Class 11 Chemistry

• Environmental Chemistry
• Filling Up Of Electrons In Different Orbitals
• Oxides Of Caron
• Structure Of Atom
• Thermodynamic Properties
• Heisenberg’s Uncertainty Principle
• Quantum Number

Environmental Components Of Earth

Earth’s environment is composed of the following four components—

1. Atmosphere,
2. Hydrosphere
3. Lithosphere and
4. Biosphere.

Among these, the first three components are abiotic while the fourth one is biotic.

1. Atmosphere: The invisible gaseous layer that surrounds and protects the Earth is called the atmosphere.
2. Hydrosphere: It includes all sources of water such as seas, oceans, rivers, fountains, lakes, polar regions, glaciers, groundwater etc.
3. Lithosphere: It comprises of the solid crust of the earth, made of rocks, forming the outer mineral cover.It includes soil, minerals, organic matter etc., and extends up to a depth of about 30 km from the earth’s surface.
4. Biosphere: It is that part of the earth where living organisms exist and interact with each other and also with the non-living components. Biosphere consists of all three zones.
5. For example: Soil, water, air etc., where living beings exist

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Atmosphere

The invisible blanket of the gaseous layer that surrounds the earth is called the atmosphere. It extends upwards to about 1600km. It is the gravitational attraction of the earth that holds this gaseous layer closely in space around the earth’s surface. The total mass of gaseous substances in the atmosphere is nearly 5.5×10¹⁵ tons.

Based on temperature gradients and altitude, the atmosphere has been divided into four distinct zones.

These are:

• Troposphere
• Stratosphere
• Mesosphere and
• Thermosphere

Different zones of atmosphere:

Again according to the proportion of different gases from the surface of the earth towards

The vacuum of interstellar space and atmosphere can be divided into two categories: 

1. Homosphere and
2. Heterosphere.

Homosphere extends from the surface of the earth upto about 100 km height. In this layer, the proportions of different gases are more or less identical. Thus, this layer is called the homosphere. The layer next to it is known as the heterosphere because the proportion of the gases in the different parts of this layer are found to be dissimilar.

Gravity holds most of the air molecules close to the earth’s surface and hence the troposphere is much more denser than the other layers. 50% of the total mass of the atmosphere exists within a height of5.5 km from the earth’s surface and 99% exists within a height of 30 km from the earth’s surface

Average gaseous composition in homosphere

Functions of gases present in the atmosphere

1. Oxygen:

• The most significant gaseous constituent of the atmosphere is oxygen. Oxygen is indispensable for any kind ofcombustion.
• Oxygen is also used for the oxidation of food taken by plants and animals to produce heat and energy.
• Oxygen is a necessary component of life as all living beings (except some microorganisms) and plants take oxygen from the atmosphere for respiration. Plants give up oxygen to the atmosphere during the process of photosynthesis.
• As a result, the balance of oxygen is maintained in the atmosphere.

2. Nitrogen:

• The major constituent of the atmosphere is nitrogen. Proteins and nucleic acids present in living bodies are nitrogenous compounds.
• But most of animals including human beings and even plants cannot utilise atmospheric nitrogen directly for the production of proteins and amino acids.
• However, some nitrogen-fixing bacteria can take nitrogen directly from the air and produce nitrate salts in the soil.
• These are used by plants in the synthesis of amino acids and nucleic acids. Herbivorous animals meet their protein demand by eating those plants. Similarly, carnivorous animals get proteins from herbivorous animals.
• After the death of plants and animals, nitrogenous compounds present in their bodies are decomposed by some bacteria releasing nitrogen gas that returns to the atmosphere.

3. Carbon dioxide (CO₂):

• Combustion of fossil fuels and carbonaceous compounds, and respiration of plants and animals increase carbon dioxide content in the atmosphere.
• Again plants, during photosynthesis, absorb carbon dioxide from the atmosphere for the preparation of food.
• As a result, the balance of carbon dioxide is maintained in the atmosphere.
• But due to excessive … combustion of carbonaceous fuels and indiscriminate deforestation, the quantity of carbon dioxide in the atmosphere is increasing constantly leading to a constant increase in the average temperature of the earth (See Greenhouse effect).

4. Ozone:

• The quantity of ozone gas present in the atmosphere is negligible.
• Almost the entire amount of ozone (=90%) is present in the stratosphere whichis about 15-35 km above the earth’s surface.
• Presence of ozone gas close to the earth’s surface hurts mankind and other animals.
• But the presence of ozone in the upper layer of tyre atmosphere is beneficial since it absorbs the harmful ultraviolet rays of the sun

Thermodynamic Properties And Thermodynamic State Of A System

Thermodynamic properties

The measurable physical quantities by which the dynamic state of a the system can be defined completely are called thermodynamic properties or variables of the system. Examples: The pressure (P), temperature (T), volume (V), composition, etc., of a system are the thermodynamic properties or variables of the system because the state of the system can be defined by these variables or properties.

The properties or variables required to define the state of a system are determined by experiment. Although a thermodynamic system may have many properties (like— pressure, volume, temperature, composition, density, viscosity, surface tension, etc.), to define a system we need not mention all of them since they are not independent If we consider a certain number of properties or variables having certain values to define the state of a system, then the other variables will automatically be fixed.

In general, to define the state of a thermodynamic system, four properties or variables are needed. These are the pressure, volume, temperature, and composition of the system. If these variables of a thermodynamic system are fixed then the other variables will also be fixed for that system.

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For a closed system of fixed composition, the state of the system depends upon the pressure (P), temperature (T), and volume (V) of the system. If these three variables of the system (P, V, T) are fixed, then other variables (like density, viscosity, internal energy, etc.) ofthe system automatically become fixed.

Thermodynamic state of a system

A system is said to be in a given thermodynamic state ifthe properties (For example pressure, volume, temperature, etc.) determining its state have definite values.

If the thermodynamic properties or variables of a thermodynamic system remain unchanged with time, then the system is said to be in thermodynamic equilibrium. A system is said to be in thermodynamic equilibrium if it attains thermal equilibrium, mechanical equilibrium, and chemical equilibrium simultaneously.

1. Thermal equilibrium: A system is said to be in thermal equilibrium if the temperature throughout the system is the same and is equal to that of its surroundings.
2. Mechanical equilibrium: If no imbalanced force exists within a system and also between the system and its surroundings, the system is said to be in mechanical equilibrium.
3. Chemical equilibrium: If the chemical composition throughout a system remains the same with time, the system is said to be in chemical equilibrium.

State function of a thermodynamic system

The state function of a thermodynamic system is a property whose value depends only on the present state of the system but not on how the system arrived at the present state. Examples: Pressure (P), volume (V), temperature (T), internal energy (E or U), enthalpy (H), entropy (S), Gibbs free energy (G), etc., of a thermodynamic system the state functions because the values of these functions depend only on the present state ofa system, not on how the system arrived at that state.

Change of a state function in a process: The state of a thermodynamic system at the beginning of a process is called its initial state and the state attained by the system after completion of the process is called its final state. Let X (like P, V, T, etc., of a system) be a state function of a thermodynamic system. The values of X at the beginning and the end of a process are X1 and respectively. So, the change in the value of X in the process, AX = X2-X1.

• Infinitesimal change in x is represented by dx and finite change in x is represented by ax. For example, the infinitesimal change in pressure (p) of a system is dp and the finite change is ap.
• If X is a state function of a thermodynamic system, then dX must be a perfect differential as the integration of dX between two states results in a definite value of X

The state function of a system is a path-independent quantity: A state function of a system depends only on the state of the system.

Consequently, the change in any state function ofa system undergoing a process depends only upon the initial and final states of the system in the process, not on the path of the process. Thus the state function of a system is a path-independent quantity.

Explanation: Suppose, a system undergoes a process in which its state changes from A (initial state) to B (final state), and because of this, the value of its state function X changes from XA (value of X at state A ) to XB (value of X at state B). The process can be carried out by following three different paths.

But the change in X, i.e., AX= (XB-XA) will be the same for all three paths. This is because all the paths have identical initial and final states and consequently X has identical initial and final values for these paths.

Example: The change in temperature of a system depends only upon the initial and the final stages of the process. It does not depend on the path followed by the system to reach the final state. So the temperature of a system is a state function. Similarly, the change of other state functions like pressure (P), volume (V), internal energy (U), enthalpy (H), entropy (S), etc., (i.e. AP, AV, AU, AH, AS, etc.) does not depend upon the path ofthe process.

Path-dependent quantity

Two terms commonly used in thermodynamics are heat (q) and work ( w). These are not the properties ofa system. They are not state functions.

Heat change or work involved in a process depends on the path of the process by which the final state of the system is achieved. Thus, heat and work are the path-dependent quantities.

In general, capital letters are used to denote the state functions (for example, P, V, T, U, etc.), and small letters are used to denote path functions (for example q, w, etc.). q and w are not state functions.

Hence, 5q or 8w (S = delta) are used instead of dq or dw. Unlike dP or dV, which denotes an infinitesimal change in P or V, 8q or 8w does not indicate such kind of change in q or w. This is because q and w like P or F are not the properties of a system. 8q and 8w are generally used to denote the infinitesimal transfer of heat and work, respectively, in a process.

Rules For Filling Up Of Electrons In Different Orbitals

The correct ground state electronic configuration of an atom is obtained on the basis of the following principles—Pauli’s exclusion principle, Hund’s rule, and the Aufbau principle.

Pauli’s exclusion principle

Principle: The knowledge of four quantum numbers is important in assigning the exact location of the electron within an atom.

After meticulous study of the line spectra of atoms, Wolfgang Pauli in 1925 proposed his exclusion principle which is widely known as Pauli’s exclusion principle.

According to this principle, no two electrons in an atom will have the same values for all four quantum numbers (n, l, m, and s).

If three of the quantum numbers of any two electrons are the same then they must differ in their fourth quantum number.

If the quantum numbers n, l, and m of two electrons have identical values, then the value of s should be different (+i for one and for the other).

Therefore, the corollary of this principle may be stated as—each orbital can accommodate a maximum of two electrons having an opposite spin.

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With the help of Pauli’s exclusion principle, the maximum number of electrons a subshell can accommodate can be calculated. For example—

s -subshell: In the case of s -subshell, 1 = 0. Therefore m = 0. Number of orbitals in s -subshell = 1.

According to Pauli’s exclusion principle, each orbital can hold a maximum number of two electrons. So, s -subshell can accommodate a maximum of 2 electrons.

p -subshell: For p -subshell, 1=1 and m = —1,0, +1. The number of orbitals in the -subshell is three (px, py, and pz ).

According to Pauli’s exclusion principle, since each orbital can hold a maximum of 2 electrons, the maximum accommodating capacity of p -subshell {i.e., three p orbitals) =3×2 = 6 electrons.

d -subshell: In the case of d -subshell, 1 = 2, m = -2, -1, 0 +1, +2. Thus, m has 5 values indicating the presence of 5 orbitals. As the maximum number of electrons that each orbital can hold is 2, the maximum number of electrons that a d -d-subshell can accommodate is 5 X 2 = 10.

f-subshell: For /-subshell, l = 3, m = -3, -2, -1, 0, +1, +2, +3. Seven values of m indicate the presence of seven orbitals. Hence the maximum number of electrons that may be present in /-subshell is 7 x 2 = 14 .

Pauli’s exclusion principle also permits the determination of the maximum number of electrons that can be present in a certain orbit or shell.

Example: For L -shell (n = 2), l has two values, i.e., 1 = 0 [ssubshell] and l = 1 [p -subshell].

The s -subshell can hold 2 electrons and p -subshell can accommodate 6 electrons. Therefore, the maximum accommodating capacity for L shell =(2 + 6) = 8 electrons.

Similarly, it can be shown that, the maximum number of electrons that can be accommodated in M-shell (n = 3) = 18 and the maximum number of electrons that may be present in IVshell (n = 4) =32.

Electron accommodating capacity of K, L, M, and JV-shell

Thus, it is seen that the maximum number of electrons accommodated in any electronic orbit with the principal quantum number’ n’ is 2n2.

Number of orbitals and electron accommodating capacity of different shells.

Hund’s multiplicity rule

This rule is helpful for deciding the mode of filling of the orbitals ofthe same energy level with electrons.

Rule: The pairing of electrons in the orbitals within the same subshell does not take place until the orbitals are singly filled up with electrons having parallel spin.

Discussion: The rule implies that orbitals with the same energy are filled up first with one electron and then the additional electron occupies the singly filled orbital orbital to form paired electrons (with opposite spin).

The energy order of the orbitals, the Aufbau principle, and the electronic configuration of atoms

The German word ‘Aufbau’ means ‘to build one by one! The Aufbau principle gives the sequence of gradual filling up of the different subshells of multi-electron atoms.

Aufbau principle:

Aufbau principle states that electrons are added progressively to the various orbitals in the order of increasing energy, starting with the orbital with the lowest energy.

Electrons never occupy the die orbital of higher energy leaving the orbital of lower energy vacant.

A study of the results of spectral analysis has led to the arrangement of the shells and subshell in the increasing order of their energies in the following sequence:

Is < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f< Sd < 6p < 7s < 5f< 6d ..

Electronic configuration always conforms to Pauli’s Exclusion Principle.

According to Hund’s rule, pairing of electrons in the orbitals within the same subshell (degenerate orbitals hating the same n ) cannot occur until the orbitals are singly filled up.

The energy of the subshell increases with an increase in the value of {n + l). In a multi-electron atom, the energy of a subshell, cannot be determined only by principal quantum number (n ), in exclusion of azimuthal quantum number (Z).

The correct order of energies of various subshells is determined by the (n + 1) rule or Bohr-Bury rule.

The implication of the rule can be better understood with the help ofthe following example.

In case of 3d -subshell, (n + Z) = (3 + 2) = 5, but for 4s -subshell, (n + Z) = (4 + 0) = 4 .

From this, it is clear that the energy of the 4s -subshell is less than that of the 3d -subshell. Hence, the electron goes to the 4s subshell first, in preference to the 3d -subshell.

If Two subshells have the same value for{n + l), then the electron enters that subshell which has a lower value of n.

For example, for 3d -subshell, {n + l) = (3 + 2) = 5 and for 4p -subshell, {n + l) = (4 + 1) = 5 In this case, the electron first enters the 3d -subshell which has a.lower value of n.

The sequence in which the subshells are filled with electrons.

The figure depicts the sequence of filling up of the subshells with electrons. The electronic configuration of any atom can be easily predicted from this diagram.

Exceptions to (n+1) rule: Exceptions to the {n + Z) rule are found to occur in the case of filling up of electrons in Lanthanum (La) and Actinium (Ac).

The values of {n + l) in the case of both the subshells 4/ and 5d (4 + 3 = 7 = 5 + 2) are found to be the same.

Similarly the values of (n +1) in the case of both the subshells 5/ and 6d (5 + 3 = 8 = 6 + 2) are equal. So, the order of energies of these subshells is 4/< 5d and 5/< 6d.

According to the (n + Z) rule, the expected electronic configuration of La (57) and Ac (89) should be [Xe]4/15d06s2 and [Rn]5/16d°7s2 respectively.

However, the electronic configuration of La and Ac are actually [Xe]4/ and [Rn]5/°6d17s2 respectively. In other words, lanthanum and actinium are exceptions to the (n + l) rule.

Method of writing electronic configuration of an atom 1) In order to express the electronic configuration of an atom, the principal quantum number (n = 1, 2, 3… etc.) is written first.

The symbol ofthe subsheU(s, p, d, f, etc.) is written to the right ofthe principal quantum number. For example, s -subshell of the first shell is expressed as Is; sand subshells of the second shell are expressed as 2s and 2p respectively.

The total number of electrons present in any subshell is then written as the right superscript of the subshell symbol.

For example, the electronic configuration, ls22s22p5 suggests that the s -subshell of the first shell contains 2 electrons, and the s, and p -subshells of the second shell contain 2 electrons and 5 electrons respectively. Thus, the total number of electrons present is equal to 9.

Examples: Electronic configuration of 17 CL atom: The atomic number of chlorine is 17. Number of electrons present in chlorine atom is 17.

Out of these 17 electrons, 2 electrons are present in the s -subshell of first shell (K-shell), 2 electrons and 6 electrons in the s – and p -subshell of the second shell (L -shell) respectively, and 2 and 5 electrons are present in the s – and p -subshell of the third shell (Mshell) respectively.

Thus, the electronic configuration of the chlorine atom is ls²2s²2p63s²3p5.

Electronic configuration of 26Fe atom: The atomic number of iron is 26. Number of electrons present in an atom of iron is 26. These 26 electrons are distributed in K, L, M, and N-shells in such a way that their electronic configuration becomes ls²2s²2p63s²3pe3de4s².

Here the symbol signifies an orbital and the arrow sign (↑) means an odd electron and the paired arrow sign (↓↑) stands for a pair of electrons with opposite spins.

Stability of half-filled or completely filled subshells The electronic configurations of some atoms have certain characteristic features.

It is seen that half-filled and completely filled subshells are more stable compared to nearly half-filled or nearly completely filled subshells.

Hence, if the (n-1)d -subshell of any atom contains 4 or 9 electrons and the ns -subshell contains 2 electrons, then one electron from the ns -subshell gets shifted to the (n-1) d subshell, thereby making a total number of either 5 or 10 electrons in it. As a result, ns -subshell is left with 1 electron instead of 2.

The extra stability of half-filled and completely filled subshells can be explained in terms of the symmetrical distribution of electrons and exchange energy.

Symmetrical distribution of electrons: The subshells with half-filled or completely filled electrons are found to have a more symmetrical distribution of electrons.

Consequently, they have lower energy which ultimately results in greater stability of the electronic configuration.

Electrons present in the same subshell have equal energy but their spatial distribution is different. As a result, the magnitude of the shielding effect of another is quite small and so, the electrons are more strongly attracted by the nucleus.

Interelectronic repulsion: Two types of interactions are possible between electrons of the same subshell due to interelectronic repulsive force.

Interaction due to electronic charge: The magnitude of the repulsive force acting between two electrons situated at n distance r from each other is inversely proportional to the square of the distance between them.

Consequently, the stability of two-electron or multi-electron ions or atoms increases with an increase in distance r. Thus, die two electrons present in the d -d-subshell prefer to be in two separate d -orbitals instead of one leading to the increased stability ofthe atom or ion.

Interaction due to rotation of electrons: Two electrons tend to remain close to each other if they have opposite spins. On the other hand, if both the electrons have parallel spin, then they prefer to remain far from each other.

The electrons occupying degenerate orbitals (orbitals of the same energy) can exchange their positions with other electrons with the same spin. In this process, exchange energy is released.

The greater the probability of exchange, the more stable the configuration. The probability of exchange is greater in the case of a half-filled or completely filled subshell.

Thus, the magnitude of exchange energy is greatest for half-filled or completely filled subshells leading to their exceptionally high stability.

This exchange energy forms the basis of Hund’s multiplicity rule. The relative magnitude of exchange energy can be calculated by the formula,

No. of exchanges \(=\frac{n !}{2 \times(n-2) !}\)

(n = number of degenerate electrons with parallel spin.)

Number of interactions in case of d4 electronic configuration

Total number of exchanges for d4 electronic configuration

=3+2+1=6

Number of interactions in case of d5 electronic configuration

Electronic configuration of ions

When an additional electron is added to an orbital of an atom, a negatively charged ion called an anion is formed while the removal of an electron from the orbital of an atom produces a positively charged ion called cation.

Electronic configuration of anions: The total number of electrons present in an anionic species is = (Z + n) where Z = atomic number and n = number of electrons gained. The electronic configuration ofthe anion is written on the basis of the total number of electrons present in it.

Examples: Fluoride ion (F-): Total number of electrons present in F- ion = (9 + 1) = 10

∴ Electronic configuration of F- ion: ls²2s²2p6

Nitride ion (N³¯ ): Total number of electrons present
in N3- ion = (7 + 3) = 10

Electronic configuration of N3- ion: ls22s22p6

Oxide ion (O²¯): Total number of electrons present in  O²¯ ion =(8 + 2) = 10.

∴ Electronic configuration of O2- ion: ls22s22p6

Sulphide Ion (S²¯) : Total number ofelectrons present in S2- ion =(1.6 + 2) = 18

Electronic Configuration of cations:

A total number of electrons present in a cationic species = (Z-n) where Z = atomic number and n = number of electrons lost.

For writing the electronic configuration of the cation, the electronic configuration of the neutral atom is written first.

Then requisite no. of electrons is removed from the outermost shell. Electrons from the ns -subshell should be removed before removing any electron from the (n- l)d -subshell.

The total number of electrons present in a cationic species = (Z-n) where Z = atomic number and n = number of electrons lost.

For writing the electronic configuration of the cation, the electronic configuration of the neutral atom is written first.

Then requisite no. of electrons is removed from the outermost shell. Electrons from the ns -subshell should be removed before removing any electron from the (n- l)d -subshell.

Examples:

Sodium ion (Na+) : Electronic configuration of \({ }_{11} \mathrm{Na}: 1 s^2 2 s^2 2 p^6 3 s^1 \text {. So, } \mathrm{Na}^{+} \text {lon: } 1 s^2 2 s^2 2 p^6\)

2. Chromium Ion (Cr3+): Electronic Configuration of

\({ }_{24} \mathrm{Cr}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1\)

⇒ \(\mathbf{C r}^{3+} \text { ion: } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3\)

Manganese ion (Mn2+): Electronic configuration of:

Mn2+ ion: 1s22s22p63s2363d5

Ferrous (Fez+) and Ferric (Fe3+) ion: Electronic

Configuration of 26Pe: ls22s2263sz3763d64s2

Ferrous ion (Fe2+): ls22s22p63s23p63d6

Similarly, ferric ion (Fe3+): ls22s22/763s23/?63d5

Cuprous (Cu+) and Cupric (Cu2+) ion: Electronic configuration of 2gCu: ls²2s²2p63s23p63dl04s1

Cu+ ion: ls22s22/763s23/763d10

similarly, cupric ion (Cu2+): 1s²2s22p63s23p63d9

Orbital angular momentum of electron = Jl(l + 1) x ( l = azimuthal quantum number).

Molecules, atoms, or ions containing one or more unpaired electrons exhibit paramagnetic properties. Paramagnetic substances are attracted by the magnetic field.

The magnetic moment of paramagnetic substances depends on the number of unpaired electrons.

Magnetic moment = Jx(x + 2) BM BM = Bohr Magneton (unit of magnetic moment) x = Number of impaired electrons.

Molecules, atoms, or ions containing an even number of electrons exhibit diamagnetic properties. Diamagnetic substances are repelled by the magnetic field.

Carbon forms three oxides,: e.g., carbon monoxide (CO), carbon dioxide (C02) and carbon suboxide (C₃O₂). Among these, the first two are important

1. Carbon Monoxide

Carbon Monoxide Laboratory preparation:

1. In the laboratory, carbon monoxide is prepared by dehydrating formic acid or oxalic acid after heating with concentrated sulphuric acid.

2.  When potassium ferrocyanide is heated with excess of the cone, sulphuric acid, and pure carbon monoxide is obtained

CO cannot be dried by concentrated sulphuric acid:

Concentrated sulphuric acid is a strong oxidising agent. Thus, when CO (a reducing agent) is passed through concentrated H₂SO₄, it is oxidised by sulphuric acid to CO₂

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Carbon Monoxide Other methods of preparation

1. From carbon:

When steam is passed over red hot coke, water gas or synthesis gas (CO + H₂) is obtained. CO is separated from the mixture by liquefaction.

When air is passed over hot coke, producer gas (CO + N₂) is formed. CO is separated by liquefaction

2 . From carbon dioxide: When CO₂ is passed over red hot carbon, zinc, Iron etc., it is reduced to CO.

CO₂ + C→ 2CO; CO₂ + Zn→CO + ZnO

CO₂ + Fe→FeO + CO

3. From metal oxides: Carbon reduces die oxides of zinc, lead or iron to produce CO.

ZnO + C→Zn + CO; Fe₂O₃ + 3C →2Fe + 3CO

4. From nickel tetracarbonyl: Pure CO is obtained when nickel tetracarbonyl vapour is heated above 150°C

Carbon Monoxide’s Physical properties

• Carbon monoxide is a colourless, tasteless, odourless gas which is lighter than air. © It is slightly soluble in water.
• It is a neutral oxide.
• It is a highly poisonous gas. If only a volume of CO is present in 10,000 volumes of air, then that air is considered to be poisonous.
• Carbon monoxide molecule with a lone pair of electrons on carbon combines with Fe-atom present in haemoglobin of the blood to form a very stable complex compound named carboxyhaemoglobin.

Hb + CO →  HbCO; (Hb =Haemoglobin) As CO is almost 100 times more rigidly bonded to Fe-atom than O₂, O₂ can no longer combine with haemoglobin.

In other words, haemoglobin fails to act as an oxygen-carrier. As a consequence, the body tissues become slackened due to lack of oxygen and ultimately causing death In case of CO poisoning, the patient should immediately be taken to an open area and artificial respiration with carbogen (a mixture of oxygen and 5-10% CO₂) should be started.

Carbon Monoxide Chemical properties

1. Combustion:

Carbon monoxide is itself a combustible gas but does not support combustion. It burns in the air with a blue flame and is oxidised to C02. Because of evolution of a large amount ofheat, CO is used as fuel.

2CO + O₂ → 2CO₂ + 135.2 kcal

The two important fuels containing carbon monoxide are water gas and producer gas. Water gas contains 50% of H₂, 40% of CO, 5% of CO₂ and 5% of CH₄ and N₂ while producer gas contains 25% of CO, 4% of CO₂,70% of N₂ and traces of H₂, CH₄ and O₂

2. Reducing property:

Carbon monoxide is a powerful reducing agent. The oxidation number of carbonin CO is +2 and the highest oxidation number of carbon is +4. So, CO tends to be oxidised and behaves as a strong reducing agent. Various metal oxides are reduced by CO to the corresponding metal.

CuO + CO→Cu + CO₂ ; PbO + CO→Pb + CP

ZnO + CO→Zn + CO₂; Fe₂O₃ + 3CO→2Fe + 3CO₂

At 90°C, CO reduces iodine pentoxide (12Os) to give violet-coloured iodine. This reaction is called the Ditte reaction.

I₂O₅+ 5CO→I₂ + 5CO₂

3. Reaction with sodium hydroxide:

Being a neutral oxide CO does not react with alkali or base under ordinary conditions. But at 200°C and under high pressure, it reacts with caustic soda solution to yield sodium formate.

4. Absorption of CO:

When CO is passed through an ammoniacal or acidified cuprous chloride solution, it gets absorbed in that solution to give a white crystalline addition compound as a precipitate. CO can be separated from a gas mixture by this process.

Cu₂Cl₂ + 2CO + 4H₂O→ 2[CuClCOH₂O]↓

The addition compound evolves CO on heating.

5. Formation of addition compounds:

1. In the presence of sunlight, CO combines directly with chlorine gas to form carbonyl chloride or phosgene gas. It is a colourless poisonous gas:

2. CO reacts with sulphur vapour to produce carbonyl sulphide.

3. CO combines with many transition metals to form metal carbonyl compounds. For example, CO reacts with nickel powder at 30-40°C under ordinary pressure to form nickel tetracarbonyl. Again, at 200°C and 100 atmosphere pressure, CO reacts with freshly reduced iron to form pentacarbonyl.

Ni + 4CO →Ni(CO)₄; Fe + 5CO→ Fe(CO)₅

6. Formation of organic compounds:

Hydrogen reacts with CO at 350°C in the presence of Ni or Pt catalyst to yield methane. If the reaction is carried out at 300°C and 200 atmospheric pressure in the presence of ZnO and Cr₂O₃ catalyst, methyl alcohol is produced. The oxidation number of carbon in CO decreases from +2 to -4 in methane and to -2 in methyl alcohol.

Therefore, in these two cases, CO exhibits its oxidising property.

Identification of carbon monoxide

1. Carbon monoxide burns in air with a blue flame and the gaseous product turns lime water milky [H₂ also burns with a blue flame but in this case, steam is formed which turns white anhydrous copper sulphate blue.

2. CO is completely absorbed by the Cu₂Cl₂ solution in a cone. hydrochloric acid or ammonium hydroxide and as a result, a white crystalline addition compound is precipitated.

3. When a filter paper soaked with a solution of platinum or palladium chloride is held in CO gas, the paper turns pink-green or black due to the reduction ofthe metal salts.

PtCl₂ + CO + H₂O→ Pt (pink-green) + CO₂ + 2HCl

PdCl₂ + CO + H₂O→Pd (black)+ CO₂ + 2HCl

4. When CO gas is passed through an ammoniacal AgNO₃ solution, the solution becomes brown

5. When a dilute solution of blood shaken with CO, is subjected to spectroscopic analysis, the observed band in the spectrum indicates the presence of CO. The presence of traces ofCOin air can be detected by this experiment.

6. The presence of a very small amount of CO in the air can be detected with the help of halamite tube or colour detector tube. When air containing CO is introduced into this tube I₂O₅ present in the tube reacts with CO to liberate I₂

Because of the violet colour of evolved I₂, the colour of the tube changes and the presence of CO in the air is indicated

I₂O₂ + 5CO→ I₂  (Ditte reaction) + 5CO₂

Structure of carbon monoxide

Both the carbon and the oxygen atoms in a CO molecule are sp -hybridised. One of |> the sp -hybrid orbital of each atom is used to form a C —O cr -bond while the other sp -orbital of each contains a lone pair of electrons. The two unhybridised 2p -orbitals of each atom are involved in the formation of two pn-pn bonds. In terms of resonance, the CO molecule can be best represented as a resonance hybrid of the following two i resonance structures(I and II).

The resonance structure (I) is relatively more stable because of the fulfilment of the octet of both atoms.

Uses Of carbon monoxide:

• CO is used as fuel in the form of producer gas or water gas.
• It is used as a reducing agent in the extraction of metals.
• It is used for the preparation of pure nickel by Mond’s process.
• It is used for the
• Preparation of methanol, methane, formic acid and synthetic petrol (Fischer-Tropsch process).

Preparation of pure nickel:

Ni(CO)₄ is prepared by the reaction between impure nickel and carbon monoxide. Ni(CO)₄ is then allowed to decompose by heating to 1.50°C to get pure nickel.

2. Carbon dioxide

Carbon dioxide Laboratory preparation:

At ordinary temperature, CO₂ is prepared in the laboratory by the action of diluting HCl on calcium carbonate (CaCO₃) or marble.

CaCO₃ + 2HCl → CaCl₂ + CO₂ ↑ + H₂O

The gas is collected in the gas jar by the upward displacement of air, as it is 1.5 times heavier than air. Carbon dioxide thus produced contains a small amount of HC1 and water vapour. The gas is then passed successively through NaHCO₃ solution and cone, sulphuric acid to remove HCl vapour and water vapour respectively.

Dilute sulphuric acid cannot be used for the preparation of CO₂ from marble or limestone:

This is because sulphuric acid reacts with CaCO₃ to produce insoluble; CaSO₄ which forms a layer of CaCO₃. This insoluble layer prevents CaCO₃ from reacting with the acid and as a result, the evolution of CO₂ ceases within a very short time

CaCO₃ + H₂SO₄ →CaSO₄+ CO₂ + H₂O

On the other hand, when dilute hydrochloric acid is, used, highly soluble calcium chloride (CaCl₂) is formed. So, the reaction proceeds without any interruption

CO₂ can be prepared by the action of dilute H₂SO₄ on Na₂CO₃:

The salt, Na₂SO₄ produced soluble in water or dilute H₂SO₄

Na₂CO₃ + H₂SO₄→ Na₂ SO₄ + CO₂ + H₂O

At ordinary temperatures, CO₂ is highly soluble in water. Therefore, it is not collected by the downward displacement of water. The solubility of CO₂ in hot water is very low and hence it can be collected over hot

Carbon dioxide Other methods of preparation:

1. From carbonate salts:

Except for alkali metal carbonates, all other carbonates undergo thermal decomposition to produce CO₂ and the oxides ofthe corresponding metals.

BaCO₃ decomposes only at very high temperatures.

Calcium carbonate or limestone is thermally decomposed (1000°C) for the preparation of carbon dioxide on a commercial scale.

2.  From bicarbonate salts:

Bicarbonates of all the elements decompose on heating with the evolution of CO₂

3. From fermentation:

A large amount of C02 is obtained as a by-product during the manufacture of ethyl alcohol by fermentation of sugar

From water gas: Water gas is industrially prepared by passing steam through a bed of white-hot coke at about 100°C.

C + H₂ O →CO + H₂  When a mixture of water gas and excess of steam is passed over (Fe₂O₃+ Cr₂ O₃) catalyst heated at 400°C, CO is oxidised to CO₂

(CO + H₂) + H₂O → CO₂ + 2H₂

The gaseous product is then passed through a solution of potassium carbonate when C02 is completely absorbed and KHCO₃ is formed. H₂ and unconverted CO pass out. When the resulting KHC03 solution is boiled, CO₂ is obtained.

K₂CO₃ + CO₂ + H₂O→ 2KHCO₃

Carbon dioxide Physical properties

• Carbon dioxide is a colourless, odourless and tasteless gas having slightly acidic properties.
• CO₂ is 1.5 times heavier than air. So, this gas often accumulates in abandoned wells or pits and because of this, severe breathing problems are caused in such places.
• By the application of pressure (nearly 40 atmospheric pressure and a temperature < 40°C), CO₂ can be easily liquefied.
• When liquid CO₂ is allowed to vaporise rapidly by releasing the pressure, it further gets cooled down and freezes like ice. This is called dry ice or cardice.
• When solid carbon dioxide is allowed to evaporate at atmospheric pressure, it gets converted into the vapour state without passing through the intermediate liquid state. Therefore, unlike ordinary ice, it does not wet the surface of the substance and because of this, it is called dry ice.
• It is highly soluble in water (1.7 cm³ of CO₂ dissolves in 1 cm³ of water). The solubility increases with an increase in pressure. Aerated waters such as soda water, lemonade etc. contain CO₂ under pressure.
• When the cork of the bottle of aerated water is opened, the pressure is released and excess CO₂ escapes in the form of bubbles. Its solubilityin water, however, decreases with temperature rise.

Carbon dioxide Chemical properties

1. Combustion:

Carbon dioxide is neither combustible nor helps in combustion. When it (heavier than air) falls on a binning substance, it removes air from the surface of the substance and thereby the substance can no longer remain in contact with air. As a result, the fire is extinguished. A burning jute stick when inserted into a jar of CO₂, extinguishes.

However, when a burning Mgribbon or metallic sodium is inserted into a CO₂ jar, it continues to bum with the separation of black carbon.

During the burning of such metals, the temperature, due to the liberation of a large amount of heat, is so high that CO₂ decomposes into carbon and O₂ and it is the oxygen which helps in the burning ofthe metals.

In these reactions, CO₂ acts as an oxidising agent and itself gets reduced to carbon. These reactions prove the existence of carbon in CO₂ It is to be noted that the oxidation number of carbon in CO₂ is +4 and this is its highest state of oxidation.

Thus, there is no possibility of an increase in its oxidation number, i.e., CO₂ cannot be further oxidised. That is why CO₂ cannot exhibit any reducing property. For the same basic reason, CO₂ is not combustible [CO, on the other hand, is combustible because in this case, the oxidation number of carbon may increase from +2 to +4 ].

2.  Acidic property:

Carbon dioxide is an acidic oxide. It dissolves in water forming an unstable dibasic acid called carbonic acid (H₂CO₃). CO₂ is, therefore, regarded as the anhydride of carbonic acid.

H₂CO₃ is known only in solution and when the solution is heated, CO₂ is evolved out The solution turns blue litmus red but it cannot change the colour of methyl orange. H₂CO₃ forms two types of salts, bicarbonates (HCO₃ ) and carbonates (CO₃^2-). Being an acidic oxide, CO₂ combines directly with strongly basic oxides such as CaO, Na₂O etc. to form their corresponding salts.

CaO + CO₂→ CaCO₂; Na₂O + CO₂ →Na₂CO₃

Reaction with alkali:

When CO₂ is passed through a strong alkaline solution of NaOH, a carbonate salt is first formed. If the passage of CO₂ is continued for a long time, white crystals of sparingly soluble sodium bicarbonate are precipitated. The bicarbonate salt decomposes on heating to form carbonate salt, CO₂and water.

2NaOH + CO₂→ Na₂ CO₃+ H₂O

Na₂CO₃ + CO₂+ H₂O →2NaHCO₃

 Rection with lime water:

When CO₂ is passed through lime water, the solution becomes milky due to the formation of white insoluble calcium carbonate. However, when an excess of CO₂ gas is passed through this milky solution, its milkiness disappears as insoluble calcium carbonate gets converted into soluble calcium bicarbonate

Ca(OH)₂ + CO₂ →CaCO₃.↓ (white) +H₂O

CaCO₃ + CO₂ + H₂O→Ca(HCO₃)O₂ (soluble)

On heating, calcium bicarbonate decomposes to form calcium carbonate, CO₂ and water and as a result, the clear solution becomes milky again.

Ca(HCO₃)₂→CaCO₃↓ + CO₂ + H₂O

3.  Manufacture of sodium carbonate:

When CO₂ gas is passed through a concentrated solution of sodium chloride (brine) saturated with ammonia at 30-40°C, white crystals of sodium bicarbonate are precipitated. The reaction occurs in two stages

NH3 + CO₂ + H₂O ⇌   NH₄HCO₃

NH₄HCO₃ + NaCI→NaHCO₃↓+ NH₄Cl

Sodium carbonate is prepared by thermal decomposition of sodium bicarbonate. The Solvay process for the manufacture of sodium carbonate is based on this reaction.

4. Production of ammonium sulphate:

This is carried out by passing CO, and NH₃ gases through a slurry of powdered gypsum (CaSO₄,  2H₂O) in water. At first, NH₃ and CO₂ react together in the presence of water to form ammonium carbonate. It then reacts with calcium sulphate (gypsum) to form calcium carbonate and ammonium sulphate by double decomposition.

2NH₃ + CO₂ + H₂O ⇌ (NH₄)₂CO₃

CaSO₄ + (NH₄)₂CO₃→ CaCO₃ ↓+ (NH₄)₂SO₄

The nitrogenous fertiliser ammonium sulphate is manufactured by using this reaction. In this process, (NH₄)₂SO₄ is produced without using H₂SO₄.

5. Production of urea:

At 200-210°C and 150 atm pressure, C02 reacts with ammonia to produce urea.

CO₂ + 2NH₃ ⇌   NH₄COONH₂ (Ammonium carbamate) ⇌   CO(NH₂)₂  (Urea) + HO₂ 

The important fertiliser, urea is manufactured on a large scale by using this reaction.

6. Photosynthesis:

Plants absorb atmospheric carbon dioxide. In the presence of chlorophyll and sunlight, the absorbed CO₂ combines with water (absorbed from the soil) to form glucose, water and oxygen. This process is called photosynthesis. In this process, CO₂ is reduced to carbohydrates by water

7. Reduction of CO₂: When CO₂ is passed over heated C, Fe, Zn etc., it is reduced to CO

Identification of carbon dioxide

• It extinguishes a burning stick.
• Lime water becomes turbid when CO₂ is passed through it. When excess of CO₂ is passed through it, the turbidity disappears but when that clear solution is boiled, the turbidity reappears.
• N₂ gas also extinguishes burning sticks but it does not turn the water milky. Again, SO₂ gas also turns lime water milky but unlike CO₂ it reacts with an acidified solution potassium dichromate and changes the colour of the solution from orange to green

Uses Of carbon dioxide

• CO₂ is used in the manufacture of sodium carbonate by the Solvay process and also for the manufacture of fertilisers such as urea, ammonium sulphate etc.
• CO₂  is used in fire extinguishers.
• It finds extensive use in the preparation of aerated waters such as soda water, lemonade etc. and baking powder.
• Solid carbon dioxide i.e., dry ice is used as a refrigerant under the commercial name drikold. Dry ice is also used for making cold baths in the laboratory by mixing it with some volatile organic solvents.
• It is extensively used as a coolant for preserving perishable articles in the food industry, for curing local burns and for surgical operations of sores.

Supercritical CO₂ :

• Supercritical CO₂ is used as a. solvent to extract organic compounds from their natural sources, for example, caffeine from coffee beans, perfumes from flowers etc.
• It is used under the name carbogen (a mixture of 95% O₂ and 5% CO₂) for the artificial respiration of patients suffering from pneumonia and affected by poisonous gases (CO poisoning).
• Liquid CO₂ is used as a substitute for chlorofluorocarbons in aerosol propellants.

Fire extinguisher

It is a specially designed metallic pressure vessel having a nozzle at one end. A glass bottle containing dilute sulphuric acid is placed inside it and the remaining portion of the vessel is filled with concentrated solution of sodium bicarbonate. When required, the glass bottle can be broken by pressing a knob fitted with the vessel at the other end.

When the glass bottle is broken, the add comes in contact with sodium bicarbonate solution and reacts to yield copious CO₂ gas. The gas, ejected under high pressure through the nozzle, falls on the burning substance and as a result, the fire gets extinguished

Na₂CO₃ + H₂SO₄ →Na₂SO₄ + CO₂ ↑+ H₂O

Baking powder

Baking powder which is used Fire extinguisher in the preparation of bread consists of a dry mixture of potassium hydrogen tartrate, NaHCO₃, tartaric acid and -starch. When this tithe comes in contact with water present in the bread, a chemical reaction leading to the formation of CO₂ occurs.

The resulting CO₂ gas evolved in the form of bubbles making the bread porous and soft. Moreover, NaHCO₃ and tartaric acid also produce CO₂ on thermal decomposition

Structure of carbon dioxide:

In a CO₂ molecule, the carbon atom is sp -hybridised whereas the oxygen atoms are sp² – hybridised. Carbon forms two σ -bonds and two pπ- bonds with two oxygen atoms. The shape of the carbon dioxide molecule is, therefore, linear. The molecule is symmetrical (the two bond moments cancel each other) and hence, it is non-polar. The C —O bond length is 1.15Å. CO₂ can be represented as a resonance hybrid of the following three structures:

Structure Of Atom Introduction

The Atomic theory of matter was first proposed by Sir John Daltonn (an English scientist) in 1808 his theory, called Dalton’s atomic theory was a landmark in the history of chemistry.

According to this theory, the atom is the smallest, indivisible, discrete particle of matter, which takes part in chemical reactions.

However, the research done by eminent scientists like J.J Thomson Goldstein, Rutherford, Chadwick, Bohr, and others towards the End Of the 19th Century and at the beginning of the 20th century has conclusively proved that atoms were no longer the smallest in divisible practice.

At present Scientists have identified about 35 different subatomic particles that may be divided under three heads which is shown in the adjacent table.

The three subatomic particles namely electrons, protons, and neutrons are the main constituents of an atom and are regarded as the fundamental particles.

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Atomic Number, Mass Number, Isotope, Isobar And Isotone

Atomic number

The atomic number indicates the total number of unit positive charges present in the nucleus of an atom.

As each proton carries a unit positive charge, the total number of protons present in the nucleus of an atom represents the atomic number of the corresponding element.

As an atom is electrically neutral, the total quantity of positive charge must be equal to that of negative charge carried by the electrons. So the number of protons is equal to the number of electrons.

Thus, the atomic number of the element = total number of protons in the nucleus ofthe atom = total number of electrons in the neutral atom

The atomic number determines the fundamental property of an element. No two elements can have the same atomic number.

Any difference in the number of electrons produces ions without altering the constitution ofthe nucleus.

In the modern version of the periodic table (both short and long), elements are arranged in the increasing order of their atomic numbers.

Mass number

Since the electrons have negligible masses, the mass of an atom is determined by the number of protons and neutrons present in the nucleus. The sum of the number of protons and neutrons present in the nucleus of an atom is called the mass number of that element.

Mass number (A) = No. of protons (Z) + No. of neutrons (N)

Example: The nucleus of a fluorine atom contains 9 protons and 10 neutrons. Therefore, the mass number of fluorine =9+10=19.

Generally, the atomic number (Z) and mass number (A) of an element are represented along with the symbol (X) ofthe element as shown below.

Here mass number A and atomic number Z are inserted as superscripts (to the left or right side) and subscripts (to the left side) of the symbol of the element respectively.

Thus, the symbol \({ }_{17}^{35} \mathrm{Cl} \text { or }{ }_{17} \mathrm{Cl}^{35}\) denotes a chlorine atom with mass number 35 and atomic number 17.

Isotope

According to Dalton’s atomic theory, all atoms of an element are identical in all respects.

However British chemist F. Soddy pointed out for the first time one limitation of this theory when he observed that the same element may have atoms with different atomic masses.

This is because atoms of the same element always contain the same number of protons but they may have different numbers of neutrons, which lead to differences in mass numbers. This phenomenon is known as isotopy.

Isotope Definition: Atoms Of the same element having the same atomic number but different mass numbers are called isotopes.

Example: Hydrogen has three isotopes, protium (H), deuterium (D), and tritium (T) with mass numbers 1, 2, and 3 respectively.

All three isotopes have the same atomic number 1, and they are represented as \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H}\) respectively.

Isotopes of other elements (some examples are given below) have no such special names; they are represented by simply indicating the values of mass number and atomic number on their symbol.

Thus isotopes ofchlorine are represented as and \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl} \text {. }\)

Characteristics: The characteristics of isotopes are—

The chemical properties of the isotopes of an element are the same. This is because the chemical properties of an element are determined by the number of electrons present in its atom, which in turn is equal to the number of protons present in the nucleus (and hence its atomic number).

However, the different isotopes of an element react at different rates. The lighter isotopes react faster and the reactions involving the heavier isotopes occur slowly.

The physical properties of the isotopes e.g., density, rate of diffusion, etc., which depend on the atomic masses are different.

All the isotopes of an element occupy the same position in the periodic table, although they have different atomic masses. The Greek word isotopes means the place (/so = same, topes = place).

Isotopes may be both radioactive and non-radioactive. The emission of one a -particle and two beta -particles from a radioactive element produce an element that occupies the same place as that of the parent element in the periodic table, although the mass number of the end (daughter) element is 4 units less than that of the parent element So, it will be an isotope of the parent element.

Example: \({ }_{92} \mathrm{U}^{238}\)(Uranium-I) and \({ }_{92} U^{234}\) (Uranlum-Il) are isotopes of the element uranium.

⇒ \({ }_{92}^{238} \mathrm{U} \stackrel{-\alpha}{\longrightarrow} \quad{ }_{90}^{234} \mathrm{Th} \quad \stackrel{-\beta}{\longrightarrow}{ }_{91}^{234} \mathrm{~Pa} \stackrel{-\beta}{\longrightarrow}{ }_{92}^{234} \mathrm{U}\)

Uranium-1 Uranium-9 Uranium-X2 Uranium-2

Different isotopes of an element may have different radioactive properties. Thus \({ }_6^{14} \mathrm{C}\) is not radioactive, while \({ }_6^{14} \mathrm{C}\) exhibits radioactivity.

Elements which do not have natural isotopes: Be-9, F-19, Na- 23, Al-27, P-31, Sc-45, Mn-55, Co-59, As-75, Y-89, Nb-93, Rh-103, 1-127, Cs-133, Pr-141, Tb-159, Ho-165, Tm-169, Au-197, Bi-209. The elements Sn and Xe have 10 and 9 isotopes respectively.

Isobar

Isobar Definition: Atoms having the same mass number but different atomic numbers are called isobars.

Example: \({ }_{18}^{40} \mathrm{Ar} \text { and }{ }_{20}^{40} \mathrm{Ca}\) are isobars. Here, Ar and Ca have the respective atomic numbers, 18 and 20.

Therefore, the number of their protons is 18 and 20 respectively, but the total number of protons and neutrons in both cases is 40. Number of neutrons in \({ }_{18}^{40} \mathrm{Ar}=(40-18)=22\) number of neutrons in \({ }_{20}^{40} \mathrm{Ca}=(40-20)=20\)

Although isobars have the same mass number, their atomic numbers are different. Thus isobars are atoms of different elements displaying different physical and chemical properties. They occupy different positions in the periodic table. Other examples of isobars are

⇒ \({ }_1^3 \mathrm{H},{ }_2^3 \mathrm{He}\)

⇒ \({ }_6^{14} \mathrm{C},{ }_7^{14} \mathrm{~N}\)

⇒ \({ }_{51}^{123} \mathrm{Sb},{ }_{52}^{123} \mathrm{Te}\)

⇒ \({ }_{88}^{228} \mathrm{Ra},{ }_{89}^{228} \mathrm{Ac},{ }_{90}^{228} \mathrm{Th}\)

⇒ \({ }_{88}^{228} \mathrm{Ra}^{228}{ }_{89}^{228} \mathrm{Ac},{ }_{90}^{228} \mathrm{Th}\)

⇒ \({ }_{82}^{210} \mathrm{~Pb},{ }_{83}^{210} \mathrm{Bi},{ }_{84}^{210} \mathrm{Po}\)

Isotone

Isotone Definition: Atoms having the same number of neutrons but a different number of protons are called isotones.

Consequently, isotones possess different mass numbers.

Example: \({ }_1^3 \mathrm{H} \text { and }{ }_2^4 \mathrm{He}\) The former contains 1 proton and 2 neutrons, while the latter contains 2 protons and 2 neutrons.

Isotopes are atoms of different elements having the same number of neutrons but different atomic numbers (number of protons).

They occupy different positions in the periodic table and have different physical and chemical properties.

Some Other examples are-

⇒ \({ }_1^3 \mathrm{H},{ }_2^4 \mathrm{He}\)

⇒ \({ }_{14}^{30} \mathrm{Si},{ }_{15}^{31} \mathrm{P},{ }_{16}^{32} \mathrm{~S}\)

⇒ \({ }_{33}^{77} \mathrm{As},{ }_{34}^{78} \mathrm{Se}\)

⇒ \({ }_6^{14} \mathrm{C},{ }_7^{15} \mathrm{~N},{ }_8^{16} \mathrm{O}\)

Comparative Study Of Isotope, Isobar And Isotone

A nuclear isomer is a diaper and isostere Amtoic (nuclides) having the same atomic number and mass number but different radioactive properties are called nuclear isomers and this phenomenon is known as nuclear isomerism.

The nuclei of a radioactive element which exist in different energy states are nuclear isomers.

Examples:

1. U-X2 (t_l/2 = 1.14 min) and U-Z (t_l/2 = 6.7 hr.)
2. ⁶⁹Zn (t_l/2 = 13.8 hr) and ⁶⁹Zn (t_l/2 = 57 min)
3. ⁸⁰Br (t_l/2 = 4.4 hr) and ⁸⁰Br (t_l/2 = 18 min)

Isodiapher Atoms m which the difference between the number of protons is the same are called isodiapherr, An atomic nuclide and the atom produced from It due to the emission of an a-particle are isodiaphers.

Example: \({ }_{92}^{238} \mathrm{U} \stackrel{-\alpha}{\longrightarrow}{ }_{90}^{234} \mathrm{Th}\)

As the difference in the number of neutrons and protons In the two atoms are the same, they are isodiaphers.

Isostere Atom molecules or ions of similar sires containing the same number of atoms 3rd valence electrons are called isosteres.

Example:

Isosleric species involving cations and neutral atoms: Ne, Na⁺, Mg2⁺, Al3⁺.

Isosteric species involving anions and neutral atoms: N₃^–, O₂^–, F^–, Ne.

Isosteric species involving cations, anions and mcutral atoms: O₂^– , F^– , Ne, Na⁺, Mg²⁺

Isosteric species involving neutral molecules, cations and anions: 1. CN^–, CO, NO⁺, N₂ 2. CO₂, N₂O, N₂ (azide), OCN^– (cyanate), SCN^– (thiocyanate).

In general isosteric molecules and Ions have the same shape. In both NO₂ and CO₂, number of atoms =3 and number of valence electrons = 16. So, they are Isosteres.

Quantum Number

Quantum Number Definition: A set of four numbers that provide complete information about any electron in an atom are known as quantum numbers.

Quantum Number Classification: The four quantum numbers are—

1. Principal Quantum Number (N)
2. Azimuthal Or Subsidiary quantum number (l)
3. Magnetic quantum number (m or m1)
4. Spin quantum number (5 or mg ).
5. To specify an electron in an atom, the following four quantum numbers should be mentioned.
6. Principal quantum number [n]

Quantum Number Origin:

1. From Bohr’s postulates, it is known that each electronic orbit surrounding the nucleus in an atom represents an energy level.
2. The average energy of the electrons revolving in a particular orbit is fixed. So, these orbits are called principal energy levels or principal quantum levels.
3. Depending on their distance from the nucleus, these orbits or principal energy levels are designated by the numbers 1,2,3, 4… etc. These numbers 1,2,3,4… etc. are called principal quantum numbers.

Quantum Number Designation: The principal quantum number is denoted by the letter ‘n ’. For AT-shell n = 1, for L -shell n = 2, for Mshell n = 3 and so on.

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Information obtained:

1. The higher the value of n, the greater the distance of the orbit from the nucleus, and hence, the greater the size ofthe orbit. Thus, r1<r2<r3< r4< …
2. The higher the value of ‘ n,’ the greater will be the electronic energy associated with the orbit.
3. Thus, El<E2<E3<E4<………..
4. A maximum number of electrons that can be accommodated in a principal quantum level n is given by the formula 2n2.
5. Limitations of 2β2 The maximum number of electrons in any orbit can never be more than 32 even if the value of n exceeds 4.
6. The outermost electronic shell does not contain more than 8 electrons.
7. The penultimate shell (i,e., the shell just preceding the outermost shell) does not contain more than 18 electrons.

Azimuthal or subsidiary quantum number

Azimuthal or subsidiary quantum number Origin: A spectrograph with high resolving [/]power has revealed that each bright line in the spectrum of atomic hydrogen consists of some closely spaced finer lines.

This fact suggests that each orbit or energy level in an atom is composed of subshells. Electrons occupying these subshells within the same -shell, exhibit slight differences in energy.

In order to explain the formation of the fine structure of spectral lines, Sommerfeld proposed
the existence of elliptical orbits, besides Bohr’s circular
orbits.

To specify the shape of the elliptical orbit, another supplementary quantum number is necessary.

This supplementary quantum number which indicates the captivity of the electronic orbit is called azimuthal or subsidiary quantum number denoted by the letter.

If the principal quantum number of any orbit is n, then the total number of subshells incorporated in that orbit will also be n.

Magnitude:

1. As per quantum mechanical calculations, the angular momentum of a moving electron in an elliptical path is given by, L = Jl(l+ 1) X.
2. This is often called orbital angular momentum.
3. The value of l determines the shape of the path. So, with the help of the principal quantum number and azimuthal quantum number, a precise idea about the size and shape of the electronic path can be obtained.
4. If n stands for the principal quantum number of an electronic orbit, the values of l will be from to (n- 1) i.e., with respect to the value of principal quantum number n, the azimuthal quantum number / may assume n number of different values including zero, e.g., for n = 4, 1=0, 1, 2 and 3.
5. To indicate the subshells within a shell, spectroscopic symbols are used instead ofthe numbers 0, 1, 2, 3 etc.
6. The symbols s, p, d,f, etc., (spectroscopic coinage) are merely the first letters ofthe words sharp, principal, diffuse, and fundamental, used extensively in spectral analysis.
7. To express the position of an electron in the atom, the principal quantum number should be written first followed by the symbol of the azimuthal quantum number to its right side, e.g., the subshells included in K, L, M, and N-shells are represented as

• The ratio of the major axis to the minor axis of an elliptical path is given by = (/ + 1)/n .
• An elliptical path for which l = (rc- 1), becomes circular e.g., in the case of 4-orbit if 1 = 3, then that orbit becomes circular. The greater the difference between the values of l and n, the larger the ellipticity of that path.
• The penetrating power and screening effect of an elliptical orbit increases on increasing the ellipticity of the orbit.
• So the penetrating and screening powers of different subshells within the same shell follow the sequence: s> p> d>f.
• Due to the difference in the internal energies of the subshells [s, p, d, f, etc.), the electrons moving in those subshells also possess different energies. Energy associated with the subshells in a particular orbit increases in the following order: s <p<d<f.

Magnetic quantum number (m or mt)

Origin:

1. Zeeman in 1896 observed that each fine line in atomic spectra splits further into finer lines in the presence of the highly powerful magnetic field.
2. In the absence of a magnetic field, such finer splitting i.e., hyperfine splitting disappears. This phenomenon is called the Zeeman effect. To explain the Zeeman effect, a third type of quantum number, known as a magnetic quantum number was introduced.

Discussion:

1. Due to the angular motion of electrons around the nucleus, a magnetic field is produced, which interacts with the external magnetic field.
2. As a result subshells of definite energy split into three-dimensional spatial regions called orbitals.
3. Magnetic quantum number (MI) signifies the orientation of the orbitals in space in which the electron exists.
4. The value of m depends on the azimuthal quantum number l.
5. For a certain value of l, m has an o total of (2Z +1 ) different values. These values may be any whole number starting from -Z to +1 (including zero).
6. For s- subshell, l = 1 and m – 1. This subshell, l = 0 and m – 0. This orbital (i.e., s-orbital). Z = 1 denotes p -subshell consisting of three orbitals which are directed along three axes.
7. These are marked as px, py, and pz orbitals which have the respective values of m = -1, 0, and + 1 . Similarly, d and /-subshells contain 5 and 7 orbitals respectively.
8. The negative values of the magnetic quantum number signify that these orbitals are inclined in the direction opposite to the magnetic field and the positive values indicate that these orbitals are inclined in the direction ofthe magnetic field.
9. shows the different directions of the d -d-subshell (Z = 2) in the magnetic field.

Orientation of different orbitals of ZV-shell (n = 4] under the influence of magnetic field.

Values of magnetic quantum number (m] for different values of azimuthal quantum number [l]

Spin quantum number [s or ms]

Uhlenbeck and Goudsmit introduced a fourth quantum number called the spin quantum number.

This is because the other three quantum numbers were not able to give sufficient explanation to the hyperfine structure of the atomic spectra.

% Just like the earth, an electron while moving around the nucleus also spins about its own axis either in a clockwise or in an anti-clockwise direction,

Each type of spin can give rise to characteristic spectral lines with the formation of a hyperfine spectrum in the spectral series.

The spin quantum number denoted by the symbol ‘s’ expresses two opposite types of spinning motions of each electron.

The spin quantum number ‘s’ can have only two values, \(+\frac{1}{2} \text { and }-\frac{1}{2}\) The positive and negative signs represent two opposite directions of spinning motion of any spinning motion of any spinning motion of electron are very often represented by two arrows pointing in opposite directions,| and.

Q A spinning electron behaves like a tiny magnet with a definite magnetic moment. The angular mentum associated with the spinning electron is given by the mathematical expression.

\(s=\sqrt{s(s+1)} \times \frac{h}{2 \pi}\)

Spin Quantum number (s) signifies the mode of Electron Spin (Clockwise or Anti-clockwise).

Heisenberg’s Uncertainty Principle

Werner Heisenberg introduced his uncertainty principle in 1927 which is a direct consequence of the dual nature of electrons.

Heisenberg’s uncertainly principle: it is impossible to measure simultaneously both the position and the momentum of a sub-atomic particle like an electron, accurately, at any instant of time.

Explanation: If at a particular moment, the uncertainty in position and the uncertainty in the momentum of a sub-atomic particle be Ax and Ap respectively, then it can be
shown that the product of these two uncertainties must be at least equal to or greater than \(\frac{h}{4 \pi}.\)

Mathematically it can be expressed as \(\Delta x \times \Delta p \frac{h}{4 \pi}\) [where h = Planck’s constant]

\(\text { or, } \Delta x \times \Delta(m v) \frac{h}{4 \pi} \quad \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi}\)

Heisenberg’s uncertainty principle

Werner Heisenberg introduced his uncertainty principle in 1927 which is a direct consequence of the dual nature of electrons.

Heisenberg’s uncertainly principle: it is impossible to measure simultaneously both the position and the momentum of a sub-atomic particle like an electron, accurately, at any instant of time.

Read and Learn More CBSE Class 11 Chemistry Notes

Explanation: If at a particular moment, the uncertainty in position and the uncertainty in the momentum of a sub-atomic particle are Ax and Ap respectively, then it can be shown that the product ofthese two uncertainties must be at least equal to or greater than \(\frac{h}{4 \pi}.\)

Mathematically it can be expressed as, \(\Delta x \times \Delta p \frac{h}{4 \pi}\) [where h = Plancks constant ….[1]

⇒ \(\text { or, } \Delta x \times \Delta(m v) \frac{h}{4 \pi} \quad \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi}\)

⇒\(\text { or, } \Delta x \times \Delta v \frac{h}{4 \pi m}\)

[since m is constant]

Suppose we are going to measure simultaneously both the position and momentum of an electron in an atom. If an attempt is made to measure the position of the electron with high accuracy, then the measured value of the momentum will be less accurate and vice versa.

It should be realized that the uncertainty principle is not due to any limitation of the measuring instrument but is the consequence of the dual nature of moving particles and electromagnetic radiation (light).

Heisenberg’s uncertainty principle rules out the concept of a fixed circular path with definite position and momentum electrons in an atom as proposed by Bohr.

It should be remembered that the uncertainty principle is applicable to the position and momentum of a particle along the same axis.

Thus, if Ax represents the uncertainty in position along the x-axis then Ap must be the uncertainty in momentum along the x-axis (let it be represented as Ap ).

Thus \(\Delta x \times \Delta p_x \frac{h}{4 \pi}\)

Similarly, \(\Delta y \times \Delta p_y \frac{h}{4 \pi} \text { and } \Delta z \times \Delta p_z \frac{h}{4 \pi}\)

The uncertainty principle can also be applied to the conjugate pair, energy, and time. If Af represents the uncertainty in measuring the lifetime of a state and AE represents the uncertainty in measuring its energy in that state, then according to uncertainty principle \(\Delta E \times \Delta t \frac{n}{4 \pi}.\)

Uncertainty Principle For Macrocophic Objects: Theoretically, the uncertainty principle holds good for objects of all sizes, but in reality, it has no significance for macroscopic objects (big objects).

To realize this, let us consider a particle of mass of 1 mg, and for this, the approximate value ofthe product of Ax and Aw is given by, \(\begin{aligned}
\Delta x \cdot \Delta v \approx \frac{h}{4 \pi m} & =\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{4 \times 3.14 \times 10^{-6} \mathrm{~kg}} \\
& =0.53 \times 10^{-28} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}
\end{aligned}\)

Thus the product of Ax and Av is extremely small. In other words, for objects of ordinary size, the uncertainties in position and momentum are very small as compared to the size of the object and the momentum of the object respectively.

Hence from the practical point of view, the values of these uncertainties may be taken as zero.

This means that the position and velocity of large objects can be measured almost accurately at any instant in time.

since in everyday life, we come across big objects only, it can be concluded that Heisenberg’s uncertainty principle has no significance in everyday life.

For a subatomic particle such as an electron, we have

\(\begin{aligned}
\Delta x \cdot \Delta v \approx \frac{h}{4 \pi m} & =\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{4 \times 3.14 \times 10^{-6} \mathrm{~kg}} \\
& =0.58 \times 10^{-4} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}
\end{aligned}\)

This value is quite large so neither the uncertainty in position nor the uncertainty in velocity of the sub-atomic particle can be neglected.

For example, if the uncertainty in the position of the electron is 10-4m, uncertainty in its velocity will be ~ 0.58 m.s-1, which is quite significant.

It is for this reason, that Bohr’s concept of a fixed circular path with a definite velocity needs modification.

Electrons cannot rollers In Ihn nucleus: The diameter of the Nucleus is of the order 10-15m. For the electron to reside within the nucleus, the maximum uncertainty in
its position should be 10-15m,i.e., Ax = 10-15m.

⇒ \(\begin{aligned}
\Delta x \times \Delta p &\frac{h}{4 \pi} \quad \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi} \text { or, } \Delta v \frac{h}{4 \pi m \times \Delta x} \\
\text { or, } \Delta v &\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{4 \times 3.14 \times 9.1 \times 10^{-31} \mathrm{~kg} \times 10^{-15} \mathrm{~m}} \\
& =5.7 \times 10^{10} \mathrm{~m} \cdot \mathrm{s}^{-1}
\end{aligned}\)

This value is much higher than the velocity of light (3 x 108m-s_1) which is not possible. Thus, an electron cannot reside within the nucleus.

Reasons for the failure of Bohr’s atomic model:

In Bohr’s model of an atom, the electron is regarded as a charged particle moving with definite velocity in a circular orbit of a definite radius.

This is not supported by Heisenberg’s uncertainty principle according to which it is impossible to determine both the velocity and the position of an electron simultaneously with certainty.

Furthermore, Bohr’s model does not take into consideration the concept of the dual nature of an electron (which is a sub-atomic particle). Due to such inherent weakness, Bohr’s atomic model lost its significance

Class 11 Chemistry First Law Of Thermodynamics

Question 1. Calculate die change in internal energy when heat released by the die system Is 200 J and work done on the die system is 120J heat absorbed and work done by the system arc 200J and 120J respectively 0heat released and work done by the system arc 200 J and 120 J respectively.
Answer: As per given data, q = -200 J and w = +120 J

∴ Change in internal energy,

AU = q + w =(- 200 + 120) J =-80 J

As per given data, q = +200 J and w = -120

∴ Change in internal energy,

AU = q+w=(200-120) J = + 80 J

As per given data, q = -200 J and w = -120 J

∴ Change in internal energy,

AU = q + w=(- 200- 120) J = -320 J

Question 2. In a process, the system performs 142 J of work and the internal energy of the system Increases by 879 J. Predict the direction of heat flow and also calculate the quantity of heat transferred.
Answer: Given, w = -142 J and AU = + 879 J

∴ AU = q+w or, 879 = q- 142

∴ q = + 1021 J

The positive value of q indicates that the system absorbs heat from the surroundings i.e., heat flows from the surroundings to the system. The amount of heat transferred in the process is 1 021 J.

Question 3. A system undergoes n process in which It gives up 300J of heat and Its internal energy decreases by 300J. Of the system and its surroundings, which one docs work in this process?
Answer: Given, A U = -300J and q = -900 J

∴ AU = q + w or, -300 = – 900 + w or, w = + 600 J

w is (+)ve. This means work is done on the system by the surroundings and the amount of work done is 600 J

Question 4. The initial volume of a gas, confined in a cylinder fitted with a piston is U.2L. The final volume of the gas becomes 33.6L after expansion against a constant external pressure of 2atm. During expansion if the gas absorbs 1kg of heat from the surroundings then what will be the change in the internal energy of that gas?
Answer: Work done, w = -Pex(V2-V t)

Given: Pgx = 2 atm; Vj = 11.2 L and V2 = 33.6 L

∴ w =-2(33.6- 11.2) = -44.8 L-atm = -44.8 X 101.3
= -4538.24 J [v 1 L-atm = 101.3 J]

So, the amount of work done by the gas is 4538.24 J.

Again, the heat (q) absorbed by the gas = 1 of = 1000J Therefore, the change in internal energy of the gas, AU = q+w =(1000-4538.24) J = -3538.24 J

Therefore, the decrease in internal energy is 3538.24 J

Question 5. An ideal gas is expanded from 1L to 6L in a closed vessel at 2 atm pressure by applying heat at a fixed temperature. Calculate the work done and heat absorbed by the gas. [Given: lLatm = 24.22 cal]
Answer: Work done, w = -PexAV = -Pex(V2- V1)

Given: Pgx = 2 atm; Vy = 1 L and V2 = 6 L

w = -2(6-1) = -10 L-atm =-242.2 cal

From the first law of thermodynamics, AU = q + w.

Since the expansion is carried out isothermally and the system is an ideal gas, the change in internal energy in the process will be equal to zero. Thus, AU = 0.

0 = q + w or, q = -w = + 242.2cal

Therefore, the amount of heat absorbed bythe gas is 242.2 cal.

Question 6. A cylinder fitted with a piston contains an ideal gas with a volume of 21L. The gas is compressed isothermally to l/3rd of its initial volume under a constant external pressure of 3 atm. Calculate q, w, and AU.
Answer: Work done, w = -Pex(V2- Vx) [Vy > V2]
Given: Pgx = 3 atm; Vy = 21L and V2 ,\(=21 \times \frac{1}{3}=7 \mathrm{~L}\)

∴ w = -3(7-21) = 42 L. atm

= 42 X 101.3 J = 4254.6 J p[since 1L.atm =101.3J]

In an isothermal process of an ideal gas, the internal energy ofthe gas does not change i.e., AU = 0

From the first law of thermodynamics, AU = q + w

∴ 0 = q + 4254.6 J or, q = -4254.6 J

Thus, in the process g=- 4254.6 J, w=+ 4254.6 J, AU=0.

Question 1. The number of isomers possible for disubstituted borazine, B₃N₃H₄X₂ is-

1. 3
2. 4
3. 5
4. 6

Answer: 2. 4

Question 2. Pentaborane-9 (B₅H₉) is an example of—

1. Arachno-borane
2. Pseudo-borane
3. Nido-borane
4. Closo-borane

Answer: 1. Arachno-borane

Question 3. Si when reacts with A forms B. A & B respectively are

1. HF, H₂SiF₄
2. HF, H₂SiF₆
3. HCl, H₂SiCl₆
4. HI, H₂SiI₆

Answer: 2. HF, H₂SiF₆

Question 4. Boric acid is a

1. Monobasic and weak Lewis acid
2. Monobasic and weak Bronsted acid
3. Monobasic and strong Lewis acid
4. Tribasic and weak Bronsted acid

Answer: 1. Monobasic and weak Lewis acid

Question 5. Which ofthe following does not exist in a free state

1. BF₃
2. BCl₃
3. BBr₃
4. BH₃

Answer: 4. BH₃

Question 6. The correct order of decreasing Lewis acid character is

1. BCl₃ > AlCl₃ > GaCl₃ > InCl₃
2. AlCl₃ > BCl₃ > InCl₃ > GaCl₃
3. AlCl₃ > GaCl₃ > BCl₃ > InCl₃
4. InCl₃ > GaCl₃ > AlCl₃ > BCl₃

Answer: 1. BCl₃ > AlCl₃ > GaCl₃ > InCl₃

Question 7. Which of the following is present in the chain structure of silicate

1. (Si₃O^2-₅)_n
2. (Si₃O^2-₃)_n
3. (SiO^4-₄)
4. Si₂O^6-₇

Answer: 2. (Si₃O^2-₃)_n

Question 8. A metal, M forms chlorides in +2 and +4 oxidation states. Which ofthe following statements about these chlorides is correct

1. MCl₂ is more volatile than MCl₄
2. Ml₂ is more ionic than MCl₄
3. MCl₂ is more soluble in any. ethanol than MCl₄
4. MCl₂ is more easily hydrolysed than MCl₄

Answer: 2. Ml₂ is more ionic than MCl₄

Question 9. The number of O-atoms that are shared per Si04 tetrahedra in silicate anion of beryl is

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

Answer: 3. 2

Question 10. Which of the following on hydrolysis produces crosslinked silicone polymer

1. R₄Si
2. RSiCl₃
3. R₂SiCl₂
4. R₃SiCl

Answer: 2. RSiCl₃

Question 11. The antidote of poisoning caused by CO is

1. Carborundum
2. Carbogen
3. Carbonic acid
4. Pure oxygen

Answer: 2. Carbogen

Question 12. Carbon suboxide in reaction with water produces

1. Oxalic acid
2. Formic acid
3. Lactic acid
4. Malonic acid

Answer: 4.  Malonic acid

Question 13. The volume of which liquid metal increases on solidification

1. Ga
2. Al
3. Zn
4. Cu

Answer: 1. Ga

Question 14. Which ofthe following reacts only with alkali

1. B₂O₃
2. Al₂O₃
3. Ga₂O₃
4. ln₂O₃

Answer: 1. B₂O₃

Question 15. Which is the strongest Lewis acid

1. BF₃
2. BCl₃
3. BBr₃
4. BI₃

Answer: 4. BI₃

Question 16. The atomic radius of Ga is slightly less than that of Al. The reason is

1. Weaker shielding effect of -electrons of Ga
2. Stronger shielding effect of s -electrons of Ga
3. Weaker shielding effect of d -electrons of Ga
4. Stronger shielding effect of d -electrons of Ga

Answer: 3. Weaker shielding effect of d -electrons of Ga

Question 17. Carbon does not form complexes, because

1. Vacant d – orbitals are absent in it
2. It is not a metal
3. Its atomic radius is small
4. It is neutral

Answer: 1. Vacant d -d-orbitals are absent in it

Question 18. Supercritical CO₂ is used as

1. Dry ice
2. Fire extinguisher
3. A solvent for the extraction of organic compounds from natural sources
4. The inert solvent in various reactions

Answer: 3. Asolvent for the extraction of organic compounds from natural sources

Question 19. The stability of the +1 oxidation state increases in the sequence—

1. Al < Ga < In < Tl
2. Tl<In<Ga<Al
3. In < Tl < Ga < Al
4. Ga<In<Al<Tl

Answer: 1. Al < Ga < In < Tl

Question 20. Which ofthe following is acidic—

1. B₂O₃
2. Al₂O₃
3. Ga₂O₃
4. ln₂O₃

Answer: 1.B₂O₃

Question 21. The correct order of first ionisation enthalpy for Gr-13 elements is

1. B > Al > Ga > In > Tl
2. B < Al < Ga < In < Tl
3. B < Al > Ga < In > Tl
4. B > Al < Ga > In < Tl

Answer: 4.  B > Al < Ga > In < Tl

Question 22. Which of the following elements is not likely to be the central atom in MF^3-₆

1. B
2. Al
3. Ga
4. In

Answer: 1. B

Question 23. The tendency of catenation in Gr-14 elements follows the order

1. C > Si > Ge > Sn
2. C>>Si > Ge s: Sn
3. Si > C > Sn > Ge
4. Ge > Sn > Si > C

Answer: 2. C>>Si > Ge s: Sn

Question 24. The repeating structural unit in silicone is—

Answer: 2

Question 25. Which of the following allotropic forms of carbon is isomorphous with crystalline silicon

1. Graphite
2. Coal
3. Coke
4. Diamond

Answer: 4.  Diamond

Question 26. The shape and hybridisation of the B-atom of BH₄^– is

1. Pyramidal, sp³
2. Octahedral, sp³d²
3. Tetrahedral, sp³
4. None of these

Answer: 3.  Tetrahedral, sp³

Question 27. Germanium is transparent in

1. Visible light
2. Infrared region
3. Ultraviolet region
4. Ultraviolet region

Answer: 2. Infrared region

Question 28. The chain length of silicone polymer can be controlled by adding

1. MeSiCl₃
2. Me₂SiCl₂
3. Me₃SiCl
4. Me₄Si

Answer: 3. Me₃SiCl

Question 29. Higher B—F (in BF₃) bond dissociation energy as compared to that of C— F (in CF₄) is due to

1. Stronger σ-bond between B and F in BF₃ as compared to that between C and F in CF₄
2. Significant pn-pn interaction between B and F in
3. BF₃ whereas there is no possibility of such interaction between C and F in CF₄
4. A lower degree of pπ-pπ interaction between B and F in BF₃ than that between C and F in CF₄ smaller size of B -atom as compared to that of C -atom

Answer: 2. Significant pn-pn interaction between B and F in

Question 30. The reaction of diborane with ammonia initially gives

1. B₂H₆.NH₃
2. Borazol
3. B₂H₆-3NH₃
4. [BH₂(NH₃)₂]⁺[BH₄]^–

Answer: 4.  [BH₂(NH₃)₂]⁺[BH₄]^–

Question 31.    X, y, Z and their respective colours are

1. X = Cu(BO₂)₂ (blue), Y = Cu(BO₂) (colourless),Z = Cu (red)
2. X = CuBOz (blue) , Y = Cu(BO₂)₂ (colourless), Z = Cu (Black)
3. X = CU(BO₂)₂ (red) , Y = CuBO₂ (blue) Z = (red)
4. X = Cu (red) , Y = Cu(BO₂)₂ (blue) , Z = CuBO₂ (colourless)

Answer: 1. X = Cu(BO₂)₂ (blue) , Y = Cu(BO₂) (colourless) ,Z = Cu (red)

Question 32. The correct formula for borax is

1. Na₂[B₄O₄(OH)₃].9H₂O
2. Na₂[B₄O₄(OH)₄].8H₂O
3. Na₂[B₄O₆(OH)₅].7H₂O
4. Na₂[B₄O₇(OH)₆]-6H₂O

Answer: 2. Na₂[B₄O₄(OH)₄].8H₂O

Question 33. Which of the following statements is correct—

1. Sn (II) and Pb (IV) salts are used as oxidants
2. Sn (II) and Pb (IV) salts are used as reductants
3. Sn (II) salts are used as oxidants and Pb (IV) salts are used as reductants
4. Sn (II) salts are used as reductants and Pb (IV) salts are used as oxidants

Answer: 4. Sn (II) salts are used as reductants and Pb (IV) salts are used as oxidants

Question 34. SiCl₄ gets readily hydrolysed but CCl₄ does not, because

1. Si can expand its octet but C does not
2. The ionisation enthalpy of C is greater than that of Si
3. C forms both double and triple bonds
4. The electronegativity of C is greater than that of Si

Answer: 1. Si can expand its octet but C does not

Question 35. PbCl₄ exists but PbBr₄ and Pbl₄ do not, because

1. Chlorine is a most electronegative element
2. Bromine and iodine are larger
3. Bromine and iodine cannot oxidise Pb to Pb⁴⁺
4. Bromine & iodine are stronger oxidants than chlorine

Answer: 3. Bromine and iodine cannot oxidise Pb to Pb⁴⁺

Question 36. Which of the following resembles CO in terms of physical properties

1. O₂
2. Cl₂
3. N₂
4. F₂

Answer: 3.N₂

Question 37. Which ofthe following statements is incorrect

1. Most of the silicones are water repellents
2. Silicones get dissociated at high temperature
3. Silicones do not get oxidised in air at high temperature
4. Silicones are good thermal and electrical insulators

Answer: 2.  Silicones get dissociated at high temperature

Question 38. Wollastonite is a

1. Three-dimensional silicate
2. Chain silicate
3. Sheet silicate
4. Cyclic silicate

Answer: 4. Cyclic silicate

Question 39. B(OH)₃ + NaOH ⇌ NaBO₂ + Na[B(OH)₄] + H₂O; The above reaction be made to proceed in the forward direction by

1. Addition of diol
2. Addition of borax
3. Addition of KHF₂
4. Addition of NaHPO₄

Answer: 1. Addition of diol

Question 40. Which ofthe following is correct

1. Al(OH)₃ is more acidic than B(OH)₃
2. B(OH)₃ is basic but Al(OH)₃ is amphoteric in nature
3. B(OH)₃ is acidic but Al(OH)₃ is amphoteric in nature
4. Both B(OH)₃ and Al(OH)₃ are amphoteric

Answer: 3. B(OH)₃ is acidic but Al(OH)₃ is amphoteric in nature

Question 41. Which of the following is correct

1. B₂H₆-2NH₃ is known as inorganic benzene
2. Boric acid is a protonic acid
3. Be exhibits coordination number = 6
4. BeCl₃ and AlCl₃ have bridged chlorine structures in the solid phase

Answer: 4. BeCl₃ and AlCl₃ have bridged chlorine structures in the solid phase

Question 42. B cannot form B³⁺ ion, because

1. Formation of B³⁺ ion requires a greater amount of energy and this cannot be obtained from lattice energy or hydration energy
2. B is a non-metal
3. B do not possess any vacant d -orbitals
4. B possess the highest melting point among its group members

Answer: 1.  Formation of B³⁺ ion requires a greater amount of energy and this cannot be obtained from lattice energy or hydration energy

Question 43. Which of the following has the minimum heat of dissociation

1. (CH₃)₃N : → BF₃
2. (CH₃)₃N :→B(CH₃)₂F
3. (CH₃)₃N :→ B(CH₃)₃
4. (CH₃)₃N :→B(CH₃)F₂

Answer: 3. (CH₃)₃N :→ B(CH₃)₃

Question 44. The correct statement concerning CO is

1. It combines with H₂O to give carbonic acid
2. It reacts with haemoglobin
3. It acts only as a reducing agent
4. It cannot form adducts

Answer: 2.  It reacts with haemoglobin

Question 45. Foamite mixture consists of

1. Al₂ (SO₄)₃ + NaHCO₃
2. Al₂(SO₄)₃ + Na₂CO₃
3. Fe₂(SO₄)₃ + Na₂CO₃
4. CuSO₄ + NaHCO₃

Answer: 1. Al₂ (SO₄)₃ + NaHCO₃

Question 46. In which of the following compounds, the 3c-2e bond is present

1. AI₂(CH₃)₆
2. In (C₆H₅)₃
3. B₂H₆
4. Al₂Cl₆

Answer: 1 and 2

Question 47. Which ofthe following oxides do not get reduced by CO

1. ZnO
2. Fe₂O₃
3. CaO
4. Na₂O

Answer: 1 and 3

Question 48. Which of the following is not isostructural with CO₂

1. SnCl₂
2. HgCl₂
3. SCl₂
4. Znl₂

Answer: 2 and 4

Question 49. C(OH₄) is unstable but Si(OH)₄ is stable. Possible reasons are

1. C — O bond energy is low
2. C — O  bond energy is high
3. Si — O bond energy is low
4. Si — O bond energy is high

Answer: 1 and 4

Question 50. Which ofthe following statements are correct

1. Fullerenes have dangling bonds
2. Fullerenes are cage-like molecules
3. Graphite is thermodynamically the most stable allotrope of carbon
4. Graphite is the purest allotrope of carbon

Answer: 2 and 3

Question 51. Boron trifluoride (BF₃) is

1. An electron-deficient compound
2. A Lewis acid
3. An ionic compound
4. Used as rocket fuel

Answer: 1 and 2

Question 52. Compounds which readily undergo hydrolysis are

1. AlCl₃
2. CCl₄
3. SiCl₄
4. PbCl₄

Answer: 1, 3 and 4

Question 53. Which of the following compounds undergo disproportionation in aqueous solution

1. TlCl₃
2. GaCl
3. InCl
4. TlCl

Answer: 2 and 3

Question 54. Me₃SiCl is used during the polymerisation of organosilicon because

1. The chain length of organosilicon polymers can be controlled by adding Me₃SiCl
2. Me₃SiCl blocks the end terminal of the silicone polymer
3. Me₃SiCl improves the quality and yield of the polymer
4. Me₃SiCl acts as a catalyst during polymerisation

Answer: 1 and 2

Question 55. Which of the following acids, on dehydration, produce oxides of carbon

1. Succinic acid
2. Propanoic acid
3. Mlonicacid
4. Formic acid

Answer: 3 and 4

Question 56. Which of the following are basic nature

1. B₂O₃
2. Tl₂O
3. ln₂O₃
4. Al₂O₃

Answer: 2 and 3

Question 57. The linear shape of CO₂ is due to

1. sp³ -hybridisation of C
2. sp -hybridisation of C
3. pπ-pπ bonding between C and 0
4. sp² -hybridisation of C

Answer: 2 and 3

Question 58. Which metallic salts exhibit the same colouration both in oxidising and reducing flame in the borax-bead test

1. Fe
2. Mn
3. Co
4. Cr

Answer: 3 and 4

Question 59. Which of the following two acidic substances react to give an alkaline solution

1. H₂B₄O₇
2. H₃BO₃
3. HF
4. KHF₂

Answer: 2 and 4

Question 60. Which of the following are the ingredients of baking powder

1. NaOH
2. Tartaric acid
3. Formic acid
4. Potassium hydrogen tartrate

Answer: 2 and 4

Question 61. Which ofthe following are sheet silicates

1. Diopside
2. Kaolinite
3. Talc
4. Beryl

Answer: 2 and 3

Question 62. Identify the correct resonating structures

1. O -C ≡ O
2. O = C = O
3. O ≡ C –  O⁺
4. ^–O – C ≡  O⁺

Answer: 2 and 4

Question 63. Which of the following species are not known

1. [SiCl₆]^2-
2. [CF₆]^2-
3. [PbCl₆]^2-
4. [SiF₆]^2-

Answer: 2 and 3

Question 64. Which of the following is correct concerning Gr-14 elements

1. Stability of dihalides: CX₂ > SiX₂ > GeX₂ > SnX₂
2. The tendency to form pπ-pπ multiple bonds increases down the group
3. The tendency of catenation decreases down the group
4. Each of them forms oxide ofthe type MO₂

Answer: 2, 3 and 4

Question 65. Which ofthe following has a bridge bond

1. Water
2. Inorganic benzene
3. Phenol
4. Diborane

Answer: 4. Diborane

Question 66. An aqueous solution of borax is

1. Neutral
2. Amphoteric
3. Basic
4. Acidic

Answer: 3. Basic

An aqueous solution of borax is basic because it is a salt of strong base (NaOH) and weak acid (H₃BO₃)

Question 67. Boric acid is polymeric due to

1. Its acidic nature
2. The presence of hydrogen bonds
3. Its monobasic nature
4. Its geometry

Answer: 2. The presence of hydrogen bonds

Boric acid is polymeric due to the presence of; hydrogen bonds

Question 68. The type of hybridisation of boron In dlborane

1. sp
2. sp²
3. sp³
4. dsp²

Answer: 2.sp²

In B₂HO, the hybridisation state of B is sp²

Question 69. Thermo dynamically the most stable form of carbon is

1. Diamond
2. Graphite
3. Fullerenes
4. Coal

Answer: 2. Graphite

Question 70. Elements of Gr-14

1. Exhibit oxidation state of +4 only
2. They exhibit oxidation state of +2 and +4
3. Form M^2-and M²⁺Ions
4. Form M²⁺ and M⁴⁺ ions

Answer: 2. They exhibit oxidation states of +2 and +4

Elements of group 14 exhibit oxidation states of +2 and +4 due to the inert pair effect