 1. The silver atom has filled d orbitals (4d ) in its ground state. How can you say that it is a transition element?
Answer: Ag has a filled 4d orbital (4d¹⁰ 5s¹) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital.

Thus, the d-orbital now becomes incomplete (4d⁹). Hence, it is a transition element.

Question 2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol^-1Why?
Answer: The extent of metallic bonding a clement undergoes decides the enthalpy of atomization.

• The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d¹⁰ 4s²), some unpaired electrons account for their stronger metallic bonding.
• Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

Question 3. Which of the 3d scries of the transition metals exhibits the largest number of oxidation states and why?
Answer: Mn (Z = 25) = 3d⁵ 4s²‘ Mm has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence. Mn exhibits the largest number of oxidation states, ranging from +2 to +7.

Question 4. The E⁰(M²⁺/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high energy and low energy)
Answer: The E⁰(M²⁺/M) value of a metal depends on the energy changes involved in the following:

1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

M_(s) → M_(g) Δ_sH(Sublimation energy)

2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

M_(g) → M²⁺_(g) Δ_iH(Ionization energy)

3. Hydration: The energy released when one mole of ions is hydrated.

M²⁺_(g)→ M²⁺_(aq) Δ_hydH(Hydration energy)

Now. copper has a high energy of atomization and low hydration energy.

Hence, the E^θ(M²⁺/M) value for copper is positive.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of (the transition elements?
Answer: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals.

• The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d⁰, d⁵, and d¹⁰. Since these states arc exceptionally stable, their ionization enthalpies are very high.
• In the case of first ionization energy. Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d⁵ ). On the other hand,  Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d¹⁰ 4s²).
• The second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out.
• Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr⁺:3d⁵ and Cu⁺:3d¹⁰). Hence, taking out one electron more from this stable configuration will require a lot of energy.

Question 6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they can oxidize the metal to its highest oxidation state.

Question 7. Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Answer: The following reactions are involved when Cr²⁺ and Fe²⁺ act as reducing agents.

→ Cr Cr²⁺ 3+ Fe₂₊ → Fe³⁺

The value is -0.41 V and is +0.77 V. This means that Cr²⁺ can be easily oxidized to Cr³⁺, but Fe does not get oxidized to Fe³⁺ easily. Therefore. Cr²⁺ is a better-reducing agent than Fe³⁺.

Question 8. Calculate the ‘spin only’ magnetic moment of M²⁺_(aq), ion (Z = 27).
Answer:

Z = 27

⇒ [ Ar] 3d⁷ 4s²

M²⁺ = [Ar] 3d⁷

i.e., 3 unpaired electrons

.’. n = 3

⇒ \(\sqrt{n(n+2)}=\mu\)

⇒ \(\sqrt{3(3+2)}=\mu\)

⇒ \(\sqrt{15}=\mu\)

⇒ \(\mu \approx 4 B M\)

Question 9. Explain why the Cu⁺ ion is not stable in aqueous solutions.
Answer: In an aqueous medium, Cu²⁺ is more stable than Cu⁺. This is because although energy is required to remove one electron from Cu⁺ to Cu²⁺, the high hydration energy of Cu²⁺ compensates for it. Therefore. Cu⁺ ions in an aqueous solution are unstable. It is disproportionate to give Cu²⁺ and Cu.

Question 10. Actinoid contraction is greater from element to element than lanthanide contraction. Why?
Answer: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanides).

Thus, the effective nuclear charge experienced by electrons in valence shells in the case of actinides is much more than that experienced by lanthanides. Hence, the size contraction in actinides is greater as compared to that in lanthanides.

Question 11. Write down the electronic configuration of:

1. Cr³⁺
2. Pm³⁺
3. Cu⁺
4. Ce⁴⁺
5. Co²⁺
6. Lu²⁺
7. Mn²⁺
8. Th⁴⁺

Answer:

1. Cr³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ Or, [AT]¹⁸ 3d³
2. Pm³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f⁴ Or, [Xe]⁵⁴ 4f⁴
3. Cu⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ Or, [Ar]¹⁸ 3d¹⁰
4. Ce⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4P⁶ 4d¹⁰ 5s² 5p⁶ Or, [Xe]⁵⁴.
5. Co²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or, [Ar ]¹⁸ 3d⁷
6. Lu²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹ Or, [Xe]⁵⁴ 4f¹⁴ 5d¹
7. Mn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or, [Ar]¹⁸ 3d⁵
8. Th⁴⁺:  1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶ Or, [Rn]⁸⁶

Question 12. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Answer: The electronic configuration of Mn²⁺ is [Ar]¹⁸ 3d⁵. (half-filled stability) The electronic configuration of Fe²⁺ is [Ar]¹⁸ 3d⁶. (After losing one electron gain half-filled stability).

Question 13. Explain briefly how the +2 state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer: The oxidation states displayed by the first half of the first row of the transition metals arc are given in the table below.

It can be easily observed that except Sc, all other metals display a +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.

1. Sc (+2) = d¹
2. Ti (+2) = d²
3. V (+2) = d³
4. Cr (+2) = d⁴
5. Mn (+2) = d⁵

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in the (+2) state also increases from Tit+2) to Mn(+2), the stability of+2 state increases (as d-orbital is becoming more and more half-filled). Mn(+2) has d electrons (that is a half-filled d shell, which is highly stable).

Question 14. To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate your answer with examples.
Answer: The elements in the first – half of the transition series exhibit many oxidation states with Mn exhibiting a maximum number of oxidation states (+2 to +7). The stability of the +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital.

Question 15. What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³, 3d⁵, 3d⁸ and 3d⁴?
Answer:

Question 16. Name the oximeter anions of (the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:

1. Vanadate, VO^–₃ → The oxidation state of V is +5.
2. Chromate. CrO^2-₄ → The oxidation state of Cr is +6
3. Permanganate, MnO^–₄ → Oxidation state of Mn is +7.

Question 17. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer: As we move along the lanthanoid series, the atomic number increases gradually by one.

• This means that the number of electrons and protons present in an atom also increases by one. As electrons are added to the same shell, the effective nuclear charge increases.
• This happens because the increase in nuclear attraction due to the addition of a proton is more pronounced than the increase in the inter-electronic repulsions due to the addition of an electron.
• Also, with the increase in atomic number, the number of electrons in the 4f orbital increases. The 4f electrons have a poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases.
• Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with an increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction

1. There is a similarity in the properties of the second and third transition series.
2. Separation of lanthanoids is possible due to lanthanide contraction.
3. It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La (OH)₃ to Lu (OH)₃)

Question 18. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Answer:

Transition elements are those elements in which the atoms or ions (in a stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between the s-block and the p-block.

Therefore, these are called transition elements. Elements such as Zn, Cd and Hg cannot be classified as transition elements because these have filled the d-subshell.

Question 19. In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:

Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n — 1 ) d^1-10 ns^0-2. The non-transition elements either do not have a d-orbital or have a filled d-orbital. therefore, the electronic configuration of non-transition elements is ns^1-2 or ns² np^1-16.

Question 20. What are the different oxidation states exhibited by the lanthanoids?
Answer: In the lanthanide series. +3 oxidation stale is most common i.e., Ln (3) compounds are predominant. However. +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Question 21. Explain giving reasons:

1. Transition metals and many of their compounds show paramagnetic behaviour.
2. The enthalpies of atomisation of the transition metals are high.
3. The transition metals generally form coloured compounds.
4. Transition metals and their many compounds act as good catalysts.

Answer:

1. Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.
2. Transition elements have a high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.
3. Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from the visible light region to promote an electron from one of the d-orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies.
   1. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation.
   2. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.
4. The catalytic activity of the transition elements can be explained by two basic facts.
   1. Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, E_a, for the reaction.
   2. Transition metals also provide a suitable surface for the reactions to occur.

Question 22. What are interstitial compounds? Why are such compounds well-known for transition metals?
Answer:

Transition metals are large and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic sizes), such as H, C, and N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

Question 23. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer:

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation stales differ by

1. (Fe²⁺ and Fe³⁺: Cu⁺ and Cu²⁺). In non-transition elements, the oxidation states differ by
2. For example. +2 and +4 or + 3 and +5. etc.

Question 24. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer:

Potassium dichromate is prepared from chromite ore (FeCr₂O₄) in the following steps.

Step (1): Preparation of sodium chromate

⇒ \(4 \mathrm{FeCrO}_4+16 \mathrm{NaOH}+7 \mathrm{O}_2 \longrightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{FeO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Step (2): Conversion of sodium chromate into sodium dichromate

⇒ \(2 \mathrm{NaCrO}_4+\text { conc. } \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}\)

Step (3): Conversion of sodium dichromate to potassium dichromate

⇒ \(\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KCl} \longrightarrow \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{NaCl}\)

Potassium dichromate being less soluble than sodium dichromate is obtained in the form of
orange-coloured crystals and can be removed by filtration.

The dichromate ion ( Cr₂O^2-₄ ) exists in equilibrium with the chromate ( Cr₂O²₄ ) ion at pH 4.

However, by changing the pH, they can be interconverted.

⇒ \(\underset{\text { Chromate ion(Yellow) }}{2 \mathrm{CrO}_4^{2-}}+2 \mathrm{H}^{+} \stackrel{\mathrm{PH}^{-4}}{\longrightarrow} \underset{\text { Dichromate ion(Orange) }}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}}+\mathrm{H}_2 \mathrm{O}\)

On increasing pH, \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}\)

Question 25. Describe the oxidation action of potassium dichromate and write the ionic equations for its reaction with:

1. Iodide
2. Iron(2) solution and
3. H₂S

Answer:

K₂Cr₂O₇ acts as a very strong oxidising agent in the acidic medium.

⇒ \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+4 \mathrm{H}_2 \mathrm{O}+3[\mathrm{O}]\)

K₂Cr₂O₇ takes up electrons to gel reduced and acts as an oxidising agent. The reaction of K Cr 0_ with iodide ion, iron (2) solution and H S are given below.

1. K₂Cr₂O₇ oxidizes iodide to iodine.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

2I^– → I₂ + 2e^– × 3

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}2+7 \mathrm{H}_2 \mathrm{O}\)

K₂Cr₂O₇ oxidizes iron (2) solution to iron (3) solution i.e., ferrous ions to ferric ions.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

Fe²⁺ →  Fe³⁺ + e^– × 6

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}\)

K₂Cr₂O₇ oxidizes H₂S to sulphur.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

H₂S →  S + 2H⁺ + 2e^– × 3

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{H}_2 \mathrm{~S}+8 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{~S}+7 \mathrm{H}_2 \mathrm{O}\)

Question 26. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

1. Iron(2) ions
2. SO₂
3. Oxalic acid?

Write the ionic equations for the reactions.

Answer:

Potassium permanganate can be prepared from pyrolusite (MnO₂). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO₃ or KCIO₄ to give K₂MnO₄.

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.

Electrolytic oxidation

⇒ \(\mathrm{K}_2 \mathrm{MnO}_4 \longleftrightarrow 2 \mathrm{~K}^{+}+\mathrm{MnO}_4^{2-}\)

⇒ \(\mathrm{H}_2 \mathrm{O} \longleftrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}\)

At the anode, manganate ions are oxidized to permanganate ions.

⇒ \(\underset{\text { Green }}{\mathrm{MnO}_4^{2-}} \longleftrightarrow \underset{\text { Purple }}{\mathrm{MnO}_4^{-}}+\mathrm{e}^{-}\)

Oxidation by chlorine

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KCl}\)

⇒ \(2 \mathrm{MnO}_4^{2-}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{MnO}_4^{-}+2 \mathrm{Cl}\)

Oxidation by ozone

⇒ \(2 \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{KMnO}_4+2 \mathrm{KOH}+\mathrm{O}_2\)

⇒ \(2 \mathrm{MnO}_4^2+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{OH}+\mathrm{O}_2\)

Acidified KMnO₄ solution oxidizes Fe (2) ions to Fe (3) ions i.e., ferrous to ferric ions.

⇒ \(2 \mathrm{MnO}_4^2+\mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{OH}+\mathrm{O}_2\)

Fe²⁺ → Fe³⁺ + e^– × 5

⇒ \(\mathrm{MnO}_{+}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}2 \mathrm{O}\)

Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.

⇒ \(\left.\mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{e} \longrightarrow \mathrm{Mn}^{2+}+3 \mathrm{H}_2 \mathrm{O}\right] \times 2\)

⇒ \(\left.2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{SO}_2+\mathrm{O}_2 \longrightarrow 4 \mathrm{H}^{+}+2 \mathrm{SO}_4^{2-}+2 \mathrm{e}\right] \times 5\)

⇒ \(2 \mathrm{MnO}_4^{-}+10 \mathrm{SO}_2+5 \mathrm{O}_2+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{SO}_4^{2-}+8 \mathrm{H}^{+}\)

Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

⇒ \(\mathrm{MnO}_4+8 \mathrm{H}^{+}+5 \mathrm{e} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \times 2\)

⇒ \(\left.\mathrm{C}_2 \mathrm{O}_4^2- \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{c}\right] \times 5\)

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_7^{2+}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Question 27. For M²⁺/M and M³⁺/M²⁺+ systems, the E^– values for some metals are as follows:

1. Cr²⁺ /Cr – 0.9 V
2. Cr³/Cr²⁺ – 0.4 V
3. Mn²⁺/Mn -1.2V
4. Mn³⁺/Mn²⁺ +1.5V
5. Fe²⁺/Fe – 0.4 V
6. Fe³⁺/Fe²⁺ + 0.8V

Use this data to comment upon:

1. The stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ and Mn³⁺ and
2. The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer:

• The E^– value for Fe³⁺/Fe²⁺ is higher than that for Cr³⁺ /Cr²⁺ and lower than that for Mn³⁺/Mn²⁺ So. the reduction of Fe³⁺ to Fe²⁺ is easier than the reduction of Cr³⁺ to Cr²⁺ but not as easy as the reduction of Mn³⁺ to Mn²⁺.
• Hence, Fe³⁺ is more stable than Mn³⁺, but less stable than Cr³⁺. These metal ions can be arranged in the increasing order of their stability as Mn³⁺< Fe³⁺ < Cr³⁺

The reduction potentials for the given pairs increase in the following order.

Mu²⁺ /Mn < Cr²⁺ /Cr < Fe²⁺/Fe

So, the oxidation of Fe to Fe²⁺ is not as easy as the oxidation of Cr to Cr²⁺ and the oxidation of Mn to Mn²⁺. Thus these metals can be arranged in the increasing order of their ability to get oxidised: Fe < Cr < Mn

Question 28. Predict which of the following will be coloured in aqueous solution. Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.
Answer: Only the ions that have unpaired electrons in the el-orbital will be coloured. The ions in which the d orbital is empty or filled will be colourless.

Question 29. Compare the stability of the +2 oxidation state for the elements of the first transition series.
Answer:

The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.

Question 30. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

1. Electronic configuration
2. Atomic and ionic sizes
3. Oxidation state
4. Chemical reactivity.

Answer:

1. Electronic configuration: The general electronic configuration for lanthanoids is [Xe]⁵⁴ 4f^1-145d^0-1 6s² and that for actinoids is [Rn]⁸⁶ 5f^1-14 6d^0-1 7s²
2. Atomic and Ionic sizes: Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.
3. Oxidation states: The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of the extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies.
4. Chemical reactivity: In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similarly to Al. Actinoids, on the other hand- are highly reactive metals, especially when they are finely divided.

Question 31. How would you account for the following:

1. Of the d⁴ species, Cr²⁺ is strongly reducing while manganese(3) is strongly oxidising.
2. Cobalt(2) is stable in an aqueous solution but in the presence of complexing reagents, it is easily oxidised.
3. The d¹ configuration is very unstable in ions.

Answer:

1. Cr²⁺ is strongly reducing in nature. It has a d⁴ configuration. While acting as a reducing agent, it gets oxidized to Cr³⁺ (electronic configuration, d³ ). This d³ configuration can be written as a t³_2g configuration, which is a more stable configuration. In the case of Mn³⁺ (d⁴), It acts as an oxidizing agent and gets reduced to Mn (d⁵). This has an exactly half-filled d-orbital and is highly stable.
2. Co(2) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co(3). Although the 3^rd ionization energy for Co is high, the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong-field ligands overcomes this ionization energy.
3. The ions in the d¹ configuration tend to lose one more electron to get into the stable d⁰ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the cl -orbital of these ions. Therefore, they act as reducing agents.

Question 32. What is meant by ‘disproportionation’? Give two examples of disproportionation reactions in an aqueous solution.
Answer:

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation-reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

For example,

⇒ \(\underset{\mathrm{Cr}(\mathrm{5})}{3 \mathrm{CrO}_4^{3-}}+8 \mathrm{H}^{+} \longrightarrow \underset{\mathrm{Cr}(\mathrm{6})}{2 \mathrm{CrO}_{+}^{2-}}+\underset{\mathrm{Cr}(\mathrm{3})}{\mathrm{Cr}^{3+}}+4 \mathrm{H}_2 \mathrm{O}\)

Cr(5) is oxidized to Cr(6) and reduced to Crd(3)

⇒ \(\underset{\mathrm{Mn}(\mathrm{6})}{3 \mathrm{MnO}_4^{2-}}+4 \mathrm{H}^{+} \longrightarrow \underset{\mathrm{Mn}(\mathrm{7})}{2 \mathrm{MnO}_4^{-}}+\underset{\mathrm{Mn}(\mathrm{4})}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{O}\)

Mn (6) is oxidized to Mn (7) and reduced to Mn (4).

Question 33. Which metal in the first series of transition metals exhibits a +1 oxidation state most frequently and why?
Answer: In the first transition series, Cu exhibits a + 1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d¹⁰. The filled el-orbital makes it highly stable.

Question 34. Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺ \ Cr³⁺ \ V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?
Answer:

Cr³⁺ is the most stable in aqueous solutions owing to a t³_2g half-filled configuration.

Question 35. Give examples and suggest reasons for the following features of the transition metal chemistry:

1. The lowest oxide of transition metal is basic, (and the highest is amphoteric/acidic.
2. A transition metal exhibits the highest oxidation state in oxides and fluorides.
3. The highest oxidation state is exhibited in oxoanions of a metal.

Answer:

1. In the ease of a lower oxide of a transition metal, the metal atom has a low oxidation stale. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.
   1. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so. they are unavailable. There is also a highly effective nuclear charge. As a result, it can accept electrons and behave as an acid.
   2. For example. Mn²O is basic and Mm₂⁷O₇ is acidic.
2. Oxygen and fluorine act as strong oxidising agents because of their high electronegatives and small sizes. Hence, they bring out the highest oxidation states from the transition metals.
   1. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF₆ and V₂O₅ the oxidation states of Os and V are +6 and +5 respectively.
3. Oxygen is a strong oxidising agent due to its high electronegativity and small size. So. oxo-anions of metal have the highest oxidation state. For example, in MnO₄^– the oxidation state of Mn is +7.

Question 36. What are alloys? Name an important alloy which contains some of the lanthanoid metals Mention its uses.
Answer:

• An alloy is a solid of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements.
• An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C. Si. Ca and Al.

Uses:

1. Mischmetal is used in cigarettes and gas lighters.
2. It is used in flame-throwing tanks.
3. It is used in tracer bullets and shells.

Question 37. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
Answer:

Inner transition elements are those elements in which the last electron enters the f-orbital. The elements in which the 4f and 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95 and 102.

Question 38. The chemistry of the actinoid elements is not as smooth as that of the Lanthanoids. Justify this statement by giving some examples of the oxidation state of these elements.
Answer:

• Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states. +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d and 6s orbitals is quite large.
• On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very low. Hence aclinoids display a large number of oxidation states. For example, uranium and plutonium display +3. +4. +5. and +7. The most common oxidation state in the case of retinoids is also +3.

Question 39. Which are the last elements in the series of actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Answer:

The last element in the actinoid series is lawrencium. Lr. Its atomic number is 103 and its electronic configuration is [Rn] 5f¹⁴ 6d¹ 7s². The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains a stable f¹⁴ configuration.

Question 40. Use Hund’s rule to derive the electronic configuration of the Ce³⁺ ion and calculate its magnetic moment based on the ‘spin-only’ formula.
Answer:

Ce³⁺:1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹

Magnetic moment can be calculated as:

⇒ \(\mu=\sqrt{n(n+2)}\)

Where, n = number of unpaired electrons.

In Ce³⁺, n = 1

Therefore, \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)

Question 41. Name the members of the lanthanoid series which exhibit a +4 oxidation state and those which exhibit a +2 oxidation state. Try to correlate this type of behaviour with the electronic configuration of these elements.
Answer:

The lanthanoids that exhibit + 2 and +4 states are shown in the given table. The atomic numbers of these elements are given in parentheses.

1. Ce after forming Ce⁴⁺ attains a stable electronic configuration of [Xe].
2. Tb after forming Tb⁴⁺ attains a stable electronic configuration of [Xe] 4f⁷
3. Eu after forming Eu²⁺ attains a stable electronic configuration of [Xe] 4f⁷
4. Yb after forming Yb²⁺ attains a stable electronic configuration of [Xe] 4f¹⁴

Question 42. Write the electronic configuration of the elements with the atomic numbers 61.91. 101. and 109.
Answer:

Question 43. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

1. Electronic configurations.
2. Oxidation states,
3. Ionisation enthalpies, and
4. Atomic sizes.

Answer:

1. In the 1^st, 2^nd and 3^rd transition series, the 3d, 4d and 5d orbitals are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations:

1. Cr(24) = 3d⁵ 4s¹
2. Cu(29) = 3d¹⁰ 4s¹

Similarly, there are exceptions in the second transition series. These are:

1. Nb(41) = 4d⁴ 5s¹
2. Mo(42) = 4d⁵ 5s¹
3. Tc (43) = 4d⁶ 5s¹
4. Ru(44) = 4d⁷ 5s¹
5. Rh(45) = 4d⁸ 5 s¹
6. Pd(46) = 4d¹⁰ 5s⁰
7. Ag(47) = 4d¹⁰ 5s¹

There are some exceptions in the third transition series as well. These are:

1. Pt(78) = 5d⁹ 6s¹
2. Au(79) = 5d¹⁰ 6s¹

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

2. In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

• However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
• The stability of the +2 and +3 oxidation states decreases in the second and third transition series, wherein higher oxidation states are more important.

For example \(\left[\begin{array}{l}
\mathrm{2} \\
\mathrm{Fe}(\mathrm{CN})_6
\end{array}\right]^{4-} \cdot\left[\begin{array}{l}
\mathrm{3} \\
\mathrm{Co}\left(\mathrm{NH}_3\right)_6
\end{array}\right]^{3+} \cdot\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^3\) are stable complexes,

but no such complexes are known for the second and third transition series such as Mo, W, Rh, and In. They form complexes in which their oxidation states are high. For example WCI_6, ReF₇, RuO₄ etc.

3. In each of the transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions.

• The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series.
• Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series.
• There are also elements in the 2′ transition series whose first ionisation enthalpies are lower than those of (the elements corresponding to the same vertical column in the 1 transition series.

4. Atomic si/.e generally decreases from left to right across a period. Now. among the three transition series, the atomic sizes of the elements in the second transition series are greater than those of the element corresponding to the same vertical column in the first transition series.

However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Question 44. Write down the number of 3d electrons in each of the following ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni ²⁺ and Cu²⁺ Indicate how would you expect the live 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer:

Question 45. What can be inferred from the magnetic moment values of the following complex species?

Answer:

Magnetic moment (μ) is given as μ = \(\sqrt{n(n+2)} B M\)

1. For value n = 1, μ = \(\sqrt{1(1+2)}=\sqrt{3}=1.732 \mathrm{BM}\)
2. For value n = 2, μ = \(\sqrt{2(2+2)}=\sqrt{8}=2.83 \mathrm{BM}\)
3. For value n = 3, μ = \(\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
4. For value n = 4, μ = \(\sqrt{4(4+2)}=\sqrt{24}=4.899 \mathrm{BM}\)
5. For value n = 5, μ = \(\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{BM}\)

1. K₄[Mn(CN)₆]

For transition metals, the magnetic moment is calculated from the spin-only formula. Therefore.

⇒ \(\mu=\sqrt{n(n+2)}=2.2 \mathrm{BM}\)

We can see from the above calculation that the given value is closest to n = 1. Also, in this complex. Mn in in the 4-2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN is a strong field ligand that causes the pairing of electrons.

2. [Fe(H₂O)₆]²⁺

⇒ \(\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}=5.3 \mathrm{BM}\)

We can see from the above calculation that the given value is closest to n = 4. Also, in this complex. Fe is in the 4-2 oxidation state. This means that Fe has 6 electrons in the d-orbital. Hence, we can say that H₂O is a weak field ligand and does not cause the pairing of electrons.

3. K₂[MnCI₄]

⇒ \(\mu=\sqrt{n(n+2)}=5.9 B M\)

• We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the 4-2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
• Hence, we can say that Cl^– is a weak field ligand and does not cause the pairing of electrons.

Chemical Kinetics

Question 1. For the reaction R → P. the concentration of a reactant changes from 0,03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:

Average rate of reaction = \(-\frac{\Delta[\mathrm{R}]}{\Delta \mathrm{t}}\)

⇒ \(-\frac{[R]_2-[R]_1}{t_2-t_1}=-\frac{0.02-0.03}{25} \mathrm{Mmin}^{-1}\)

⇒ \(-\frac{0.02-0.03}{25} \mathrm{Mmin}^{-1}\)

⇒ \(4 \times 10^{-4} \mathrm{M} \mathrm{Min}^{-1}\)

⇒ \(\frac{4 \times 10^{-4}}{60} \mathrm{Ms}^{-1}\)

⇒ \(6.67 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\)

Question 2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Answer:

⇒ \(-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{2} \frac{[\mathrm{A}]_2-[\mathrm{A}]_1}{\mathrm{t}_2-\mathrm{t}_1}\)

⇒ \(-\frac{1}{2} \frac{[0.4-0.5]}{10}=-\frac{1}{2} \frac{-0.1}{10}\)

⇒ 0.005 mol L^-1 min^-1 = 5 × 10^-3 M min^-1

Question 3. For a reaction, A + B → Product; the rate law is given by, r = K[ A]^1/2[B]² . What is the order of the reaction?
Answer:

The order of the reaction, \(\frac{1}{2}+2=2 \frac{1}{2}=2.5\)

Question 4. The conversion of molecules X to Y follows second-order kinetics. If the concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer:

The order of reaction is defined as the sum of the powers of concentrations in the rate law.

The rate of second-order reaction can be expressed as rate = k[A]²

The reaction X → Y follows second-order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k [ X ]²

Let [X] = a mol L^-1 . then equation (l) can be written as:

Rale = k [a]² = ka²

If the concentration of X is increased to three times, then [X] = 3a mol L^-1

Now, the rate equation will be ;

Rate = k [3a]² = 9(ka²)

Hence, the rate of formation will increase by 9 times,

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. A first-order reaction has a rate constant of 1.15 × 10^-3 S^-1. How long will 5g of this reactant take to reduce to 3g?
Answer:

We know that for a 1^st order reaction

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]}=\frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}\)

⇒ \(\frac{2.303}{1.15 \times 10^{-3}} \times 0.2219=444.38 \mathrm{~s}=444 \mathrm{~s}\)

Question 6. The time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If decomposition is a first-order reaction, calculate the rate constant of the reaction.
Answer:

We know that fora 1^st order reaction, \(t_{1 / 2}=\frac{0.693}{k}\)

It is given that t_1/2 = 60 min

⇒ \(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{60}=0.01155 \mathrm{~min}^{-1}=1.155 \times 10^{-2} \mathrm{~min}^{-1}\)

Question 7. What will be the effect of temperature on the rate constant?
Answer:

The rate constant of a reaction is nearly doubled with a 10°C rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by the Arrhenius equation, k =Ae^-Ea/RT

Where A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_a is the activation energy

Question 8. The rate of the chemical reaction doubles for an increase of 10 K in an absolute temperature from 298 K. Calculate E_a.
Answer:

It is given that

T₁ = 298 K

T₂ = (298 + 10) K = 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁= k and that k₂ = 2k

Also, R = 8.314 JK^-1 mol^-1

Now, substituting these values in the equation:

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]\)

We get:

⇒ \(\log \frac{2 k}{k}=\frac{E_a}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right] \Rightarrow \log 2=\frac{E_2}{2.303 \times 8.314}\left[\frac{10}{298 \times 308}\right]\)

⇒ \(E_a=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log 2}{10}\)

⇒ 52897.78 J mol^-1 = 52.9 kJ mol^-1

Question 9. The activation energy for the reaction 2HI_g → H₂+ I_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactant having energy equal to or greater than the activation energy.
Answer:

The fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

⇒ \(\mathrm{x}=\mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}\)

⇒ In x = -Ea/RT

⇒ \(\log x=\frac{-E_a}{2.303 R T}\)

⇒ \(\log x=\frac{209500 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 581}=-18.8323\)

Now, x = Anti log (-18.8323)

= 1.471 × 10^-19

Question 10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
Answer:

1. 3NO(g) → N₂O(g) ; Rate = k[NO]²
2. H₂O₂(aq) + 31 (aq.) + H⁺ → 2H₂O(I) + I^–₃ ; Rate = k[H₂O)][I^–]
3. CH₃CHO(g) → CH₄(g) + CO(g) : Rate = k[CH₃CHO]^3/2
4. C₂H₅Cl(g) → C₂H₂(g) + HCI(g) : Rate = k[C₂H₅Cl]

Answer:

Given rate = k[NO]²

Therefore, the order of the reaction = 2

Dimension of k = \(\frac{\text { Rate }}{[\mathrm{NO}]^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{-2} \mathrm{~L}^{-2}}=\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)

Give rate = k[H₂O][I^–]

Therefore, the order of the reaction = 2

Dimension of k = \(\frac{\text { Rate }}{\left[\mathrm{H}_2 \mathrm{O}_2\right]\left[\mathrm{I}^{-}\right]}=\frac{\mathrm{molL}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{molL}^{-1}\right)\left(\mathrm{molL}^{-1}\right)}=\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)

Given rate = k[CHCHO]^3/2

Therefore, order of reaction = \(\frac{3}{2}\)

⇒ \(\text { Dimension of } \mathrm{k}=\frac{\text { Rate }}{\left[\mathrm{CH}_3 \mathrm{CHO}\right]{\frac{3}{2}}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{\frac{3}{2}}}=\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol}^{\frac{3}{2}} \mathrm{~L}^{-\frac{3}{2}}}=\mathrm{L}^{\frac{1}{2}} \mathrm{~mol}^{-\frac{1}{2}} \mathrm{~s}^{-1}\)

Given rate- k[C₂H₅Cl]

Therefore, the order of the reaction = 1

⇒ \(\text { Dimension of } \mathrm{k}=\frac{\text { Rate }}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}\right]}=\frac{\mathrm{molL}^{-1} \mathrm{~s}^{-1}}{\mathrm{~mol} \mathrm{~L}^{-1}}=\mathrm{s}^{-1}\)

Question 11. For the reaction: 2A + B → A₂B the rate = k [A][B]² with k = 2.0 × 10^-6 mol^-2 L²S^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1. [B] = 0.2 mol L^-1. Calculate the rate of reaction after fAJ is reduced to 0.06 mol L^-1.
Answer:

The initial rate of the reaction is Rate = k[A][B]² = (2.0 × 10^-6 mol^-2 L² S^-1) (0.1 mol L^-1) (0.2 mol L^-1)² = 8.0 × 10^-9 mol ^-2 L² s^-1 when [A] is reduced from 0.1 mol L^-1 to 0.06 mol^-1, the concentration of A reacted = (0.1 -0.06) mol L^-1 =0.04 mol L^-1.

Therefore, concentration of B reacted \(=\frac{1}{2} \times 0.04 \mathrm{~mol} \mathrm{~L}^{-1}=0.02 \mathrm{~mol} \mathrm{~L}^{-1}\)

Then, a concentration of B is available. |B] = (0.2 – 0.02) mol L^-1 = 0. 1 8 mol L^-1 After [A] is reduced to 0.06 mol L^-1. the rate of the reaction is given by.

Rate = k(A][B]² = (2.0 × 10^-6 mol^-2 L² s^-1) (0.06 mol L^-1) (0.18 mol L^-1)² = 3.88 × 10 mol L^-1s^-1.

Question 12. The decomposition of NH₃, on the platinum surface is an order reaction. What is the rate of production of N₂, and H₂ if k = 2.5 × 10^-4 mol liter s^-1?
Answer:

2NH₃ → N₂ + 3H₂

⇒ \(\text { Rate of reaction }(\mathrm{r})=-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}\)

Rate (r) = k[NH₃]° = k

(∵ zero order reaction)

= 2.5 × 10^-4

⇒ \(\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\mathrm{r}=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~s}^{-1}\)

⇒ \(\frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=3 \mathrm{r}=3 \times 2.5 \times 10^{-4}=7.5 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{sec}^{-1}\)

Question 13. The decomposition of dimethyl ether leads to the formation of CH_4, H_2, and CO, and the reaction rate is given by: Rate = k [ CH₃OCH₃]^3/2 The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether i.c., Rate = k[P_CH3OCH3],^3/2 If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Answer:

⇒ \(\mathrm{CH}_3 \mathrm{OCH}_3 \longrightarrow \mathrm{CH}_4+\mathrm{CO}+\mathrm{H}_2\)

Rate = k(CH₃OCH₃]^3/2; = k[P(CH₃OCH₃)]^3/2

unit of rate = bar min^-1

⇒ \(\text { Unit of } \mathrm{K}=\frac{\text { Rate }}{\left[\mathrm{P}_{\mathrm{CH}_3 \mathrm{OCH}_3}\right]^{3 / 2}}=\frac{\text { bar } \mathrm{min}^{-1}}{\text { bar }^{3 / 2}}=\mathrm{bar}^{-1 / 2} \mathrm{~min}^{-1}\)

Question 14. Mention the factors that affect the rate of a chemical reaction.
Answer:

The important factors on which the rate of a chemical reaction depends are

1. Nature of the reacting species.
2. Concentration of the reacting species.
3. The temperature at which a reaction proceeds.
4. The surface area of the reactants,
5. Presence of a catalyst

Question 15. A reaction is of second order concerning a reactant. How is its rate affected if the concentration of the reactant is (1) doubled and (2) reduced to half?
Answer:

Given rate (r₀ ) = K [A]²

If [A] is doubled: \(r_1=k[2 A]^2\)

r₁ = 4r₀

If [A] is reduced to half: \(\mathrm{r}_2=\mathrm{k}\left[\frac{\mathrm{A}}{2}\right]^2\)

⇒ \(r_2=\frac{1}{4} r_0\)

Question 16. What is the effect of temperature on the rate constant, of a reaction? How can this temperature effect on the rate constant be represented quantitatively?
Answer:

The rate constant is nearly doubled with a rise in temperature by 100C for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation, k = Ac^-Ea/RT

⇒ \(2.303 \log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{E}_a}{\mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right]\)

where K is the rate constant at the temperature

A is the Arrhenius parameter,

R is the gas constant,

T is the temperature,

and E_a is the energy of activation which is always positive.

Question 17. In a pseudo-first-order hydrolysis of ester in water, the following results were obtained:

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo-first-order rate constant for the hydrolysis of ester.

Answer:

The average rate of reaction between the time interval 30 to 60 seconds

⇒ \(=\frac{\mathrm{d}[\text { Ester }]}{\mathrm{dt}}=-\frac{(0.17-0.31)}{60-30}=\frac{-(-0.14)}{30}=4.67 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

For a pseudo-first-order reaction.

⇒ \(\text { For } t=30 s, \quad k_1=\frac{2.303}{30} \log \frac{0.55}{0.31}=1.911 \times 10^{-2} \mathrm{~s}^{-1}\)

⇒ \(\text { For } \mathrm{t}=60 \mathrm{~s}, \quad \mathrm{k}_2=\frac{2.303}{60} \log \frac{0.55}{0.17}=1.957 \times 10^{-2} \mathrm{~s}^{-1}\)

⇒ \(\text { For } \mathrm{t}=90 \mathrm{~s}, \quad \mathrm{k}_3=\frac{2.303}{90} \log \frac{0.55}{0.85}=2.075 \times 10^{-2} \mathrm{~s}^{-1}\)

Then, average rate constant, k = \(\frac{k_1+k_2+k_3}{3}\)

⇒ \(=\frac{\left(1.911 \times 10^{-2}\right)+\left(1.957 \times 10^{-2}\right)+\left(2.075 \times 10^{-2}\right)}{3}=1.98 \times 10^{-2} \mathrm{~s}^{-1}\)

Question 18. A reaction is first order in A and second order in B:

1. Write the differential rate equation.
2. How is the rate affected when the concentration of B is tripled?
3. How is the rate affected when the concentration of both A and B are doubled?

Answer:

Rate = k [A]¹[B]²

r₀= K[A]¹[B]²

r₁,= k [A]¹[3 B]²

r₁, = 9 × r0

r₀ = K [A]¹[B]²

r₀ = k [2A]¹[2B]²

r₂ = – 8 × r₀

Question 19. In a reaction between A and B the initial rate of reaction (r₀) was measured for different initial concentrations A and B as given below:

What is the order of the reaction concerning A and B?

Answer:

Let the order of the reaction concerning A be x and B be y.

Therefore, r₀ = k [A]^x[B]^y 

5.07 x 10^-5 =k [0.20]^x[10.30]^y _______ 1

5.07 x 10^-5 = k [0.20]^x[0.10]^y _______ 2

1.43 x 10^-4 = k [0.40]^x[0.05]^y _______ 3

Dividing equation (1) by (2), we obtain

⇒ \(\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{\mathrm{k}[0.20]^{\mathrm{x}}[0.30]^y}{\mathrm{k}[0.20]^{\mathrm{x}}[0.10]^y}\)

⇒ \(I=\frac{[0.30]^y}{[0.10]^y}\)

⇒ \(\left(\frac{0.30}{0.10}\right)^0=\left(\frac{0.30}{0.10}\right)^y\)

⇒ y = 0

Dividing equation (3) by (1), we obtain

⇒ \(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^x[0.05]^y}{k[0.20]^x[0.30]^y}\)

⇒ \(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^x}{[0.20]^x}\)

[Since y = 0 [0.05]y =[0.30] = 1]

⇒ 2.821 =2^x

⇒ log 2.821 = x log 2 (Taking log on both sides)

⇒ \(x=\frac{\log 2.821}{\log 2}=1.496=1.5\)

Hence, the order of the reaction concerning A is 1.5, and concerning B is zero.

Question 20. The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D

Determine the rate law and the rate constant for the reaction.

Answer:

Let the order of the reaction concerning A be x and for B be y.

Therefore, the rate of the reaction is given by,

Rate = k[A]^x[B]^y

According to the question.

6.0 × 10^-3 = k[0.1]^x [0.1]^y

7.2 × 10^-2 = k[0.3]^x[0.2]^y

2.88 × 10^-1 = k[0.3]^x [0.4]^y

2.40 × 10^-2 = k[0.4]^x [0.1]^y

Dividing equation (4) by (1), we obtain

⇒ \(\frac{2.4 \times 10^{-2}}{6.0 \times 10^{-3}}=\frac{k[0.4]^x[0.1]^y}{k[0.1]^x[0.1]^y}\)

⇒ \(4=\frac{[0.4]^x}{[0.1]^x} \Rightarrow 4=\left(\frac{0.4}{0.1}\right)^x \Rightarrow(4)^1=4^x \Rightarrow x=1\)

Dividing equation (3) by (2), we obtain

⇒ \(\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}=\frac{k[0.3]^x[0.4]^y}{k[0.3]^x[0.2]^y}\)

⇒ \(4=\left(\frac{0.4}{0.2}\right)^y \Rightarrow 4=2^y \Rightarrow 2^2=2^y \Rightarrow y=2\)

Therefore, the rate law is

⇒ \(\text { Rate }=k[A][B]^2 \Rightarrow k=\frac{\text { Rate }}{[A][B]^2}\)

From experiment I, we obtain

⇒ \(\mathrm{k}=\frac{6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From experiment 2, we obtain

⇒ \(\mathrm{k}=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From exp. 3, we obtain,

⇒ \(\mathrm{k}=\frac{2.88 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

From exp. 4, we obtain,

⇒ \(\mathrm{k}=\frac{2.40 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}=6.0 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~min}^{-1}\)

Therefore, the rate is constant.

k = 6.0 L² mol^-2 min^-1.

Question 21. The reaction between A and B is first order concerning A and zero order concerning B. Fill in the blanks in the following table:

Answer:

The given reaction is of the first order concerning A and zero order concerning B.

Therefore, the rate of the reaction is given by.

Rate = k [A]¹[B]⁰

Rate = k [A]

From experiment 1. we obtain

2.0 × 10^-2 mol L^-1 min^-1 = k (0.1 mol L^-1)

k = 0.2 min^-1

From experiment 2. we obtain

4.0 x 10^-2 mol L^-1 min^-1 = 0.2 min^-1 [A]

[A] = [0.2] mol L^-1

From experiment 3, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1 = 0.08 mol L^-1 min^-1

From experiment 4, we obtain

2.0 x 10^-2 mol L^-1 mim^-1 = 0.2 min^-1 [A]

[A] = 0. 1 mol L^-1

Question 22. Calculate the half-life of a first-order reaction from their rate constants given below:

1. 200 s^-1
2. 2 min^-1
3. 4 year^-1

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{200 \mathrm{~min}^{-1}}=\mathbf{3 . 4} \times 10^{-3} \sec \text { (approx.) }\)

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{2 \mathrm{~min}^{-1}}=\mathbf{0 . 3 5} \mathbf{~ m i n} \text { (approx.) }\)

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{4 \text { year }^{-1}}=1.733 \times 10^{-1} \text { years (approx.) }\)

Question 23. The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Answer:

Here, \(k=\frac{0.693}{t_2}=\frac{0.693}{5730} \text { years }^{-1}\)

⇒ \(t=\frac{2.303}{k} \log \frac{a}{(a-x)}\)

⇒ \(\mathrm{t}_{1 / 2} \text { of }{ }^{14} \mathrm{C}=5730 \mathrm{yr} \text {; }\)

Also, a = 1 00, (a – x) = 80

⇒ \(\mathrm{t}=\frac{2.303 \times 5730}{0.693} \log \frac{100}{80}=1845 \mathrm{yr} \)

Question 24. The experimental data for the decomposition of N₂O₃

⇒ \(\left[2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\right]\) in the gas phase at 3 18 K are given below:

1. Plot [N₂O₅] against t.
2. Find the half-life period for the reaction.
3. Draw a graph between log [N₂O₅] and t.
4. What is the rate law?
5. Calculate the rate constant.
6. Calculate the half-life period from k and compare it with (2).

Answer:

Time corresponding to the concentration, \(\frac{1.630 \times 10^2}{2} \mathrm{molL}^{-1}=81.5 \mathrm{~mol} \mathrm{~L}^{-1}\) is the half life. From the graph, the half-life is obtained as 1450 s.

The given reaction of the first order as the plot, log [N₂O₅] v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate= k [N₂O₅]

From the plot, log [N₂O₅] v/st, we obtain

⇒ \(\text { Slope }=\frac{-2.46-(-1.79)}{3200-0}=\frac{-0.67}{3200}\)

Again .slope of the line of the plot log [N,05] v/s t is given by \(-\frac{k}{2.303}\)

Therefore, we obtain,

⇒ \(-\frac{k}{2.303}=-\frac{0.67}{3200}\)

k = 4.82 × 10^-4,s^-1

Half-life is given by,

⇒ \(t_{1 / 2}=\frac{0.639}{k}=\frac{0.693}{4.82 \times 10^{-4}} \mathrm{~s}\)

⇒ 1.438 x 10%- 1438s

This value, 1438 s. is very close to the value that was obtained from the graph.

Question 25. The rate constant for a first-order reaction is 60 s^-1 How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?
Answer:

It is known that,

⇒ \(t=\frac{2.303}{k} \log \frac{a}{(a-x)}\)

⇒ \(\text { If } a=1 \text { then }(a-x)=\frac{1}{16}\)

⇒ \(\mathrm{t}=\frac{2.303}{60 \mathrm{~s}^{-1}} \log \frac{1}{1 / 16}=\mathbf{0 . 0 4 6 2} \mathrm{s}\)

Question 26. During a nuclear explosion, one of the products is ⁹⁰Sr with a half-life of 28.1 years. If I pg of 90Sr was absorbed in the bones of a newborn baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
Answer:

Here, \(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{28.1} y^{-1}\)

It is known that,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]} \quad \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]} \Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log [R])\)

⇒ \(\log [R]=-\frac{10 \times 0.693}{2.303 \times 28.1} \Rightarrow[R]=\text { anti } \log (-0.1071)\)

= anti log(T.8929) = 0.74814 μg

Therefore, 0.7814 μg of ⁹⁰Sr will remain after 10 years,

Again,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]} \Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]} \Rightarrow \log [R]=-\frac{60 \times 0.693}{2.303 \times 28.1}\)

[R] = antilog (-0.6425) = 0.2278 μg

Therefore, 0.2278 μg of ⁹⁰Sr will remain after 60 years.

Question 27. For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
Answer:

For a first-order reaction, the time required for 99% completion is

⇒ \(t_1=\frac{2.303}{k} \log \frac{100}{100-99}=\frac{2.303}{k} \log 100=2 \times \frac{2.303}{k}\)

For a first-order reaction, the time required for 90% completion is

⇒ \(t_2=\frac{2.303}{k} \log \frac{100}{100-90}=\frac{2.303}{k} \log 10=\frac{2.303}{k}\)

Therefore, t₁ = 2t₂

Hence, the time required for 99% completion of a first-order reaction is twice the time required for the completion of 90% of the reaction.

Question 28. A first-order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Answer:

For a first-order reaction,

⇒ \(t=\frac{2.303}{k} \log \frac{[R]_0}{[R]}\)

⇒ \(\mathrm{k}=\frac{2.303}{40 \mathrm{~min}} \log \frac{100}{100-30}=\frac{2.303}{40 \mathrm{~min}} \log \frac{10}{7}=8.918 \times 10^{-3} \mathrm{~min}^{-1}\)

Therefore. t_1/2 of the decomposition reaction is

⇒ \(t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{8.918 \times 10^{-3}} \min =77.7 \mathrm{~min} \text { (approx.) }\)

Question 29. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

Answer:

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

(CH₃)₂ CHN = NCH (CH₃)₂(g → N₂(g) + C₆H₁₄(g)

After time, t, total pressure, P_t = (P₀– p) + p + p

P_t = P₀+p

⇒  p = P_t-P₀

Therefore, P₀ – p = P₀ (P_t – P0) = 2P₀ – P₁.

For a first-order reaction,

⇒ \(k=\frac{2.303}{t} \log \frac{P_0}{P_0-x}=\frac{2.303}{t} \log \frac{P_0}{2 P_0-P_t}\)

When t = 360 s,

⇒ \(\mathrm{k}=\frac{2.303}{360 \mathrm{~s}} \log \frac{35.0}{(2 \times 35.0-54.0)}=2.175 \times 10^{-3} \mathrm{~s}^{-1}\)

When t = 720 s,

⇒ \(\mathrm{k}=\frac{2.303}{720 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-63.0}=2.235 \times 10^{-3} \mathrm{~s}^{-1}\)

Hence, the average value of the rate constant is

⇒ \(\mathrm{k}=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} \mathrm{~s}^{-1}=2.205 \times 10^{-3} \mathrm{~s}^{-1}\)

Question 30. The following data were obtained during the first-order thermal decomposition of SO₂Cl₂, at a constant volume.

⇒ \(\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

Calculate the rate of the reaction when the total pressure is 0.65 atm.
Answer:

The thermal decomposition of SO₂Cl₂, at a constant volume, is represented by the following equation.

⇒\(\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

After time, t, total pressure, P, = (P0- p) + p + p

⇒ P₁ = P₀+P

⇒ P = P₁-P₀

Therefore, P₀– p = P₀– (P_t – P₂) = 2P₀– P₁,

For a first-order reaction,

⇒ \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\mathrm{x}}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_{\mathrm{t}}}\)

When t = 100 s, \(\mathrm{k}=\frac{2.303}{100 \mathrm{~s}} \log \frac{0.5}{2 \times 0.5-0.6}=2.231 \times 10^{-3} \mathrm{~s}^{-1}\)

When P_t = 0.65 atm, P₀ + p- 0.65

p = 0.65- P₀ = 0.65- 0.5 = 0. 15 atm

Therefore, when the total pressure is 0.65 atm, the Pressure of SOCl₂

P_SO2Cl2= P₀– P = 0-5 – 0.15 = 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k (p_SO2Cl2) = (2.23 × 10^-3 s^-1 ) (0.35 atm) = 7.8 × 10^-4 atm s^-1

Question 31. The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between In k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30° and 50°C.
Answer:

For the given data, we obtain

The slope of the line.

⇒ \(\frac{y_2-y_1}{x_2-x_1}=12.301 K\)

According to Arrhenius’s equations. Slope = \(-\frac{E_a}{R}\)

⇒ \(\mathrm{E}_{\mathrm{a}}=- \text { Slope } \times \mathrm{R}=-(-12.301 \mathrm{~K}) \times\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)=102.27 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Again, In k = In \(A-\frac{E_a}{R T}\)

In A = In \(k+\frac{E_a}{R T}\)

When T = 273 K,

In k =-7,147

Then, In A = \(-7.147+\frac{102.27 \times 10^3}{8.314 \times 273}=37.911\)

Therefore, A = 2.91 x 1016

When T = 30 + 273 K = 303 K

⇒ \(\frac{1}{\mathrm{~T}}=0.0033 \mathrm{~K}=3.3 \times 10^{-3} \mathrm{~K}\)

Then, at \(\frac{1}{\mathrm{~T}}=3.3 \times 10^{-3} \mathrm{~K} \text {,}\)

In k = -2.8

Therefore, k = 6.08 x 10~2 s-1

Again, when T = 50 + 273 K = 323 K.

⇒ \(\frac{1}{\mathrm{~T}}=0.0031 \mathrm{~K}=3.1 \times 10^{-3} \mathrm{~K}\)

Then, at \(\frac{1}{\mathrm{~T}}=3.1 \times 10^{-3} \mathrm{~K} \text {, }\)

In k = -0.968

Therefore, k = 0.38s-1

Question 32. The rate constant for the decomposition of hydrocarbons is 2.418 x 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 KJ/mol, what will be the value of the pre-exponential factor?
Answer:

According to the Arrhenius equation,

⇒ \(\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}} \Rightarrow{In} \mathrm{k}={In} \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\)

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T}\)

⇒ \(\log \mathrm{A}=\log \mathrm{k}+\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)

⇒ \(=\log \left(2.418 \times 10^{-5} \mathrm{~s}^{-1}\right)+\frac{179.9 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1} \times 546 \mathrm{~K}}\)

= (0.3855 – 5) + 1 7.2082 = 12,59 17

Therefore, A = antilog ( 12.5917) = 3.9 x 10¹² s^-1

Question 33. Consider a certain reaction A → Products with k = 2.0 x 10^-2 s ^-1. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol L^-1.
Answer:

Since the unit of k is s^-1. the given reaction is a first-order reaction.

Therefore, \(k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]}\)

⇒ \(2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}} \log \frac{1.0}{[\mathrm{~A}]}\)

⇒ \(2.0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2.303}{100 \mathrm{~s}}(-\log [\mathrm{A}])\)

⇒ \(\log [A]=\frac{-2.0 \times 10^{-2} \times 100}{2.303}\)

⇒ \([A]={anti} \log \left(\frac{-2.0 \times 10^{-2} \times 100}{2.303}\right)=0.135 \mathrm{~mol} \mathrm{~L}^{-1}\)

Hence, the remaining concentration of A is 0,135 mol L^-1.

Question 34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3,00 hours. What fraction of the sample of sucrose remains after 8 hours?
Answer:

For a first order reaction, k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)

It is given that, t_1/2 = 3.00 hours

Therefore, k = \(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{3} h^{-1}=0.231 h^{-1}\)

Then, \(0.231 \mathrm{~h}^{-1}=\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}\)

⇒ \(\log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}=\frac{0.23 \mathrm{Ih}^{-1} \times 8 \mathrm{~h}}{2.303}\)

⇒ \(\frac{[R]_0}{|R|}={anti} \log (0.8024)\)

⇒ \(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}=6.3445\)

⇒ \(\frac{[\mathrm{R}]}{[\mathrm{R}]_0}=0.1576 \approx 0.158\)

Hence, the fraction of the sample of sucrose that remains after 8 hours is 0.158.

Question 35. The decomposition of hydrocarbon follows the equation k = (4.5 ¹ 10¹¹ s^-1) e^{-28000 K/T.} Calculate E,
Answer:

The given equation is k = (4.5 × 10¹¹ s^-1) e^{-28000 K/T} ______ 1

Arrhenius equation is given by, k – Ae^-Ea/RT ______ 2

From equation (1) and (2). we obtain

⇒ \(\frac{E_a}{R T}=\frac{28000}{T}\)

E₂ = R × 28000 K = 8.3141 K^-1 mol^-1 × 28000 K = 2327921 J mol^-1 = 232.792 kJ mol^-1

Question 36. The following equation gives the rate constant for the first-order decomposition of H2O2: log k = 14.34- 1.25 × 10⁴ K/T. Calculate E for this reaction and at what temperature will its half-period be 256 minutes?
Answer:

Arrhenius equation is given by.

⇒ \(\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_a / \mathrm{RT}}\)

⇒ \(\text { In } k={In} A-\frac{E_a}{R T}\)

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\) ______ 1

The given equation is, log k = 14.34 – 1.23 x 10⁴ K/T _______ 2

From equation (1) and (2), we obtain

⇒ \(\frac{\mathrm{E}_{\mathrm{i}}}{2.303 \mathrm{RT}}=\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)

E_a = 1 .25 x 1 0⁴ K × 2.303 xR = 1 .25 × 10⁴ K × 2.303 × 8.314 K^-1 mol^-1

= 239339.3 J mol^-1 = 239.34 kJ mol^-1

Also, when t_1/2= 256 minutes,

⇒ \(\mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{256}=2.707 \times 10^{-3} \mathrm{~min}^{-1}=4.51 \times 10^{-5} \mathrm{~s}^{-1}\)

It is also given that, log k = 14.34- 1.25 x 10⁴ K/T

⇒ \(\log \left(4.51 \times 10^{-5}\right)=14.34-\frac{1.25 \times 10^4}{T}\)

⇒ \(0.654-5=14.34-\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}\)

⇒ \(\frac{1.25 \times 10^4 \mathrm{~K}}{\mathrm{~T}}=18.686\)

⇒ \(\mathrm{T}=\frac{1.25 \times 10^4 \mathrm{~K}}{18.686}=668.95 \mathrm{~K}=669 \mathrm{~K}\)

Question 37. The decomposition of A into product has a value of k as 4.5 x 103s_1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10⁴ s^-1?
Answer:

From Arrhenius equation, we obtain,

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

⇒ \(\log \frac{1.5 \times 10^4}{4.5 \times 10^3}=\frac{6.0 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}\left(\frac{\mathrm{T}_2-283}{283 \mathrm{~T}_2}\right)\)

⇒ \(0.5229=3133.627\left(\frac{\mathrm{T}_2-283}{283 \mathrm{~T}_2}\right)\)

⇒ \(\frac{0.5229 \times 283 \mathrm{~T}_2}{3133.627}=\mathrm{T}_2-283\)

⇒ 0.9528 T₂ = 283

⇒ T₂ = 297.019 K = 297 K = 24° C

Hence, k would be 1.5 x 10⁴ s^-1 at 24° C

Question 38. The time required for 10% completion of a first-order reaction at 298K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10¹⁰ s^-1. Calculate k at 318 K and F.
Answer:

For a first-order reaction, \(t=\frac{2.303}{k} \log \frac{a}{a-x}\)

⇒ \(\text { At } 298 \mathrm{~K}, \mathrm{t}_1=\frac{2.303}{k_1} \log \frac{100}{90}=\frac{0.1054}{k_1}\)

⇒ \(\text { At } 308 \mathrm{~K}, \mathrm{t}_2=\frac{2.303}{\mathrm{k}_2} \log \frac{100}{75}=\frac{0.2877}{\mathrm{k}_2}\)

According to the question, t = t’

⇒ \(\frac{0.1054}{k_1}=\frac{0.2877}{k_2} \Rightarrow \frac{k_2}{k_1}=2.73\)

From the Arrhenius equation, we obtain

⇒ \(\log \frac{k_2}{k_1}=\frac{E_1}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

⇒ \(\log (2.73)=\frac{E_a}{2.303 \times 8.314}\left(\frac{308-298}{298 \times 308}\right)\)

⇒ \(\mathrm{E}_{\mathrm{a}}=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.73)}{308-298}\)

⇒ 76640.26 J mol^-1 = 76.64 Id mol^-1

To calculate k at 318 K,

It is given that, A = 4 x 10^-1 s^-1, T = 318 K

Again, from the Arrhenius equation, we obtain

⇒ \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T}=\log \left(4 \times 10^{10}\right)-\frac{76.6 \times 10^3}{2.303 \times 8.3\lceil 4 \times 318}\)

= (0.6021 + 10)- 12.5876 =- 1.9855

Therefore, k = Antilog (- 1.9855) = 1.05 x 10^-1 s^-1

Question 39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:

From Arrhenius equation, we obtain,

⇒ \(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)

It is given that, k₂ = 4k₁

T₁ = 293 K: T₂ = 313 K

Therefore, \(\log \frac{4 k_1}{k_1}=\frac{E_a}{2.303 \times 8.314}\left(\frac{313-293}{293 \times 313}\right)\)

⇒ \(0.6021=\frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313}\)

⇒ \(\mathrm{E}_{\mathrm{a}}=\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\)

⇒ 52863.33 J mol^-1 = 52.86 kJ mol^-1

Hence, the required energy of activation is 52.86 kJ mol^-1

Electrochemistry

Question 1. How would you determine the standard electrode potential of the system Mg²⁺/Mg?
Answer:

The standard electrode potential of Mg²⁺ | Mg can be measured concerning the standard hydrogen electrode, represented by Pt(s). H₂(g) (1 atm) | H⁺(aq) (1 M). A cell, consisting of Mg I MgSO₄. (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.

⇒ \(\mathrm{Mg}\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{H}(\mathrm{g}, 1 \text { bar), } \mathrm{Pt}(\mathrm{s})\)

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

⇒ \(E^{(-)}=E_R^{(-)}-E_L^{(-)}\)

Here, the E^–_R for the standard hydrogen electrode is zero.

Therefore, E^– = 0 – E^–_L = E^–_L

Question 2. Can you store copper sulphate solution in a zinc pot?
Answer:

No. Zinc is more reactive than copper. Zinc reacts with copper sulphate and displaces copper. Zn + CuSO₄+ ZnSO₄ + Cu

Question 3. Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Answer:

⇒ \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\)

⇒ \(E_{\mathrm{oxi}}^0=-0.77 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{red}}^0=+0.77 \mathrm{~V}\)

Substances that are stronger oxidizing agents and have greater reduction potential than +0.77 V will oxidize

Fe²⁺ for example., Br₂, Cl₂ and F₂

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Answer:

2H⁺ + 2e^– → H₂

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\mathrm{E}_{\mathrm{cell}}^o=0, \mathrm{P}_{\mathrm{H}_2}=1 \mathrm{~atm},\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=10^{-10}\)

⇒ \(E_{\text {cell }}=-\frac{0.059}{2} \log \frac{1}{\left(10^{-10}\right)^2}=-0.59 \mathrm{~V}\)

Question 5. Calculate the emf of the cell in which the following reaction takes place:

Nit(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s)

Given that \(E_{\text {cell }}^{\circ}=1.05 \mathrm{~V}\)

Answer:

Applying the Nernst equation we have:

⇒ \(E_{\text {(cell) }}=E_{\text {cell }}^{(-)}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\)

⇒ \(1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^2}\)

⇒ \(1.05-0.02955 \log \frac{0.16}{0.000004}\)

= 1.05 -0.02955 log 4 x 104

= 1.05 – 0.02955 (log 10000 + log 4)

= 1.05 – 0.02955 (4 + 0.6021)

= 1.05 -0.02955 (4.6021)

= 1.05-0.14 = 0.91 V

Question 6. The cell in which the following reaction occurs:

⇒ \(\left.2 \mathrm{Fe}^{3+} \text { (aq. }\right)+2 \mathrm{I}^{-} \text {(aq.) } \rightarrow 2 \mathrm{Fe}^{2+} \text { (aq.) }+\mathrm{I}_2 \text { (s) has } \mathrm{E}_{\text {cell }}^o=0.236 \mathrm{~V} \text { at } 298 \mathrm{~K}\)

Calculate the standard Gibbs energy and equilibrium constant of the cell reaction.

Answer:

⇒ \(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}=-2 \times 96500 \times 0.236 \text { joule } \mathrm{mol}^{-1}=-45.55 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Δ_rG^o = -2.303 RT log K_c

(R = 8.314 J K^-1 mol^-1)

-45.55 = -2.303 x 8.314 x 298 log K_c

or log K_c = 7.981

K_c = antilog (7.98 1)

or K_c = 9.6 x 10⁷

Question 7. Why does the conductivity the a solution decrease with dilution?
Answer:

The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) per unit volume decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

Question 8. Suggest a way to determine the Δ^o_m, value of water.
Answer:

Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows :

Δ⁰_m(H2O) = Δ⁰_H=Δ_OH^–

⇒\(\left(\Delta_{\mathrm{H}^{+}}^0+\Delta_{\mathrm{Cl}^{-}}^0\right)+\left(\Delta_{\mathrm{Na}}^0+\Delta_{\mathrm{OH}^{-}}^0\right)-\left(\Delta_{\mathrm{Na}^{-}}^0+\Delta_{\mathrm{Cl}^{-}}^0\right)\)

⇒ \(\Delta_{\mathrm{m}\left(\mathrm{H}_2 \mathrm{O}\right)}^0=\Delta_{\mathrm{m}(\mathrm{HCl})}^0+\Delta_{\mathrm{m}(\mathrm{NaOH})}^0-\Delta_{\mathrm{m}(\mathrm{NaCl})}^0\)

Hence, by knowing the values of HCl, NaOH. and NaCl, the value of water can be determined.

Question 9. The molar conductivity of 0.025 mol/L methanoic acid is 46.1 S cm²/mol. Calculate the degree of dissociation and dissociation constant. Given: λ⁰_H = 349.6 and λ⁰_HCCO^– = 54.6

λ⁰_HCCO =λ⁰_H + λ⁰_HCCO^– = 349.6 + 54.6 = 404.2

⇒ \(\alpha=\frac{\lambda_{\mathrm{m}}^{\mathrm{c}}}{\lambda_{\mathrm{m}}^{\mathrm{o}}}=\frac{46.1}{404.2}=0.114\)

⇒ \(K_a=\frac{C \alpha^2}{1-\alpha}=\frac{0.025 \times 0.114 \times 0.114}{1-0.114}=3.67 \times 10^{-4}\)

Question 10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Answer:

I = 0.5 A. t = 2 hours = 2 × 60 × 60s = 7200 s Q = It = 0.5A × 7200 s = 3600C

∵ 96500 coulomb are equivalent of 6.023 x 10²³ number of electrons

∴ 3600 coulomb are equivalent to \(\frac{6.023 \times 10^{23} \times 3600}{96500}\)

= 2.246 x 10²² electrons

Question 11. Suggest a list of metals that are extracted electrolytically.
Answer: Alkali metals such as Na, K, etc., alkaline earth metals such as Mg, Ca, etc., and aluminum.

Question 12. Consider the reaction, Cr₂O^2-₇+ 14H⁺+ 6e^– → 2Cr³⁺ +7H₂O What is the quantity of electricity in coulombs needed to reduce one mole of Cr₂O^2-₇?
Answer:

To reduce 1 mole of Cr₂O^2-₇. the required quantity of electricity will be:

6F = 6 x 96487 coulomb = 578922 C

Question 13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:

lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO,v) as the cathode, and a 38% solution of sulphuric acid (H₂SO₄) as an electrolyte.

When the battery is in use, the following cell reactions take place:

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging, PbSO₄ (s) present at the anode and cathode is converted into Pb(s) and PbO₂(s) respectively.

Question 14. Suggest two materials other than hydrogen (that can be used as fuels in fuel cells.
Answer: Methane and methanol.

Question 1. Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg, and Zn

Answer:

The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Ah Zn, Fe, Cu

Question 2. Given the standard electrode potentials,

1. K⁺/K = -2.93V, Ag⁺/Ag = 0.80 V
2. Hg²⁺/Hg = 0.79V
3. Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = – 0.74V A

Arrange these metals in their increasing order of reducing power.
Answer:

The lower the reduction potential, the higher the reducing power.

The given standard electrode potentials increase in the order of

K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Question 3. Depict the galvanic cell in which the reaction \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) takes place. Further show:

1. Which of the electrodes is negatively charged?
2. The carriers of the current in the cell?
3. Individual reaction at each electrode.

Answer:

The galvanic cell in which the given reaction takes place is depicted as:

⇒ \(\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}^{2+}{ }_{\text {(aq) }} \| \mathrm{Ag}_{(\mathrm{aq})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\)

Zn electrode (anode) is negatively charged.

Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

The reaction taking place at the anode is given by,

⇒ \(\mathrm{Zn}_{(\mathrm{s})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)

The reaction taking place at the cathode is given by,

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(\mathrm{s})}\)

Question 4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place :

⇒ \(2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Cd}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Cd}(\mathrm{s})\)

⇒ \(\mathrm{E}_{\mathrm{Cr}^3 / \mathrm{Cr}}^{\ominus}=-0.74 \mathrm{~V} ; \quad E_{\mathrm{Cd}^2 / \mathrm{Cd}}^{\ominus}=-0.40 \mathrm{~V}\)

⇒ \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\)

⇒ \(\mathrm{E}_{\mathrm{Fe}^+3 / \mathrm{Fe}^+2}^{\ominus}=-0.77 \mathrm{~V} ; \quad E_{\mathrm{Ag}^+ / \mathrm{Ag}}^{\ominus}=-0.80 \mathrm{~V}\)

Calculate the Δ_rG^– and equilibrium constant of the reactions.

Answer:

The galvanic cell of the given reaction is depicted as:

⇒ \(\mathrm{Cr}_{(\mathrm{s})}\left|\mathrm{Cr}^{3+}{ }_{(\mathrm{aq})} \| \mathrm{Cd}^{2+}{ }_{(\mathrm{at})}\right| \mathrm{Cd}_{(\mathrm{s})}\)

Now, the standard cell potential is

⇒ \(\mathrm{E}_{\text {cell }}^{\ominus}=\mathrm{E}_{\mathrm{R}}^{\ominus}-\mathrm{E}_{\mathrm{L}}^{\ominus}\)

= -0.40- (-0.74) = +0.34 V

⇒ \(\Delta_r G^{\ominus}=-n F E_{\text {cell }}^{\ominus}\)

In the given equation,

n = 6

F = 96487 C mol^-1

⇒ \(\mathrm{E}_{\text {cell }}^{\ominus}=+0.34 \mathrm{~V}\)

Then, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-6 \times 96487 \mathrm{C} \mathrm{mol}^{-1} \times 0.34 \mathrm{~V}\)

= -196833.48 CV mol^-1

= -196833.48 J mol^-1

= -196.83 kJ mol^-1

Again, \(\Delta_r \mathrm{G}^{\ominus}=-\mathrm{RT} \ln \mathrm{K}\)

⇒ \(\Delta_r G^{\ominus}=-2.303 R T \log K\)

⇒ \(\log K=-\frac{\Delta_{\mathrm{r}} \mathrm{G}}{2.303 \mathrm{RT}}=\frac{-196.83 \times 10^3}{2.303 \times 8.314 \times 298}=34.496\)

K = antilog (34.496) = 3.13 x 10 34

⇒ \(\mathrm{E}_{\mathrm{Fe}^+3 / \mathrm{Fe}^+2}^{\ominus}=-0.77 \mathrm{~V} ; \quad E_{\mathrm{Ag}^+ / \mathrm{Ag}}^{\ominus}=-0.80 \mathrm{~V}\)

The galvanic cell of the given reaction is depicted as:

⇒ \(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\left|\mathrm{Fe}_{(\mathrm{aq})}^{3+}\right| \| \mathrm{Ag}_{(\mathrm{aq})}^{+} \mid \mathrm{Ag}_{(\mathrm{s})}\)

Now, the standard ceil potential is

⇒ \(\mathrm{E}_{\mathrm{cell}}^{\ominus}=\mathrm{E}_{\mathrm{R}}^{\ominus}-\mathrm{E}_{\mathrm{L}}^{\ominus}\)

= 0.80- 0.77- 0.03 V

Here, n = 1,

Then, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-n F E_{\mathrm{cell}}^{\ominus}\)

= -1 x 96487 C mor’ x 0.03 V.

= -2894,61 J mol-1

= -2.89 kJ mol-1

Again, \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=2.303 \mathrm{RT}{Iog}\mathrm{k}\)

⇒ \(\log K=\frac{\Delta_r G}{2.303 R T}=\frac{-2894.61}{2.303 \times 8.314 \times 298}=0.5073\)

∴ K = antilog (0.5073) = 3.2 (approximately)

5. Write the Nernst equation and emf of the following cells at 298K:

⇒ \({Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\)

⇒ \(\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_2(\mathrm{~g})(\text { lbar }) \mid \mathrm{Pt}(\mathrm{s})\)

⇒ \({Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.020 \mathrm{M})\right| \mathrm{H}_2(\mathrm{~g}) \text { (Ibar) } \mid \mathrm{Pt}(\mathrm{s})\)

⇒ \({Pt}(\mathrm{s})\left|\mathrm{Br}_2(1)\right| \mathrm{Br}^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M})\left|\mathrm{H}_2(\mathrm{~g})(1 \mathrm{bar})\right| \mathrm{Pt}(\mathrm{s})\)

Answer:

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\ominus}-\frac{0.0591}{\mathrm{n}} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)

⇒ \(\{0.34-(-2.36)\}-\frac{0.0591}{2} \log \frac{.001}{.0001}\)

⇒ \(2.7-\frac{0.0591}{2} \log 10\)

⇒ 2.7 -0.02955

⇒ 2.67 (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\ominus}-\frac{0.0591}{\mathrm{n}} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\{0-(-0.44)\}-\frac{0.0591}{2} \log \frac{0.001}{1^2}\)

⇒ 0.44- 0.02955(-3)

⇒ 0.52865

⇒ 0.53 V (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(\{0-(-0.14)\}-\frac{0.0591}{2} \log \frac{0.050}{(0.020)^2}\)

⇒ 0.14-0.0295 x log 125

= 0.14-0.062

= 0.078 V

= 0.08 V (approximately)

For the given reaction, the Nernst equation can be given as:

⇒ \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\odot}-\frac{0.0591}{\mathrm{n}} \log \frac{1}{\left[\mathrm{Br}^{-}\right]^2\left[\mathrm{H}^{+}\right]^2}\)

⇒ \(=(0-1.09)-\frac{0.0591}{2} \log \frac{1}{(0.010)^2(0.030)^2}\)

⇒ \(1.09-0.02955 \times \log \frac{1}{0.00000009}\)

⇒ \(1.09-0.02955 \times \log \frac{1}{9 \times 10^{-8}}\)

⇒ 1.09-0.02955 x log(1.11 x 107)

⇒ -1.09-0.02955 (0.0453 + 7)

⇒ -1.09-0.208

⇒ -1.298 V

Question 6. In the button cells widely used in wanted and other devices, the following reaction takes place:

⇒ \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2(\mathrm{O})(l) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})\)

Determine Δ_rG^o and E^– for the reaction.

⇒ \(\mathrm{Zn}_{(\mathrm{s})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)

⇒ \(\mathrm{E}^{\ominus}=0.76 \mathrm{~V}\)

⇒ \(\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{l}_{(0)}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}_{(\mathrm{g})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} ; \mathrm{E}^{\ominus}=0 .{344 \mathrm{~V}}\)

⇒ \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Ag}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{f})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} ; \mathrm{E}^{\ominus}=1.104 \mathrm{~V}\)

∴ E^– = 1.104 V

We know that,

⇒ \(\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-\mathrm{nF} \mathrm{E}^{\ominus}\)

= -2 × 96487 × 1.04

= -213043.296 J

= -213.04 kJ

Question 7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

⇒ \(\mathrm{G}=\kappa \frac{\mathrm{A}}{\mathrm{l}}=\kappa \cdot 1=\kappa \quad(\text { Since } \mathrm{A}=1,=1)\)

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume (that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length is 1.

⇒ \(\Delta_{\mathrm{m}}=\kappa \frac{\mathrm{A}}{1}\)

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

⇒ \(\Lambda_m=\kappa V\)

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

The variation of Δ_m with √c for strong and weak electrolytes is shown in the following plot:

Question 8. The conductivity of the 0.20 M solution of KCI at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.
Answer:

Given, K = 0.0248 S cm^-1: c = 0.20 M

∴ Molar conductivity,

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa \times 1000}{c}=\frac{0.0248 \times 1000}{0.2}=124 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)

Question 9. The resistance of a conductivity cell containing 0.00 1 M KCI solution at 298 K is 1500Ω. What is the cell constant if the conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10^-3 S cm^-1?
Answer:

Given, Conductivity, K = 0.146 x 10^-3 S cm^-1

Resistance, R = 1500Ω

∴ Cell constant =K × R

= 0. 1 46 × 10^-3 × 1500 = 0.219 cm^-1

Question 10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and are results are given below:

Calculate Δ_m for all concentrations and draw a plot between Δ_m and C1/2. Find the value of Δ^o_m

Answer:

Given,

K = 1.237 × 10^-2 Sm^-1, c = 0.001 M

Then, K = 1.237 × 10^-1 S cm^-1 c^1/2= 0.03 1 6 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa}{\mathrm{c}}=\frac{1.237 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}}{0.001 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=123.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

K = 1185 × 10^-2 s m^-1, c =0.010 M

Then, ic = 11.85 x 10^-4 S cm^-1, c^1/2 = 0.1 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{c}}=\frac{11.85 \times 10^{-4} \mathrm{Scm}^{-1}}{0.010 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=118.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

K = 23.15 × 10^-2 S m^-1, c =0.020 M

Then, k = 23.15 × 10^-4 S cm^-1, cm^1/2 = 0. 1414 M^1/2

⇒ \(\Delta_m=\frac{K}{c}=\frac{23.15 \times 10^{-4} \mathrm{Scm}^{-1}}{0.020 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=115.8 \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

k = 55.53 × 10^-2 Sm^-1 ,c = 0.050 M

Then, k = 55.53 x 10^-4 S cm^-1, c^1/2 = 0.2236 M^1/2

⇒ \(\Delta_{\mathrm{m}}=\frac{\kappa}{\mathrm{c}}=\frac{55.53 \times 10^{-4} \mathrm{Scm}^{-1}}{0.050 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^2}{\mathrm{~L}}=111.11 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Given,

k= 106.74 × 10^-2 Sm^-1, c = 0.100M

Then, k = 106.74 × 10^-4 Sm^-1, c^1/2 = 0.3162 M^1/2

⇒ \(\Delta_m=\frac{K}{c}=\frac{106.74 \times 10^{-4}}{0.100 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=106.74 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Now, we have the following data.

Since the line interrupts Δ_m at 124.0 S cm² mol^-1 , Δ^o_m = 1 24.0 S cm² mol^-1

Question 11. The conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if Δ^o_m for acetic acid is 390.5 S cm2 mol^-1, what is its dissociation constant?
Answer:

Given, K = 7.896 x 10^-5 S cm^-1

C = 0.00241 mol L^-1

Then, molar conductivity,

⇒ \(\Lambda_m=\frac{\kappa}{c}=\frac{7.896 \times 10^{-5} \mathrm{Scm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^3}{\mathrm{~L}}=32.76 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Again, = 390.5 S cm mol^-1

Now, \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{mi1}}^0}=\frac{32.76 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}{390.5 \mathrm{Scm}^2 \mathrm{~mol}^{-1}}=0.084\)

∴ Dissociation constant,

⇒ \(\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)}=\frac{\left(0.00241 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.084)^2}{(1-0.084)}=1.86 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

Question 12. How much charge is required for the following reductions:

1 mol of Al³⁺ to Al.

1 mol of Cu²⁺ to Cu.

1 mol of MnO^–₄ to Mn²⁺

Answer:

Al³⁺ + 3e^– → Al

∴ Required charge = 3F

= 3 × 96487 C = 289461 C

Cu²⁺+ 2e^– → Cu

∴ Required charge = 2F

= 2 × 96487 C = 1 92974 C

MnO^–₄ → Mn²⁺

i.e., Mn⁷⁺ + 5e^– → Mr²⁺

∴ Required charge = 5F

= 5 × 96487 C = 482435 C

Question 13. How much electricity in terms of Faraday is required to produce

1. 20.0 g of Ca from molten CaCI₂
2. 40.0 g of Al from molten Al₂O₃.

Answer:

According to the question.

⇒ \(\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{40 \mathrm{~g}}{\mathrm{Ca}}\)

Electricity required to produce 40 g of calcium = 2F

Therefore, electricity required to produce 20 g of calcium = \(\frac{2 \times 20}{40} \mathrm{~F}=1 \mathrm{~F}\)

According to the question

⇒ \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{27 \mathrm{~g}}\)

Electricity required to produce 27 g of Al = 3F

Therefore, electricity required to produce 40 g of Al = \(\frac{3 \times 40}{27} \mathrm{~F}=4.44 \mathrm{~F}\)

Question 14. How much electricity is required in coulomb for the oxidation of

1. 1 mol of I H₂O to O₂.
2. 1 mol of FeO to Fe₂,O₃

Answer:

According to the question,

⇒ \(\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2\)

Now, we can write:

⇒ \(\mathrm{O}^{2-} \longrightarrow \frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)

Electricity required for the oxidation of 1 mol of H₂O to O₂ = 2F = 2 × 96487 C = 192974 C

According to the question,

⇒ \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1}\)

Electricity required for the oxidation of 1 mol of FeO to Fe₂O₃, = 1F = 96487 C

Question 15. A solution of Ni. (NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Nt is deposited at the cathode?
Answer:

Given,

Current = 5A

Time = 20×60= 1200 s

∴ Charge = current x time

= 5 x 1200 = 6000 C.

According to the reaction,

⇒ \(\mathrm{Ni}_{(\mathrm{iq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{58.7 g}{\mathrm{Ni}_{(\mathrm{s})}}\)

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C = \(\frac{58.71 \times 6000}{2 \times 96487} \mathrm{~g}=1.825 \mathrm{~g}\)

Hence, 1.825 g of nickel will be deposited at the cathode

Question 16. Three electrolytic cells A, B. C containing solutions of ZnSO₄, AgNO, and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B, How long did the current flow? What mass of copper and zinc were deposited?
Answer:

According to the reaction:

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \underset{108 \mathrm{~g}}{\mathrm{Ag}_{(\mathrm{s})}}\)

i.e., 108 g of Ag is deposited by 96487 C

Therefore, 1 .45 g of Ag will be deposited by = \(\frac{96487 \times 1.45}{108} C=1295.43 \mathrm{C}\)

Given, Current = 1.5 A

⇒ \(\text { Time }=\frac{1295.43}{1.5} \mathrm{~s}=863.6 \mathrm{~s}=864 \mathrm{~s}=14.40 \mathrm{~min}\)

Again,

⇒ \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{63.5 \mathrm{~g}}{\mathrm{Cu}_{(\mathrm{s})}}\)

2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1 295.43 C of charge will deposit = \(\frac{63.5 \times 1295.43}{2 \times 96487} \mathrm{~g}\) = 0.426 g of Cu

Given,

⇒ \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \underset{65.4 \mathrm{~g}}{\mathrm{Zn}_{(\mathrm{s})}}\)

2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1 295.43 C of charge will deposit = \(\frac{65.4 \times 1295.43}{2 \times 96487} \mathrm{~g}=0.439 \mathrm{~g} \text { of } \mathrm{Zn}\)

Question 17. Using the standard electrode potentials given in the Table predict if the reaction between the following is feasible:

1. Fe³⁺(aq) and I^– (aq )
2. Ag³⁺ (aq) and Cu(s)
3. Fe³⁺(aq) and Br'(aq)
4. Ag(s) and Fe³⁺(aq)
5. Br₂(aq) and Fe²⁺(aq).

Answer:

Since E^– for the overall reaction is positive, the reaction between Fe³⁺_(aq), and I^–_(aq) is feasible.

Since E^– for the overall reaction is positive, the reaction between Ag⁺_(aq) and Cu_(s) is feasible.

Since E^– for the overall reaction is negative, the reaction between Fe³⁺_(aq), and Br_(aq) is feasible.

Since E^– E for the overall reaction is negative, the reaction between Ag_(s) and Fe²⁺_(aq) is not feasible.

Since E^– for the overall reaction is positive, the reaction between Br_2(aq) and Fe²⁺_(aq) is feasible.

Question 18. Predict the products of electrolysis in each to the following:

1. An aqueous solution of AgNO₃, with silver electrodes.
2. An aqueous solution of AgNO₃, with platinum electrodes.
3. A dilute solution of H₂SO₄ with platinum electrodes.
4. An aqueous solution of CuCl₂ with platinum electrodes.

Answer:

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \Lambda \mathrm{g}_{(\mathrm{b})} ; \mathrm{E}^{\ominus}=0.80 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of EM takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At anode: Ag Ag⁺ + e^–

The Ag anode is attacked by NO₃^–, ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag⁺

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; \mathrm{E}^{\ominus}=0.80 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of EM takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At anode: Ag →  Ag⁺ + e^–

Since Pt electrodes arc inert, the anode is not attacked by NOT ions. Therefore, OFT or NOT ions can be oxidized at the anode. But OIF ions have a lower discharge potential and get preference and decompose to liberate O2.

⇒ \(2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} ; \mathrm{E}^0=-1.23 \mathrm{~V}\)

At the cathode, the following reduction occurs to produce H₂(g)

⇒ \(\mathrm{H}_{(aq)}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2(g)}\)

At the anode, the following processes are possible.

⇒ \(2 \mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}: \mathrm{E}^{(-)}=+1.23 \mathrm{~V}\)

⇒ \(2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \longrightarrow \mathrm{S}_2 \mathrm{O}_{6(\mathrm{aq})}^{2-}+2 \mathrm{e}^{-} ; \mathrm{E}^{\ominus}=+1.96 \mathrm{~V}\)

For dilute sulphuric acid, reaction (t) is preferred to produce O₂ gas. But for concentrated sulphuric acid, a reaction occurs.

At cathode: The following reduction reactions complete to take place at the cathode.

⇒ \(\mathrm{Cu}_{(\mathrm{ap})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{g})} ; \mathrm{E}^{\ominus}=0.34 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} ; \mathrm{E}^0=-0.83 \mathrm{~V}\)

The reaction with a higher value of E^–  takes place at the die cathode. Therefore, the deposition of copper will take place at the cathode:

At anode:

The following oxidation reactions are possible at the anode.

⇒ \(\mathrm{Cl}_{\text {(aq) }}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_{2(8)}+\mathrm{e}: \mathrm{E}^{\ominus}=-1.36 \mathrm{~V}\)

⇒ \(2 \mathrm{H}_2 \mathrm{O}_{(1)} \longrightarrow \mathrm{O}_{2(\mathrm{p})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}: \mathrm{E}^{\ominus}=-1.23 \mathrm{~V}\)

At the anode, the reaction with a lower value of E^– is preferred. However due to the over-potential of oxygen, Cl^– gets oxidized at the anode to produce Cl₂ gas.

Important Questions for Class 12 Chemistry Chapter 1 Solutions

Question 1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCI₄) if 22g of benzene is dissolved in 122 g of carbon tetrachloride.
Answer:

Total mass = 22 + 1 22 = 144 g

Mass % of benzene = \(\frac{22 \times 100}{144}=15.28\)

Mass 7r of CCI = \(\frac{122}{144} \times 100=84.72\)

Question 2. Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride.
Answer:

Mass of benzene = 30 g

Mass of carbon tetrachloride = 100- 30 = 70g

Molar mass of benzene (C₆H₆) = 78,

Molar mass of carbon tetrachloride = 154

Moles of benzene = \(\frac{w}{m}=\frac{30}{154}=\) = 0.385

Moles of carbon tetrachloride = \(\frac{w}{m}\)

Total moles = 0.385 + 0.455 = 0.840

⇒ Mole fraction of benzene = \(\frac{\text { moles of benzene }}{\text { total moles }}=\frac{0.385}{0.840}=0.458\)

Mole fraction of carbon tetrachloride = 1 -0.458 = 0.542

Question 3. Calculate the molarity of each of the following solutions:

1. 30 g of Co(NO₃)₂ . 6H2O in 4.3 L of solution.
2. 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Answer:

1. Molar mass of Co(NO₃)₂. 6H₂O = 310.7

⇒ Molarity = \(\frac{w}{m \times V(\text { lit })}=\frac{30}{310.7 \times 4.3}=0.022\)

2. M₁V₁ = M₂V₂

M₁ = 0.5 V₁ = 30 mL. M₂ = ?. V₂ = 500mL

0.5 × 30 = M₂ × 500

or M₂ = 0.03

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Calculate the mass of urea ( NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Answer:

0.25 molal aqueous solution means

Moles of urea = 0.25, Mass of water = 1 kg = 1000 g

0.25 mole urea = 0.25 × 60 = 15 gram

Total mass of solution = 1000+ 15 = 101 5g = 1.015 kg

∵ 1.015 kg solution contains urea = 15 g

∴ 2.5 kg solution contains urea = \(\frac{15 \times 2.5}{1.015}=36.94 \mathrm{gram}\)

Question 5. Calculate

1. Molality,
2. Molarity,
3. Mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1 .202 g/mL.

Answer:

20% mass/mass aqueous KI means

Mass of KI = 20 g, Mass of water = 80g,

Mass of solution = 100 g

For molality:

Molar mass of KI = 39 + 127 = 166

⇒ \(\text { Molality }=\frac{w}{m \times \text { mass of solvent in } \mathrm{kg}}=\frac{20}{166 \times 80 \times 10^{-3}}=1.51\)

For Molarity:

⇒ \(\text { Volume of solution }=\frac{\text { mass }}{\text { density }}=\frac{100}{1.202}=83.2 \mathrm{ml}\)

⇒ \(\text { Molarity }=\frac{w}{\mathrm{~m} \times \mathrm{V}(\text { lit })}=\frac{20}{166 \times 83.2 \times 10^{-3}}=1.448\)

⇒ \(\text { Moles of } \mathrm{KI}=\frac{\mathrm{w}}{\mathrm{m}}=\frac{20}{166}=0.120\)

⇒ \(\text { Moles of water }=\frac{w}{m}=\frac{80}{18}=4.444\)

Total moles = 0.120 + 4.444 = 4.564

⇒ \(\text { Mole fraction of } \mathrm{KI}=\frac{0.120}{4.564}=0.0263\)

Question 6. H₂S, a toxic gas with a rotten egg smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant. Mass of solvent (water) = 1kg = 1000g.
Answer:

Solubility = 0. 195 moles in one kg of water

Mass of solvent (water) = 1 kg = 1000g

⇒ \(\text { Moles of water }=\frac{w}{m}=\frac{1000}{18}=55.55\)

Total moles = 0. 1 95 + 55.55

Mole fraction of H₂S(x) in solution \(=\frac{0.195}{0.195+55.55}=0.0035\)

The pressure of H₂S at STP = 0.987 bar.

Since partial pressure of the gas is given,

⇒ P_H2S = K_H × X

or 0.987 = K_H × 0.0035

or K_H = 282 bar

Question 7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atmospheric CO₂ pressure at 298 K.
Answer:

K_H = 1.67 x 10⁸Pa

⇒ P_Co2 = 2.5 atm = 2.5 x 101325 Pa

⇒ P_Co2 = K_H × X_Co2

or 2.5 x 101325 = 1.67 x 10⁸ × X_Co2

or X_Co2 = 1.517 x 10^-3

⇒ X_Co2 = \(\frac{{ }^n \mathrm{CO}_2}{{ }^n \mathrm{H}_2 \mathrm{O}+{ }^n \mathrm{CO}_2}=\frac{{ }^n \mathrm{CO}_2}{{ }^n \mathrm{H}_2 \mathrm{O}}=1.517 \times 10^{-3}\)

n_Co2 is neglected in the denominator in comparison to n_H2O.

Water = 500 ml, = 500 g,

⇒ \(\text { Moles of water }=\frac{1 \times 8.314 \times 10^3 \times 310}{185000}\)

⇒ \(\frac{{ }^n \mathrm{CO}_2}{{ }^n \mathrm{H}_2 \mathrm{O}}=1.517 \times 10^{-3}\)

or, \(\frac{{ }^n \mathrm{CO}_2}{27.78}=1.517 \times 10^{-3}\)

or n_Co2 = 42.14 x 10^-3

Mass of CO₂ = moles x molar mass

= 42.14 x 10^-3 × 44 = 1.854 gram

Question 8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition in the vapour phase.
Answer:

P⁰_A = 450 mm. P⁰_B= 700 mm, P_Total = 600 mm

A solution contains two liquids.

Applying Raoulfs law liquids: P_s = P⁰_A X_A + P⁰_BX_B

or P_s = P⁰_AX_A = P⁰_B(1-X) [X_A + X_B =1]

or P_s = P⁰_B= X_A(P⁰_A-P⁰_B)

600=700+X_A (450-700)

or, \(X_A=\frac{100}{250}=0.40\)

⇒ X_B = 1-X_A = 1- 0.40 = 0.60

⇒ P_A = P⁰_AX_A = 450 × 0.40 = 180 mm

⇒ P_B = P⁰_BX_B = 700 × 0.60 = 420 mm

Total vapour pressure = 180 + 420 = 600 mm

Mole fraction of A in vapour phase = \(\frac{P_A}{P_{\text {total }}}=\frac{180}{600}=0.30\)

Mole fraction of B in vapour phase = \(\frac{P_B}{P_{\text {total }}}=\frac{420}{600}=0.70\)

Question 9. The vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 grams of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Answer:

A solution contains a non-volatile solid. Our aim is to calculate Ps and \(\frac{P_A^o-P_S}{P_S}=\frac{W \times M}{m \times W}\)

P⁰_A = 23.8 w = 50.. W = 850. P_s = ?. M = 18 for H₂O, m= 60 for urea

⇒ \(\frac{\mathrm{P}_{\mathrm{A}}^o-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{\mathrm{w} \times \mathrm{M}}{\mathrm{m} \times \mathrm{W}}\)

or \(\frac{23.8-P_S}{P_S}=\frac{50 \times 18}{60 \times 850}=\frac{3}{170}\)

or P_s = 23.387

Again, relative lowering in V.P.

⇒  \(\frac{\mathrm{P}_{\mathrm{A}}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{A}}^0}=\frac{23.8-23.387}{23.8}=0.017\)

Question 10. The boiling point of water at 750 mm is 96.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? K = 0.52 K kg/mol.
Answer:

ΔT_b = 100- 96.63 = 3.37

w = ?, W = 500 g, M = 342 for sucrose

K_b = 0.52 K kg/mol

⇒ \(\Delta \mathrm{T}_{\mathrm{b}}=\frac{\mathrm{K}_{\mathrm{b}} \times \mathrm{W} \times 1000}{\mathrm{M} \times \mathrm{W}}\)

or \(3.37=\frac{0.52 \times w \times 1000}{342 \times 500}\)

or w= 1108.2 g

Question 11. Calculate the mass of ascorbic acid (vitamin C, C₆H₈O₆) to be dissolved in 75g acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg/mol.
Answer:

⇒ \(\Delta \mathrm{T}_f=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w} \times 1000}{\mathrm{M} \times \mathrm{W}}\)

The formula for lowering in freezing point or melting point is the same.

ΔT_f = 1.5°, w = ?. W = 75g

M for ascorbic acid = 176

⇒ \(1.5=\frac{3.9 \times w \times 1000}{176 \times 75}\)

or w= 5.08 g

Question 12. Calculate the osmotic pressure in pascals excited by a solution prepared by dissolving 1.0 g of polymer of molar mass 185000 in 450 mL of water at 37°C.
Answer:

Since mass of solute is given, πV = \(\frac{w R T}{m}\)

π = ?, V = 0.450 litre, w = Ig, m = 185000,

R = 8.314 kPa LK^-1 mol^-1

= 8.314 kPa LK^-1 mol^-1

⇒ \(\pi \times 0.450=\frac{1 \times 8.314 \times 10^3 \times 310}{185000}\)

or π = 30.96 Pa

Question 13. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:

Homogeneous mixtures of two or more than two components are known as solutions. There are three types of solutions.

1. Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
2. Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas. liquid, or solid. For example, a solution of ethanol in water is a liquid solution.
3. Solid solution: The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Question 14. Give an example of a solid solution in which the solute is a gas.
Answer: A solution of hydrogen in palladium is a solid solution in which the solute is a gas.

Question 15. Define the following terms:

1. Mole fraction
2. Molality
3. Molarity
4. Mass percentage

Answer:

Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.

i.e. Mole fraction of a component \(=\frac{\text { Number of moles of the component }}{\text { Total number of moles of all components }}\)

The mole fraction is denoted by ‘X’.

If in a binary solution, the number of moles of the solute and the solvent are n and n respectively, then the mole fraction of the solute in the solution is given by.

⇒ \(x_A=\frac{n_A}{n_A+n_B}\)

Similarly, the mole fraction of the solvent in the solution is given as:

⇒ \(x_B=\frac{n_A}{n_A+n_B}\)

Molality: Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:

⇒ \(\text { Molality }(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)

Molarity: Molarity (M) is defined as the number of moles of the solute dissolved in
one Litre of solution. It is expressed as:

⇒ \(\text { Molarity }(M)=\frac{\text { Moles of solute }}{\text { Volume of solvent in Litre }}\)

Mass percentage: The mass percentage of a component of a solution is defined as the mass of the solute in grams present in l00g of the solution. It is expressed as:

Mass % of a component \(=\frac{\text { Mass of component in solution }}{\text { Total mass of solution }} \times 100\)

Question 16. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?
Answer:

Concentrated nitric acid used in laboratory work is 68% nitric acid bv mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.

Molar mass of nitric acid (HNO₃) = 1 × 1 + 14 + 3 × 16 = 63 g mol^-1

Then, number of moles of HNO₃ = \(\frac{68}{6.3} \mathrm{~mol}=1.079 \mathrm{~mol}\)

Given, Density of solution = 1.504 g mL^-1

∴ Volume of 100 g solution = \(\frac{100}{1.504} \mathrm{~mL}\) = 66.49 mL = 66.49 x 10^-3

Molarity of solution = \(\frac{1.079}{66.49 \times 10^{-3} \mathrm{~L}}=16.23 \mathrm{M}\)

Question 17. A solution of glucose in water is labelled as 10% (w/w), what would be the molality and mole fraction of each component in the solution? If the density of the Solution is 1.2 g mL^-1, then what shall
be the molarity of the solution?
Answer:

10% w/w solution of glucose in water means that 10 g of glucose is present in lOOg of the solution i.e., 10 g of glucose is present in ( 100- 1 0) g = 90 g of water.

Molar mass of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^-1

Then, number of moles of glucose = \(\frac{10}{180} \mathrm{~mol}=0.056 \mathrm{~mol}\)

∴ Molality of solution = \(\frac{0.056}{0.09 \mathrm{~kg}}=0.62 \mathrm{~m}\)

Number of moles of water = \(\frac{90 \mathrm{~g}}{18 \mathrm{gmol}^{-1}}=5 \mathrm{~mol}\)

⇒ Mole fraction of glucose (x_g) = \(\frac{0.056}{0.056+5}=0.011\)

And, mole fraction of water x_w = 1-X_g = 1- 0.011 = 0.989

If the density of the solution is 1.2 g mL^-1. then the volume of the 100 g solution can be given as:

⇒ \(\mathrm{V}=\frac{100 \mathrm{~g}}{1.2 \mathrm{~g} \mathrm{~mL}^{-1}}=83.33 \mathrm{~mL}=83.33 \times 10^{-3} \mathrm{~L}\)

Molarity of the solution = \(\frac{0.056 \mathrm{~mol}}{83.33 \times 10^{-3} \mathrm{~L}}=0.67 \mathrm{M}\)

Question 18. How many mL of 0.1 M HCl are required to react completely with a 1 g mixture of Na₂CO₃ and NaHCO₃. containing equimolar amounts of both?
Answer:

Let the amount of Na₂CO₃. in the mixture be x g.

Then, the amount of NaHCO₃. in the mixture is (1 – x) g,

The molar mass of Na₂CO₃ = 2 × 23 +1 × 12 + 3  × 16= 106 g mol^-1

Number of moles Na₂CO₃ = \(\frac{x}{106} \mathrm{~mol}\)

Molar mass of NaHCO₃ = 1 x 23 + 1 x 1 x 12 + 3 x 16 = 84g mol^-1

Number of moles NaHCO₃ = \(\frac{1-x}{84} \mathrm{~mol}\)

According to the question. \(\frac{x}{106}=\frac{1-x}{84}\)

⇒ 84 x = 106 – 106x

⇒ 190x = 106

⇒ x = 0.0053

Therefore, number of moles of Na₂CO₃ = \(\frac{0.5579}{106} \mathrm{~mol}=0.0053 \mathrm{~mol}\)

And. number of moles of NaHCO₃ = \(\frac{1-0.5579}{84}=0.0053 \mathrm{~mol}\)

HCl reacts with Na₂CO₃ and NaHCO₃ according to the following equation.

⇒ \(\underset{2 \mathrm{~mol}}{2 \mathrm{HCl}}+\underset{\text { Imol }}{\mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow 2 \mathrm{NaCl}}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

⇒ \(\underset{\text { 1 mol }}{\mathrm{HCl}}+\underset{\text { 1 mol }}{\mathrm{NaHCO}_3} \longrightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

1 mol of Na₂CO₃ reacts with 2 mol of HCl

Therefore. 0.0053 mol of Na₂CO₃ reacts with HCl = 2 × 0.0053 mol = 0.0106 mol.

Similarly. 1 mol of NaHCO₃ reacts with I mol of HCl.

Therefore. 0.0053 mol of NaHCO reacts with 0.0053 mol of HCl.

Total moles of HCI required = (0.0106 + 0.0053) mol =0.0159 mol

0.1 mol of HCl is preset in 1000 ml. of the solution.

Therefore 0.0159 mol of HCI is present in the solution

⇒ \(\frac{1000 \times 0.0159}{0.1} \mathrm{~mol}=159 \mathrm{~mL}\)

Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 mixture of NaCO₃ and NaHCO₃. containing equimolar amounts of both.

Question 19. A solution is obtained by mixing 300 g of 259f solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer: The total amount of solute present in the mixture is given by

⇒ \(300 \times \frac{25}{100}+400 \times \frac{40}{100}=75+160=235 \mathrm{~g}\)

Total amount of solution = 300 + 400 = 700 g

Therefore, the mass percentage (w/w) of the solute in the resulting solution

⇒ \(\frac{235}{700} \times 100 \%=33.57 \%\)

And, mass percentage (w/w) of the solvent in the resulting solution.

= (100-33.57%) = 66.43%

Question 20. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?
Answer:

Molar mass of ethylene glycol (C₂H₄ (OH)₂ = 2 × 12 + 6 × 1 + 2 × 16 = 62 gmo^-1

Number of moles of ethylene glycol = \(\frac{222.6 \mathrm{~g}}{62 \mathrm{gmol}^{-1}}=3.59 \mathrm{~mol}\)

Therefore, molality of the solution = \(\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}}=17.95 \mathrm{~m}\)

Total mass of the solution = (222,6 + 200) g = 422.6 g

Given,

Density of the solution = 1.072 g mL^-1

∴ Volume of the solution = \(\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}=394.22 \mathrm{~mL}=0.3942 \times 10^{-3} \mathrm{~L}\)

⇒ Molarity of the solution = \(\frac{3.59 \mathrm{~mol}}{0.39422 \times 10^{-3} \mathrm{~L}}=9.11 \mathrm{M}\)

Question 21. A sample of drinking water was found to be severely contaminated with chloroform (CHCIJ supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

1. Express this in percent by mass
2. Determine the molality of chloroform in the water sample.

Answer:

15 ppm (by mass) means 15 parts per million ( 106) of the solution

Therefore, percent by mass = \(\frac{15}{10^6} \times 100 \%=15 \times 10-4 \%\)

Molar mass of chloroform (CHCl₃) = 1 × 12 + 1 × 1 + 3 × 35.5 = 119.5 g mol^-1

Now. according to the question.

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g chloroform is present in ( 10⁶– 15) ≈ 10⁶g of water

∴ Molality of the solution = \(\frac{\frac{15}{119.5} \mathrm{~mol}}{10^6 \times 10^{-3} \mathrm{~kg}}=1.26 \times 10^{-4} \mathrm{~m}\)

Question 22. What role does the molecular interaction play in a solution of alcohol and water?
Answer:

• In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions.
• As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape.
• This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Question 23. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:

The solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.

Gas + Liquid → Solution + Heat

Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

Question 24. State Henry’s law and mention some important applications.
Answer: Henry’s Law states that the partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law’ can be expressed as:

p = K_HX

where K_H is Henry’s law constant

Some important applications of Henry’s law are mentioned below.

• Bottles are sealed under high pressure to increase the solubility of CO₂ in soft drinks and soda water.
• Henry’s Jaw states that the solubility of gases increases with an increase in pressure, Therefore, when a scuba diverts deep into the sea, the increased sea pressure causes the nitrogen present in the air to dissolve in his blood in great amounts.
• As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood.
• This results in the blockage of capillaries and leads to a medical condition known as ’bends’ or ’decompression sickness’.
• Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends. The concentration of oxygen is low in the blood and tissue of people living at high altitudes such as climbers.
• This is because, at high altitudes, the partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and prevents them from thinking clearly. These are symptoms of anoxia.

Question 25. The partial pressure of ethane over a solution containing 6.56 x 10^-1 g of ethane is 1 bar. If the solution contains 5.00 x 10^-2 g of ethane, then what shall be the partial pressure of the gas?
Answer: Molar mass of ethane (C₂H₆) = 2 × 12+ 6 × 1 = 30 g mol^-1

∴ Number of moles present in 6.56 x 10^-3 g pf ethane = \(\frac{6.56 \times 10^{-3}}{30}=0.218 \times 10^{-3} \mathrm{~mol}\)

Let the number of moles of the solvent be x.

According to Henry’s law, p = K_Hx

⇒ \(\text { 1bar }=K_H \cdot \frac{0.218 \times 10^{-3}}{0.218 \times 10^{-3}+x}\)

⇒ \(1 \text { bar }=K_H \frac{0.218 \times 10^{-3}}{x}\left(\text { Since } x>>0.218 \times 10^{-3}\right)\)

⇒ \(K_{11}=\frac{x}{0.218 \times 10^{-3}} \text { bar }\)

Number of moles present in 5.00 × 10^-2 g of ethane \(\frac{5.00 \times 10^{-2}}{30} \mathrm{~mol}=1.67 \times 10^{-3} \mathrm{~mol}\)

According to Henry’s law, p = K_Hx

⇒ \(\frac{x}{0.218 \times 10^{-3}} \times \frac{1.67 \times 10^{-3}}{\left(1.67 \times 10^{-3}\right)+x}\)

⇒ \(\frac{x}{0.218 \times 10^{-1}} \times \frac{1.67 \times 10^{-3}}{x}\) (Since, x >> 1 .67 x 10^-3)

= 0.764 bar

Hence, the partial pressure of the gas shall be 7.66 bar

Question 26. What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_solH related to positive and negative deviations from Raoult’s law?
Answer:

According to Raoult’s law. the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

• The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressure either higher or lower than that predicted by Raoult’s law.
• If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

Vapour pressure of a two-component solution showing positive deviation from Raoult’s law.

Vapour pressure of a two-component solution showing negative deviation from Raoult’s law

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solutions is zero.

A_solH = 0

In the case of solutions showing positive deviations, absorption of heat takes place,

A_solH = Positive

In the case of solutions showing negative deviations, the evolution of heat takes place.

A_solH = Negative

Question 27. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:

Here, Vapour pressure of the solution at normal boiling point ( p₁) = 1.004 bar

Vapour pressure of pure water at normal boiling point ( P⁰₁ ) = 1.01 3 bar

Mass of solute, (w₂) = 2g

Mass of solvent (water), (w₁) = 98 g

Molar mass of solvent (water), ( M₁) = 18 g mol^-1

According to Raoult’s law,

⇒ \(\frac{p_1^0-p_1}{p_1^0}=\frac{w_2 \times M_1}{M_2 \times w_1}\)

⇒ \(\frac{1.013-1.004}{1.013}=\frac{2 \times 18}{\mathrm{M}_2 \times 98}\)

⇒ \(\frac{0.009}{1.013}=\frac{2 \times 18}{\mathrm{M}_2 \times 98}\)

⇒ \(\mathrm{M}_2=\frac{1.013 \times 2 \times 18}{0.009 \times 98}\) = 41.35g mol^-1

Hence, (lie molar mass of the solute is 41.35 g mol^-1

Question 28. Heptane and octane form an ideal solution. At 373 K. the vapour pressure of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer:

Vapour pressure of heptane (P⁰₁) = 105.2 kPa

Vapour pressure of octane (P⁰₂) = 46.8 kPa

We know that,

Molar mas of heptane (C₇H₁₆ ) = 7 x 12 + 16 x 1 = 100 g mol^-1

∴ Number of moles of heptane = \(\frac{26}{100}=0.26 \mathrm{~mol}\)

∴ Number of moles of octane = \(\frac{35}{114} \mathrm{~mol}=0.31 \mathrm{~mol}\)

Mole fraction of heptane, x₁ = \(\frac{0.26}{0.26+0.31}=0.456\)

And, mole fraction of octane, x₂ = 1 – 0.456 = 0.544

Now, partial pressure of heptane, p₁ = x₁P⁰₁ = 0.456 x 105.2 = 47.97 kPa

Partial pressure of octane, p₂ = x₂p⁰₂ = 0.574 x 46.8 = 26.86 kPa

Hence, the vapour pressure of a solution. P_Total = p₁ + p₂ = 47.97 + 26.86 = 74.83 kPa

Question 29. The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water). Molar mass of water = 18 g mol^-1

∴ Number of moles present in 1000 g of water = \(\frac{1000}{18}=55.56 \mathrm{~mol}\)

Therefore, the mole fraction of the solute in the solution is x2 = \(\frac{1}{1+55.56}=0.0177\)

It is given that.

Vapour pressure of water, P⁰₁ = 12.3 kPa

Applying the relation.

⇒ \(\frac{p_1^0-p}{p_1^0}=x_2\)

⇒ \(\frac{12.3-\mathrm{p}_1}{12.3}=0.0177\)

⇒ 12.3 -P₁ = 0.2177

⇒ p₁ = 12.0823 = 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

Question 30. Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in in 114 g octane to reduce its vapour pressure to 80%.
Answer:

Let the vapour pressure of pure octane be p⁰₁.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is

⇒ \(\frac{80}{100} p_1^{0}=0.8 p_1^0\)

Molar mass of solute, M₂ = 40 g mol^-1

Mass of octane, w₁ = 114 g

Molar mass of octane, (C₈H₁₈)M₁ = 8 x 12 + 18 x 1 = 114 g mol^-1

Applying the relation,

⇒ \(\frac{p_1^0-p_1}{p_1^0}=\frac{w_2 \times M_1}{M_2 \times w_1}\)

⇒ \(\frac{p_1^0-0.8 p_1}{p_1^0}=\frac{w_2 \times 114}{40 \times 114}\)

⇒ \(\frac{0.2 \mathrm{p}_1^0}{\mathrm{p}_1^0}=\frac{w_2}{40}\)

⇒ \(0.2=\frac{W_2}{40}\)

W₂ = 8g

Hence, the required mass of the solute is 8g.

Question 31. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour the pressure becomes 2.9 kPa at 298 K. Calculate:

1. The molar mass of the solute
2. Vapour pressure of water at 298 K.

Answer:

1. Let, the molar mass of the solute be M g mol^-1

Now, the no, of moles of solvent (water), n1 = \(\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=5 \mathrm{~mol}\)

And, the no. of moles of solute, n2 \(\frac{30 \mathrm{~g}}{\mathrm{M} \mathrm{mol}^{-1}}=\frac{30}{\mathrm{M}} \mathrm{mol}\)

p₁ = 2.8 kPa

Applying the relation: \(\frac{p_1^0-p}{p_1^0}=\frac{n_2}{n_1+n_2}\)

⇒ \(\frac{p_1^0-2.8}{p_1^0}=\frac{\frac{30}{M}}{5+\frac{30}{M}}\)

⇒ \(\frac{\mathrm{p}_1^0}{2.8}=\frac{5 \mathrm{M}+30}{5 \mathrm{M}}\)

After the addition of 18 g of water:

⇒ \(n_1=\frac{90+18 g}{18}=6 \mathrm{~mol}\)

P₁ = 2.9 kPa

Again, applying the relation: \(\frac{p_1^0-p_1}{p_1^0}=\frac{n_2}{n_1+n_2}\)

⇒ \(\frac{p_1^0-2.9}{p_1^0}=\frac{\frac{30}{M}}{6+\frac{30}{M}}\)

⇒ \(\frac{p_1^0}{2.9}=\frac{6 M+30}{6 M}\)

Dividing equation (1) by (2), we have:

⇒ \(\frac{2.9}{2.8}=\frac{\frac{5 M+30}{5 M}}{\frac{6 M+30}{6 M}}\)

⇒ \(\frac{2.9}{2.8} \times \frac{6 \mathrm{M}+30}{6}=\frac{5 \mathrm{M}+30}{5}\)

⇒ 2.9 x 5 x (6M + 30) = 2.8 x 6 x (5M + 30)

⇒ 87 M + 435 = 84M + 504

⇒ 3M = 69

⇒ M = 23u

Therefore, the molar mass of the solute is 23 g mol^-1.

2. Putting the value of ‘M’ in equation (1), we have:

⇒ \(\frac{\mathrm{p}_1^0}{2.8}=\frac{5 \times 23+30}{5 \times 23}\)

⇒ \(\frac{p_1^0}{2.8}=\frac{145}{115}\)

⇒ \(p_1^0=3.53\)

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

Question 32. A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.
Answer: Here, ΔT_f = (273.1 5- 271)K = 2.15 K

Molar mass of sugar (C₁₂H₂₂O₁₁) =12 × 12 + 22 × 1 + 11 × 16 = 342 g mol^-1

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100- 5)g = 95g of water.

Now, number of moles of cane sugar = \(\frac{5}{342} \mathrm{~mol}=0.0146 \mathrm{~mol}\)

Therefore, molality of the solution, m = \(\frac{0.0146 \mathrm{~mol}}{0.095 \mathrm{~kg}}\) = 0.1537 mol kg^-1

Applying the relation.

ΔT_f = K_f × m

⇒ \(\mathrm{K}_{\mathrm{f}}=\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}=\frac{2.15 \mathrm{~mol}}{0.1537 \mathrm{~mol} \mathrm{~kg}^{-1}}\) = 13.99K kg mol^-1

Molar mass of glucose (C₆H₁₂O₆) = 6 × 12+12 × 12 × 1 + 6 × 16= 180 g mol^-1.

5% glucose in water means 5g of glucose is present in (100- 5) = 95 g of water.

∴ Number of moles of glucose = \(\frac{5}{180} \mathrm{~mol}=0.0278 \mathrm{~mol}\)

Therefore, molality of the solution, \(\mathrm{m}=\frac{0.0278 \mathrm{~mol}}{0.095 \mathrm{~kg}}=0.2926 \mathrm{mpl} \mathrm{kg}^{-1}\)

Applying the relation,

ΔT_f = K_f × m

⇒ 13.99 K kg mol^-1 × 0.2926 mol kg^-1

⇒ 4.09 K (approximately)

Hence, the freezing point of the 5% glucose solution is (273. 1 5- 4.09) K = 269.06 K

Question 33. Two elements A and B form compounds having the formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆). 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1,3K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate the atomic masses of A and B.
Answer:

We know that M₂ = {latex]\frac{1000 \times w_2 \times k_f}{\Delta T_f \times w_1}[/latex]

Then, M_AB2 = \(\frac{1000 \times 1 \times 5.1}{2.3 \times 20}=110.86 \mathrm{~g} \mathrm{~mol}^{-1}\)

M_AB4 = \(\frac{1000 \times 1 \times 5.1}{1.3 \times 20}=196.15 \mathrm{~g} \mathrm{~mol}^{-1}\)

Now, we have the molar masses of AB₂ and AB₄ as 110.87 g mol^-1 and 196.15 g mol^-1 respectively.

Let the atomic masses of A and B be x and y respectively.

Now, we can write:

x + 2y = 110.87 ____ 1

x + 4y= 196.15 _____ 2

Subtracting equation 1 from 2, we have

2y = 85.28

y = 42.64

Putting the value of ‘y’ in equation (1), we have

x + 2 × 42.64= 110.87

x = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

Question 34. At 300 K. 36g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:

Here, T = 300 K:

p = 1.52 bar:

R = 0.083 bar L K^-1 mol^-1

Applying the relation.

π = CRT

⇒ \(\mathrm{C}=\frac{\pi}{\mathrm{RT}}=\frac{1.52}{0.083 \mathrm{bar}, \mathrm{L} \cdot \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}=0.061 \mathrm{~mol}\)

Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

Question 35. Suggest the most important type of intermolecular attractive interaction in the following pairs.

1. N-hexane and n-oetane
2. I₂ and CCI₄
3. NaClO₄, and water (H₂O)
4. Methanol and acetone
5. Acetonitrile (CH₃CN) and acetone (C₃H₆O).

Answer:

1. Van der Wall’s forces of attraction.
2. Van der Wall’s forces of attraction.
3. Ion-dipole interaction.
4. Dipole – dipole interaction.
5. Dipole – dipole interaction

Question 36. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:

N-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of
a polar solute in the n-octane.

The order of increasing polarity is: Cyclohexane < CH₃CN < CH₃OH < KCI

Therefore, the order of increasing solubility is: KCI < CH OH < CH₃CN < Cyclohexane.

Question 37. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water.

1. Phenol
2. Toluene
3. Formic acid
4. Ethylene glycol
5. Chloroform
6. Pentanol,

Answer:

1. Phenol (C₆H₅OH) has the polar group -OH and non-polar group -C₆H₅. Thus, phenol is partially soluble in water.
2. Toluene (C₆H₅CH3.) has no polar group. Thus, toluene is insoluble in water
3. Formic acid (HCOOH) lias the polar group -OH and can form an H-bond with water. Thus, formic acid is highly soluble in water.
4. Ethylene glycol has a polar -OH group and can form an H-bond. Thus, it is highly soluble in water
5. Chloroform is insoluble in water.
6. Pentanol (C₅H₁₁OH) has a polar -OH group, but it also contains a very bulky non-polar -C₅H₁₁ group. Thus, pentanol is partially soluble in water.

Question 38. If the density of some lake water is 1.25 g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.
Answer:

Number of moles present in 92 g of Na⁺ ions

⇒  \(\frac{92 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=4 \mathrm{~mol}\)

Therefore, the molality of Na⁺ ions in the lake

⇒  \(\frac{4 \mathrm{~mol}}{1 \mathrm{~kg}}=4 \mathrm{~m}\)

Question 39. If the solubility product of CuS is 6 × 10⁶, calculate the maximum molarity of CuS in aqueous solution.
Answer:

Solubility product of CuS, K_sp = 6 × 10^-16

Let s be the solubility of CuS is mol L^-1

⇒ \(\mathrm{CuS} \longleftrightarrow \mathrm{Cu}_{\mathrm{S}}^{2+}+\mathrm{S}_{\mathrm{S}}^{2-}\)

Now, K_sp = [Cu²⁺][S²⁺]

= S × S = S²

Then, we have, K = S² = 6 x 10^-16

⇒ \(\mathrm{s}=\sqrt{6 \times 10^{-16}}=2.45 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}\)

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 x 10^-8 mol L^-1

Question 40. Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Answer:

6.5 g of aspirin (C₉H₈O₄) is dissolved in 450 g of acetonitrile (CH₃CN)

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage of C₉H₈O₄ = \(\frac{6.5}{456.5} \times 100 \%=1.424 \%\)

Question 41. Nalorphcne (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphine generally given is 1.5 mg. Calculate the mass of the 1.5 x 10 ³ m aqueous solution required for the above dose.
Answer:

The molar mass of nalorphine (C₁₉H₂₁NO₃) is given as:

19 × 12 + 21 × 1 + 1 × 14 + 3 × 16 = 311 g mol^-1

In 1.5 × 10^-3 m aqueous solutions of nalorphine,

1 kg (1000 g) of water contains 1.5 × 10^-3  mol = 1.5 × 10^-3  × 311 g = 0.4665 g nalorphine

Therefore, total mass of the solution = (1000 + 0.4665) g = 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphine is 1000.4665 g.

Therefore, the mass of the solution containing 1.5 mg of nalorphine is:

⇒ \(\frac{1000.4665 \times 1.5 \times 10^{-3}}{0.4665} \mathrm{~g}=3.22 \mathrm{~g}\)

Hence, the mass of aqueous solution required is 3.22 g

Note: There is slight variation in this answer and the one given in the Ncert textbook

Question 42. Calculate the amount of benzoic acid (C₆H₁₂COOH) required for preparing 250 mh. of 0.15 M solution in methanol.
Answer:

0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of benzoic acid.

Therefore. 250 mL of solution contains = \(\frac{0.15 \times 250}{1000} \mathrm{~mol} \text { of benzoic acid }\)

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 x 1 + 2 × 16 = 122 g mol^-1

Hence, required benzoic acid = 0.0375 mot × 122 g mol^-1 = 4.575 g

Question 43. The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:

• Among H, Cl, and F, H is the least electronegative while F is the most electronegative. Then, F can withdraw electrons towards itself more than Cl and H.
• Thus, trifluoroacetic acid can easily lose H⁺ ions i.e. trifluoroacetic acid ionizes to the largest extent. Now. the more ions produced, the greater the depression of the freezing point. Hence, the depression in the freezing point increases in the order.

Acetic acid < trichloroacetic acid < trifluoroacetic acid

Question 44. Calculate the depression in the freezing point of water when 10 g, CH₃ CH₂CHClCOOH is added to 250 g of water, K _a = 1.4 x 10^-3, K_f = 1.86 K kg mol^-1.
Answer:

Molar mass of CH₃CH₂CHCICOOH = 15 + 14+ 13 + 35.5 + 12 + 16 + 16 + 1 = 122.5 g mol^-1

No. of moles present in 10 g CM CH CHCICOOH = \(\frac{10 \mathrm{~g}}{122.5 \mathrm{~g} \mathrm{~mol}^{-1}}=0.0816 \mathrm{~mol}\)

It is given that 10 g CH₃CH₂CHCICOOH is added to 250 g of water.

The molality of the solution. = \(\frac{0.0186}{250} \times 1000=0.3265 \mathrm{~mol} \mathrm{~kg}^{-1}\)

Let a be the degree of dissociation of CH₃CH₂CHCICOOH

CH₃CH₂CHCICOOH undergoes dissociation according to the following equation:

CH₃CH₂CHCICOOH ↔ CH₃CH₂CHClCOO^– + H⁺

⇒ \(K_a=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^2}{(1-\alpha)}\)

Since a is very small concerning 1, 1 – α ≈ 1

Now, ⇒  K = Cα²

⇒ \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{1.4 \times 10^{-3}}{0.3264}}=0.0655 \quad\left(\mathrm{~K}_{\mathrm{a}}=1.4 \times 10^{-3}\right)\)

Again, CH₃CH₂CHCICOOH ↔ CH₃CH₂CHCICOO^– + H⁺

Total mass of equilibrium = I – α + α + α = 1 + α

⇒ \(\mathrm{i}=\frac{1+\alpha}{1}=1+\alpha=1+0.0655=1.0655\)

Hence, the depression in the freezing point of water is given as:

ΔT_f = i.K_fm = 1.0655 × 1.86 kg mol^-1 × 0.364 mol kg^-1=0.65 K

Question 45. 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t a Hoff factor and dissociation constant of fluoro acetic acid.
Answer:

It is given that:

w₁ = 500 g,

w₂ = 19.5 g

K_f = 1.86 K kg mol^-1

ΔT_f = 1 K

We know that: \(\mathrm{M}_2=\frac{\mathrm{K}_f \times \mathrm{w}_2 \times 1000}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_1}\)

⇒ \(=\frac{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 19.5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{500 \mathrm{~g} \times 1 \mathrm{~K}}\) = 72.54mol^-1

Therefore, observed molar mass of CH₂FCOOH, (M₂)obs = 72.54 mol

The calculated molar mass of CH₂FCOOH is:

(M₂)_cal = 14 +19 + 12 + 16+ 16+ 1 = 78 g mol^-1

Therefore, van’t Hoff factor,

⇒ i = \(\frac{\left(\mathrm{M}_2\right)_{\text {cat }}}{\left(\mathrm{M}_2\right)_{\text {obo }}}=\frac{78 \mathrm{~g} \mathrm{~mol}^{-1}}{72.54 \mathrm{~g} \mathrm{~mol}^{-1}}=1.0753\)

Let a be the degree of dissociation of CH₂FCOOH

⇒ \(\mathrm{CH}_2 \mathrm{FCOOH} \quad \longleftrightarrow \quad \mathrm{CH}_2 \mathrm{FCOO}^{-}+\mathrm{H}^{+}\)

Total = C (1 + α)

⇒ \(\mathrm{i}=\frac{\mathrm{C}(1+\alpha)}{\mathrm{C}}\)

⇒ i = 1 + α

⇒ α = i – 1 = 1.0753 – 1 = 0.0753

Now. the value of K_a is given as:

⇒ \(\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_2 \mathrm{FCOO}^{-}\right]\left[\mathrm{H}^{-}\right]}{\left[\mathrm{CH}_2 \mathrm{FCOOH}\right]}=\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}=\frac{\mathrm{C}^2}{1-\alpha}\)

Take the volume of the solution as 500 mL.

We have a concentration: of 19.5 M

weight = 19.5 g

Concentration = \(\frac{\frac{19.5}{78}}{500} \times 1000=0.5 \mathrm{M}\)

Therefore, K_a = \(\frac{\mathrm{C} \alpha^2}{1-\alpha}\)

⇒ \(\frac{0.5 \times(0.0753)^2}{1-0.0753}=\frac{0.5 \times 0.00567}{0.9247}=0.00307 \text { (approximately) }\)

= 3.07 x 10^-3

Question 46. The vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:

Vapour pressure of water, p⁰₁ = 17.535 mm of Hg

Mass of glucose, w₂ = 25 g

Mass of water, w₁ = 450 g

We know that.

Molar mass of glucose (C₆H₁₂O₆).M₂ = 6 x 12 + 12 x 1 + 6 x 16 = 180 g mol^-1

The molar mass of water. M₁ = 18 g mol^-1

Then, number of moles of glucose, n₂ = \(\frac{25}{180 \mathrm{~g} \mathrm{~mol}^{-1}}=0.139 \mathrm{~mol}\)

And. number of moles of water, n₁ = \(\frac{450 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=25 \mathrm{~mol}\)

We know that,

⇒ \(P_1=P^0 \times\left[\frac{n_1}{n_1+n_2}\right]\)

⇒ \(P_1=17.535\left[\frac{25}{25+0.139}\right]=\frac{17.535 \times 25}{25.139}\)

= 17.535 -p₁ = 0.097

= p₁ = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.

Question 47. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10^-3 mm Hg. Calculate the solubility of methane in benzene at 298 Kelvine 760 mm Hg.
Answer:

Here, p = 760 mm Hg;

k_H = 4.27 x 10⁵ mm Hg

According to Henry’s law,

p = K_HX \(\Rightarrow \quad x=\frac{p}{k_{H}}=\frac{760 \mathrm{~mm} \mathrm{Hg}}{4.27 \times 10^5 \mathrm{~mm} \mathrm{Hg}}\)

= 177.99 x 10^-5 = 178 x 10^-5 (approximately)

Hence, the mole fraction of methane in benzene is 178 x 10^-5

Question 48. 100g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:

Number of moles of liquids A, n_A= \(\frac{100}{140} \mathrm{~mol}=0.714 \mathrm{~mol}\)

Number of moles of liquids B, n_B = \(\frac{1000}{180} \mathrm{~mol}=5.556 \mathrm{~mol}\)

Then, the mole fraction of A

⇒ \(x_A=\frac{n_A}{n_A+n_B}=\frac{0.714}{0.714+5.556}=0.114\)

And, mole fraction of B, x_B = 1 – 0.1 14 = 0.886

Vapour pressure of pure liquid B, p⁰_B = 500 torr

Therefore, the vapour pressure of liquid B in the solution,

p_B = P⁰_B x_B = 500 x 0.886 a 443 torr

Total vapour pressure of the solution, p_Total = 475 torr

∴ Vapour pressure of liquid A in the solution,

P_A = P_Total – P_B = 475 – 443 = 32 toor

Now, \(p_A=p_A^0 x_A \quad \Rightarrow \quad p_A^0=\frac{p_A}{x_A}=\frac{32}{0.114}=280.7\)

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Question 49. The vapour pressure of pure acetone and chloroform at 328 K is 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solutions over the entire range of composition, plot p_total  P_chloroform p_acetone as a function of x_acetone. The experimental data observed for different compositions of the mixture is:

From the question, we have the following data

Answer:

It can be observed from the graph that the plot for the plural of the solution curves downwards. Therefore, the solution shows a negative deviation from the ideal behaviour.

Question 50. Benzene and toluene form ideal solutions over the entire range of compositions. The vapour pressure of pure benzene and toluene at 300 K is 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer:

Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1 = 78 g mol^-1

Molar mass of toluene = 7 × 12 + 8 × 1 = 92 g mol^-1 

Now, no. of moles present in 80 g of benzene = \(\frac{80}{78} \mathrm{~mol}=1.026 \mathrm{~mol}\)

And. no. of moles present in 100 g of toluene = \(\frac{100}{92} \mathrm{~mol}=1.087 \mathrm{~mol}\)

Mole fraction of benzene, x_b = \(\frac{1.026}{1.026+1.087}=0.486\)

And, mole fraction of toluene, x₁ = 1 – 0.486 = 0.514

It is given that vapour pressure of pure benzene, p⁰₁ = 50.71 mm of Hg

and vapour pressure of pure toluene, p⁰₁ = 32.06 mm of Hg

Therefore, the partial vapour pressure of benzene,

p₁ = x₁ × p₁ = 0.486 x 50.71 = 24.645 mm of Hg

And, partial vapour pressure of toluene,

p₁ = x₁ × p₁ = 0.5 14 × 32.06 = 16.479 mm of Hg

Hence, the mole fraction of benzene in the vapour phase is given by:

⇒ \(\frac{p_b}{p_b+p_1}=\frac{24.645}{24.645+16.479}=\frac{24.645}{41.124}=0.599=0.6\)

Question 51. The air is a mixture of several gases. The major components are oxygen and nitrogen with an approximate proportion of 20% to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K Henry’s law constants for oxygen and nitrogen are 3,30 x 10⁷ mm and 6.51 x 10⁷ ram respectively. Calculate the composition of these gases in water.
Answer:

Percentage of oxygen (O₂) in air = 20%

Percentage of nitrogen (N₂) in air= 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm. that is. (10 x 760) mm Hg = 7600 mm Hg

Partial pressure of oxygen, \(\mathrm{p}_{\mathrm{O}_2}=\frac{20}{100} \times 7600 \mathrm{~mm} \mathrm{Hg}=1520 \mathrm{~mm} \mathrm{Hg}\)

Partial pressure of nitrogen, \(p_{\mathrm{N}_2}=\frac{79}{100} \times 7600 \mathrm{~mm} \mathrm{Hg}=6004 \mathrm{~mm} \text { of } \mathrm{Hg}\)

Now, according to Henry’s law : p = K_Hx

For oxygen: p_O2 = K_H.x_O2

⇒ \(\mathrm{x}_{\mathrm{O}_2}=\frac{\mathrm{P}_{\mathrm{O}_5}}{\mathrm{~K}_{\mathrm{H}}}=\frac{1520 \mathrm{~mm} \text { of } \mathrm{Hg}}{3.30 \times 10^7 \mathrm{~mm} \text { of } \mathrm{Hg}}\left(\text { Given } \mathrm{K}_{\mathrm{H}}=3.30 \times 10^7 \mathrm{~mm} \text { of } \mathrm{Hg}=4.61 \times 10^5\right. \text {. }\)

For nitrogen. p_N2 = K_H.x_N2

⇒ \(x_{\mathrm{N}_2}=\frac{\mathrm{P}_{\mathrm{N}_1}}{\mathrm{~K}_{\mathrm{H}}}=\frac{6004 \mathrm{~mm} \mathrm{Hg}}{6.51 \times 10^7 \mathrm{~mm} \mathrm{Hg}}=9.22 \times 10^{-5}\)

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^-5 and 9.22 × 10^-5 respectively.

Question 52. Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litres of water such that its osmotic pressure is 0.75 atm at 27°C.
Answer:

We know that, \(\pi=i \frac{n}{V} R T\)

⇒ \(\pi=i \frac{w}{M V} R T \Rightarrow w=\frac{\pi M V}{i R T}\)

⇒ π = 0.75 atm

⇒ V = 2.5 L

⇒ i = 2.47

⇒ T = (27 + 273) K = 300 K

Here,

⇒ R = 0.082 L atm K^-1 mol^-1

⇒ M = 1 × 40 + 2 × 35.5 = 111 g mol^-1

Therefore, w = \(\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}=3.42 \mathrm{~g}\)

Hence, the required amount of CaCl₂ is 3.42g.

Question 53. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer: When K₂SO₄ is dissolved in water. K+ and SO₂^4- ions are produced

⇒ \(\mathrm{K}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_4^{2-}\)

Total number of ions produced = 3

∴ i = 3

Given, w = 25 mg = 0.025 g

V = 2 L

T = 25°C = (25 + 273) K = 298 K

Also, we know that :

R = 0.0821 L atm K^-1 mol^-1

M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^-1

Applying the following relation.

⇒ \(\pi=i \frac{n}{v} R T\)

⇒ \(\mathrm{i} \frac{\mathrm{W}}{\mathrm{M}} \frac{1}{\mathrm{v}} \mathrm{RT}=3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298=5.27 \times 10^{-3} \mathrm{~atm}\)