Class 11 Chemistry

Heat Of Reaction

Heat Of Reaction Definition

The amount of heat absorbed or evolved in a reaction when the stoichiometric number of moles of reactions indicated by the balanced chemical equation, is completely converted into products at given conditions is called the heat of reaction at that conditions.

At a particular temperature, the heat of a reaction depends on the conditions under which a reaction is occurring. Generally, chemical reactions are carried out either at constant pressure or at constant volume. Accordingly, the heat of reaction is of two types, namely Heat of reaction at constant volume and heat of reaction at constant pressure.

The heat of reaction at constant volume

The amount of heat absorbed or evolved in a reaction when the stoichiometric number of moles of reactants, indicated by the balanced chemical equation of the reaction, is completely converted into products at a fixed temperature and volume is called the heat of reaction at constant volume.

Explanation: Let us consider a chemical reaction that is occurring at constant temperature and volume.

⇒ \(a A+b B \rightarrow c C+d D\)

According to die first law of thermodynamics, if a reaction occurs at constant volume then the heat change (qv) is equal to the change in internal energy (AU) ofthe system (provided only pressure-volume work is performed), qv= ΔU

Therefore, for the reaction

⇒ \(q_V=\Delta U=\Sigma U_{\text {products }}-\Sigma U_{\text {reactants }}=\Sigma U_P-\Sigma U_R\)

where ΣUp = total internal energy of products and I UR = total internal energy of reactants.

∴ \(q_V=\Sigma U_p-\Sigma U_R=\left(c \bar{U}_C+d \bar{U}_D\right)-\left(a \bar{U}_A+b \bar{U}_B\right)\)

Where \(\bar{U}_A, \bar{U}_B, \bar{U}_C \text { and } \bar{U}_D\) die molar internal energies of A. B. Cand D, respectively. Thus, the heat of reaction for a reaction at constant temperature and volume is the difference between the total internal energy of tire products and the total internal energy of the reactants.

Example:

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

For the above reaction, the heat of the reaction at a constant volume:

⇒ \(q_V=\left[\bar{U}\left(\mathrm{CO}_2, g\right)+2 \bar{U}\left(\mathrm{H}_2 \mathrm{O}, l\right)\right]-\left[\bar{U}\left(\mathrm{CH}_4, g\right)+2 \bar{U}\left(\mathrm{O}_2, g\right)\right]\)

Where

⇒ \(\bar{U}\left(\mathrm{CO}_2, \mathrm{~g}\right), \bar{U}\left(\mathrm{H}_2 \mathrm{O}, l\right), \bar{U}\left(\mathrm{CH}_4, g\right) \text { and } \bar{U}\left(\mathrm{O}_2, g\right)\)

Are die molar internal energies of CO,(g), H₂O(I), CH₄(g) respectively.

Heat of reaction at constant pressure:

The amount of heat absorbed or evolved in a reaction when the stoichiometric number of moles of reactants, indicated by a balanced chemical equation of the reaction, is completely converted into products at constant pressure, and temperature is called the heat of reaction at constant pressure.

The heat of reaction at constant pressure Explanation:

Let us consider a reaction that is occurring at constant temperature and pressure: aA + bB →+ cC + dD. According to the first law of thermodynamics, if a reaction occurs at constant pressure, then the heat change in the reaction is equal to the change in enthalpy of the system (provided only pressure-volume work is performed). Thus, qp = ΔH. Therefore, for the reaction,

⇒ \(q_P=\Delta H=\Sigma H_{\text {products }}-\Sigma H_{\text {reactants }}=\Sigma H_P-\Sigma H_R \text { ; }\)

where Hp = total enthalpy ofthe products and HR = total enthalpy of the reactants

∴ \(q_P=\Sigma H_P-\Sigma H_R=\left(c \bar{H}_C+d \bar{H}_D\right)-\left(a \bar{H}_A+b \bar{H}_B\right)\)

where, \(\bar{H}_A, \bar{H}_B, \bar{H}_C \text { and } \bar{H}_D\) are the molar enthalpies of A, B, C, and D, respectively.

Thus, for a reaction, the heat of the reaction at constant temperature and pressure is equal to the difference between the total enthalpy of products and the total enthalpy of reactants.

Example:

For the Reaction, \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\) the heat ofreaction at constant pressure is given by;

⇒ \(q_P=2 \bar{H}\left(\mathrm{H}_2 \mathrm{O}, l\right)-\left[2 \bar{H}\left(\mathrm{H}_2, \mathrm{~g}\right)+\bar{H}\left(\mathrm{O}_2, \mathrm{~g}\right)\right]\)

Where \(\bar{H}\left(\mathrm{H}_2 \mathrm{O}, l\right), \bar{H}\left(\mathrm{H}_2, \mathrm{~g}\right) \text { and } \bar{H}\left(\mathrm{O}_2, \mathrm{~g}\right)\) are the molar enthalpies of H₂O(g) , H₂(g) and O₂(g) , respectively.

Relation between heat of reaction at constant volume [q_v) and heat of reaction at constant pressure (q_p)

If a reaction is carried out at a fixed pressure than the heat of the reaction, qp = change in enthalpy in the reaction, ΔH. …………………(1)

If the reaction is carried out at constant volume then the heat of the reaction, qv = change in internal energy in the reaction, ΔU …………………(2)

If the changes in internal energy and volume in a reaction occurring at constant pressure are AUp and AV, respectively, then according to the relation H = U + PV

⇒ \(\Delta H=\Delta U_P+\Delta(P V)=\Delta U_P+P \Delta V\)

[As pressure (P) is constant, Δ(PV) = PΔV

∴ \(q_P=\Delta H=\Delta U_P+P \Delta V\) …………………(3)

since qp = ΔH

Subtracting equation [2] from [3], we obtain

∴ \(q_P-q_V=\Delta U_P+P \Delta V-\Delta U=\left(\Delta U_P-\Delta U\right)+P \Delta V\)

At a particular temperature, the difference between AUp (change in internal energy at constant pressure) and AU (change in internal energy at constant volume) is very small. Therefore, it is considered that

⇒ \(\Delta U_P \approx \Delta U.\)

Since q_p -q_v = PΔV

Or, ΔH-ΔU = PΔV ………………………….(4)

In the case of solids and liquids:

For reactions involving only solids and liquids, the change in volume (ΔV) of the reaction system is negligibly small. So, in such reactions, the heat of the reaction at constant pressure (qp or ΔH) becomes equal to the heat of the reaction at constant volume (qV or ΔU).

In the case of gases:

For a reaction involving gaseous substances

Example:

⇒  \(A(g)+B(g) \rightarrow C(g) \text { or } A(g)+B(g) \rightarrow C(I)\),

Or \(A(s) \rightarrow B(s)+C(g)\), etc.]

The change in volume (Δ V) of the reaction system may be sufficiently high. We can determine the value of PΔV as illustrated below

Let us consider a gaseous reaction that occurs at constant pressure (P) and temperature (T).

Suppose, n₁ and V₁ are the total number of moles and the total volume of the reactant gases, respectively, and n₂ and IA, are the Here, total number of moles and the total volume of the product gases respectively. If the gases are assumed to behave ideally, then for gaseous reactants PV₁ = n₁RT., and gaseous products PV₂ = n₂RT.

∴ \(P\left(V_2-V_1\right)=\left(n_2-n_1\right) R T \text { or, } P \Delta V=\Delta n R T\)

Substituting ΔnRT for PΔV into equation (4) we obtain

⇒  \(q_p-q_V=\Delta n R T \text { and } \Delta H-\Delta U=\Delta n R T\)

∴ \(\left[\boldsymbol{q}_p=\boldsymbol{q}_V+\Delta n \| T\right] \text { and }[\Delta I=\Delta U+\Delta n \ RT] \cdot \cdot \cdot \cdot \cdot [5]\)

Using equation (5), ΔH (or qp) can be calculated from the known value of U (or Δv), and ΔU (or qV) can be calculated from the known value of AII (or ΔyU). If

Δn=0; ΔH=ΔU;Δn>0,ΔH>ΔU;Δn>0,ΔH<ΔU.

Comparison between the values of ΔH and ΔU for gaseous reaction having different Δn values:

Standard State And Standard Reaction Enthalpy

Enthalpy change in a reaction depends on the conditions of temperature, pressure, and physical states of the reactants and products. To compare enthalpies of different reactions, we define a set of conditions called standard state, at which the values of AH for different reactions are calculated.

Standard State:

The standard state ofa substance is defined as the most stable and purest state of that substance at the temperature of interest and atm pressure. In the definition of the standard state, temperature is not specified like pressure (1 atm). If the temperature is not mentioned, then 25 °C (298.15 K) is taken as a reference temperature. However, this does not mean 25 °C is the standard temperature. A pure substance can have different standard states depending on the temperature of interest, but in each of these states, the pressure is always 1 atm.

Examples:

• The standard state of liquid water at a particular temperature means H₂O(l) at that temperature and 1 atm pressure.
• The standard state of ice at a particular temperature means pure H₂O(s) at that temperature and atm pressure.
• The standard state of liquid ethanol at 25 °C means C₂H₅OH(l) at 25 °C and 1 atm pressure.

Standard enthalpy of reaction:

The standard enthalpy change of a reaction is defined as the enthalpy change that occurs when the stoichiometric number of moles of reactants, indicated by the balanced chemical equation of the reaction, is completely converted into products at a particular temperature and l atm (i.e., at the standard state).

The standard enthalpy of reaction at a particular temperature ( T K) is denoted by \(\Delta H_T^0\). The superscript ‘0’ indicates the standard state, and the subscript T indicates the temperature in the Kelvin scale.

Explanation:

⇒ \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(l)\)

For this reaction \(\Delta H_{2 \mathrm{qg}}^0=-2220 \mathrm{~kJ}\) indicating that at.

• 298 K temperature and atm pressure, if lmol of propane (C₃H₈) and mol of O₂ react completely to form 3 mol of CO, and 4 mol of water at the same temperature and pressure (i.e, 29S K and atm respectively), then 2220 kj of heat will be evolved.
• Alternatively, it can be said that at 298K temperature and 1 atm pressure, the change in enthalpy for the following reaction is =-2220 kJ.

Factors affecting the reaction enthalpy

1. Physical states of reactants and products:

During the change of physical states of a substance (like solid, solid→ liquid, liquid→ vapor, etc.) heat is either absorbed or evolved. Thus, the value of the heat of the reaction or enthalpy of the reaction depends upon the physical states ofthe reactants and products. For example, in the following two reactions, due to the different physical states ofthe products,

The values ofthe heat of the reaction are different:

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-571.6 \mathrm{~kJ}\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(g) ; \Delta H=-483.6 \mathrm{~kJ}\)

2. The condition under which the reaction is conducted:

At a particular temperature, a chemical reaction can be conducted either at constant pressure or at constant volume. If a reaction occurs at constant pressure, then the heat of reaction (q_p)

⇒ \(=H_{\text {product }}-H_{\text {reactant }}=\Delta H\)

Occurs at constant volume, then the heat of reaction (q_p) — The relation between ΔH and ΔH is ΔH = ΔH + PΔV The quantity PAV indicates pressure-volume work.

• So, the difference between ΔH and AH is equal to the pressure-volume work involved during the reaction.
• If the volume of the reacting system remains fixed (ΔV = 0), then pressure-volume work = 0 and ΔH = ΔH.
• If the volume ofthe reacting system changes (ΔV=0), then, the pressure-volume work, PΔV≠0 and ΔH≠ΔU.

3. Allotropic forms of the reacting elements:

As the different allotropic forms have different enthalpies, the value of reaction enthalpy depends upon the allotropic forms of the reactants.

For example, the enthalpies of the reaction are different for the oxidation of graphite and diamond (two allotropic forms of carbon)

⇒ \(\mathrm{C}(\text { graphitè, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

ΔH=-393.5 kJ

⇒\(\mathrm{C}(\text { diamond, } s)+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)\)

Δ H =-395.4 kJ

4. Amount of the reactants:

As enthalpy is an extensive property, the magnitude of enthalpy change in a reaction (ΔH) is proportional to the amount of reactants undergoing the reaction.

For example, in the reactions given below, the magnitude of ΔH (reaction enthalpy) for reaction (1) is twice that for the reaction (1). This is because the total number of moles of reactants in the reaction (2) is twice as many as that in the reaction (1).

⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H=-285.8 \mathrm{~kJ}\) ……………………………..(1)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l), \Delta H=-571.6 \mathrm{~kJ}\) ……………………………..(2)

5. Temperature:

Temperature has a significant effect on the reaction enthalpy of a reaction. The extent of temperature dependence of the reaction enthalpy depends on the nature of the reaction.

Heat Of Reaction Numerical Examples

Question 1. The value of ΔH for the given reaction at 298K is — 282.85 kj. mol^-1. Calculate the change in internal energy: \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)
Answer:

We know, ΔH = ΔH + ΔnRT where, Δn = (total number of moles of the gaseous products) — (total number of moles of the gaseous reactants)

For the given reaction, ,\(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

As per given data, ΔH = -282.85 kj-mol^-1 & T = 298 K

∴ \(-282.85=\Delta U+\left[\left(-\frac{1}{2}\right) \times 8.314 \times 10^{-3} \times 298\right]\)

∴ \(\Delta U=-281.61 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 2. The bond energy of a diatomic molecule IN given as the change in internal energy due to dissociation of that molecule. Calculate the bond energy of O₂. Given: \(\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) ; \Delta H=498.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}, T=298 \mathrm{~K}\)
Answer:

In the dissociation reaction of O₂ molecules, Δn = 2-1 = 1.

We know, ΔH = ΔU + ΔnRT

As given, ΔH = 498.3 kj.mol^-1 , T = 298 K

∴ \(498.3=\Delta U+\left(1 \times 8.314 \times 10^{-3} \times 298\right)\)

or, ΔU= 495.8J

∴ Bond energy of O₂ molecule = 495.8 kj.mol^-1

Question 3. Calculate the values of ΔH and ΔH in the vaporization of 90 g of water at 100°C and 1 atm pressure. The latent heat of vaporization of water at the same temperature and pressure = 540 cal g^-1.
Answer:

⇒ \(90 \mathrm{~g} \text { of water }=\frac{90}{18}=5 \mathrm{~mol} \text { of water. }\)

Vaporisation of water:

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Therefore, in the vaporization of1 mol of water, Δn = +1.

Hence, for the vaporization of 5 mol water, Δn = +5.

So, the amount of heat required to vaporize 90 g (5 mol) of water =540 x 90 = 48600 cal.

As the vaporization process occurs at constant pressure (1 atm), the heat absorbed = enthalpy change.

∴ The change in enthalpy in the vaporization of 90g of water, ΔH = 48600 cal

∴ The change in internal energy in the vaporization of 90gofwater, ΔH = ΔH-ΔnRT

= 48600- 5 × 1.987 × (273 + 100) = 44894.24 cal .

Question 4. Assuming the reactant and product gases obey its ideal gas law, calculate the change in internal energy (AE) at 27°C for the given reaction:
Answer:

⇒ \(\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ;\)

In tlui rimdlon, Δn= 2-1=1.

We know, ΔH = ΔU+ΔuRT

As given, ΔH = 498.3 kj. mol^-1 , T = 298 K

∴ 498.3 = ΔH + (1 × 8.314 × 10-3 × 298) or, ΔH = 495.8 kJ

The bond energy of the O₂ molecule = 495.8 kj.mol^-1

In the reaction , Δn =2- (1 + 3)= -2

we know ΔH+ ΔnRT

Or, 337= ΔU-2 × 1.987 × 10^-3 × 300

Or, ΔU =-335.8kcal

Balancing Of Chemical Equations Involving Redox Reactions

Redox reaction can be balanced with the help of two methods. These are the

1. Ion-electron method
2. Oxidation number method.

Ion-electron method

Jade and Lamer in 1927 introduced this method. In the ion-electron method, only the molecules and ions which participate in the chemical reaction are shown.

In balancing redox reactions by this method the following steps are followed: 

• The reaction is written in ionic form.
• The reaction is divided into two half-reactions with the help of ions and electrons. One half-reaction is for oxidation reaction and the other half-reaction is for reduction reaction.
• While writing the oxidation reaction, the reducing agent and the oxidised substance are written respectively on the left and right of an arrow signing are written respectively on the left and right of the arrow sign
• To denote the loss of electrons in oxidation half-reaction, the number of electrons (s) is written on the right of the arrow sign (→). While writing the reduction half-reaction, the number of electrons (s) gained is written on the left arrow sign ( →).

Thus, the oxidation half-reaction is:

Reducing agent – Oxidised substance +ne [where n = no. of electron (s) lost in oxidation reaction] Thus reduction half-reaction is Oxidising agent + ne Reduced substance [where n = no. of electron(s) gained reduction reaction] 2Cr³⁺

Then each half-reaction is balanced according to the following steps:

In each of the half-reactions, the number of atoms other than H and O -atoms on both sides ofthe arrow sign is balanced.

If a reaction takes place in an acidic medium, for balancing the number of H and O-atoms on both sides of the arrow sign, H₂O or H⁺ is used. First, oxygen atoms are balanced by adding H₂O molecules to the side that needs O-atoms.

Then to balance the number of H-atoms, two H⁺ ions (2H⁺) for each molecule of water are added to the opposite side (i.e., the side deficient in hydrogen atoms) of the reaction occurs in an alkaline medium, for balancing the H and O -atoms, H₂O or OH^– ion is used.

• Each excess oxygen atom on one side of the arrow sign is balanced by adding one water molecule to the same side and two ions to the other side.
• If the hydrogen atom is still not balanced, it is then balanced by adding one OH^– for every excess hydrogen atom on the side of the hydrogen atoms and one water molecule on the other side of the arrow sign in a half-reaction, both H⁺ and OH^– ions cannot participate.
• The charge on both sides of each half-reaction is balanced. This is done by adding an electron to that side which is a deficient negative charge.
• To equalise the number of electrons of the two half-reactions, any one of the reactions or both reactions should be multiplied by suitable integers.
• Now, the two half-reactions thus obtained are added. Cancelling the common term(s) on both sides, the balanced equation is obtained.

Examples:

1. In the presence of H₂SO₄, potassium dichromate (K₂Cr₂O₂) and ferrous sulphate (FeSO₄) react together to produce ferric sulphate [Fe₂(SO₄)₃] and chromic sulphate [Cr₂(SO₄)₃].

Reaction:

K₂Cr₂O₇ + FeSO₄ + H₂SO₄→ K₂SO₄ + Cr₂(SO₄)₃ + Fe₂(SO₄)₃ + H₂O

The reaction can be expressed in ionic form as:

Cr³⁺ + Fe³⁺ + H₂O

Oxidation half-reaction: Fe²⁺→Fe³⁺+ e ……………………(1)

Reduction half-reaction: Cr₂O₇^2-+ Cr³⁺ ……………………(2)

1. Balancing the Cr -atom: Cr₂O₇^2-– Cr³+ 7H₂O

2.To equalise the number of O -atoms on both sides, 7 water molecules are to be added to the right side. 2Cr³⁺ + 7H₂O

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

∴ One water molecule is required for each O-atom.

3. To balance H-atoms on both sides, 14H+ ions are to be added to the left side.

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

∴ 2H⁺ ions are required for each water molecule.

4. For equalising the charge on both sides, 6 electrons are to be added to the left side.

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

Now, for balancing the number of electrons in oxidation and reduction half-reactions, the balanced oxidation half-reaction is multiplied by 6 and the balanced reduction half-reaction by 1. Then these two equations are added.

This balanced equation has been expressed in ionic form. This equation can be represented in molecular form as:

6FeSO₄ + K₂Cr₂O₇ + 7H₂SO₄ → 3Fe₂(SO₄)₃ + Cr₂(SO₄)₃+ K₂SO₄+ 7H₂O

∴ For 2H⁺ ions, one H₂SO₄ molecule is required

2. In presence of H₂SO₄, KMnO₄ and FeSO₄ react together to produce MnSO₄ and Fe₂(SO₄)₃.

Reaction: KMnO₄ + FeSO₄ + H₂SO₄ → K₂SO₄ + MnSO₄+ Fe₂(SO₄)3 + H₂O

The equation can be expressed in ionic form as:

MnO₄ + Fe²⁺ + H⁺→-Mn²⁺ + Fe³⁺ + H₂O

Oxidation half-reaction: Fe²⁺ — Fe³⁺ + e ………………………..(1)

Reduction half-reaction: MnO₄^– + 8H⁺ + 5e → Mn²⁺ + 4H₂O………………………..(2)

To balance the number of electrons lost in the oxidation half-reaction, the oxidation half-reaction is multiplied by 5 and then the two reactions are added.

As one Fe₂(SO₄)₃ molecule contains two Fe -atoms, the equation is multiplied by 2

10Fe²⁺ + MnO₄^– + 16H⁺ →10Fe³⁺ + 2Mn²⁺ + 8H₂O

This is the balanced equation in ionic form. This equation when expressed in molecular form becomes

10FeSO₄ + 2KMnO₄+ 8H₂SO₄ →  5Fe₂(SO₄)₃ + 2MnSO₄ + 8H₂O

Equalising the number of atoms of different elements and the sulphate radicals we get,

10FeSO₄+ 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O

This is a balanced equation of the given reaction in molecular form.

3. In the reaction between K₂Cr₂O₇ acidified with dilute H₂SO₄  and KI, Cr₂(SO₄)₃ and I₂ are formed.

K₂Cr₂O₇+KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O

The equation can be expressed in ionic form as— Cr₂O₇^2-+I^– + H⁺ — Cr³⁺ + I₂ + H₂O

Oxidation half-reaction: 2I^–→ I₂ + 2e……………..(1)

Reduction half-reaction: Cr₂O₇ ^2-+ I^–+14H⁺ + 6e — 2Cr³⁺ + 7H₂O ……………………….(2)

To balance the electrons, equation (1) is multiplied by 3 and added to equation (2). Thus the equation stands as—

This is the balanced equation of the reaction in ionic form. The above ionic reaction can be expressed in molecular form as follows

6KI + K₂Cr₂O₇ + 7H₂SO₄→3I₂ + Cr₂(SO₄)₃ + 7H₂O

Equalising the number of atoms of potassium and sulphate radical on the left and right sides, we have,

6KI + K₂Cr₂O₇+ 7H₂SO₄ → 3I₂ + Cr₂(SO₄)₃ + 4K₂SO₄ + 7H₂O

4. In the reaction between KMnO₄, acidified with dilute H₂SO₄ and oxalic acid (H₂C₂O₄), MnSO₄ and CO₂ were produced.

Reaction: KMnO₄ + H₂C₂O₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + CO₂ + H₂O

Oxidation half-reaction: C₂O₄→  2CO₄^2-+ 2e ……………………..(1)

Reduction half-reaction: MnO₄^– +8H⁺+ 5e → Mn²⁺+ 4H₂O……………………..(2)

Now, multiplying equation (1) by 5 and equation (2) by 2 and then adding them, we get,

5C₂O₄ ^2-+ 2MnO₄ + 16H⁺+10CO₂ + 2Mn²⁺ + 8H₂O

This Is the balanced equation of the given reaction in molecular form.

5H₂C₂O₄ + 2KMnO₄+ 3H₂SO₄→ 10CO₂ + Mn²⁺ +8H₂O

Equalising the number of atoms of potassium and sulphate radical we get

5H₂C₂O₄ + 2KMnO₄ + 3H₂SO₄ → 10CO₂ + 2MnSO₄ + K₂SO₄ + 8H₂O

5. In NaOH solution, Zn reacts with NaNO₃ to yield Na₂ZnO₂, NH₃ and H₂O.

Reaction:

Zn + NaNO₃ + NaOH — Na₂ZnO₂ + NH₃ + H₂O The equation can be expressed in ionic form as—

Zn + NO^– + OH^–→ ZnO_2^– →+ NH₃ +H₂O

Oxidation half-reaction: Zn + 4OH^–→ ZnO₂^– + 2H₂O + 2C ……………………..(1)

Reduction half-reaction: NO₃ + 6H₂O + 8C — NH₃ + 9OH^– ……………………..(2)

Now multiplying equation (1) by 4 and then adding to equation (2), we get,

4Zn + 16OH- + NO, + 6H₂O→ 4ZnO₂ + NH₃ + 90H- + 8H₂O

Or, 4Zn + 7OH^–+ NO^–₃→ 4ZnO₂^– + NH₃ + 2H₂O

It is the balanced equation of the reaction in ionic form. Expressing the above equation in molecular form

4Zn + 7NaOH + NaNO₃→ 4Na₂ZnO₂ + NH₃ + 2H₂O

It is the molecular form of the balanced equation of the reaction.

6. In the presence of HNO₃, sodium bismuthatic (NaHO₃) reacts with Mn(NO₃)₂ to produce coloured sodium permanganate (NaMnO₄) and itself gets reduced to bismuth nitrate.

Reaction: NaBIO₃ + Mn(NO₃) + UNO₂ →NaMnO₄ + Bi(NO₃) + H₂O The equation can be expressed in ionic form as— BIO₂ + Mn²⁺ → Bl³⁺ + MnO₄ + H₂O

Oxidation half-reaction: Mn²⁺ + 4H₂O → MnO₄^–+8H⁺→+5e ……………………(1)

Reduction half-reaction: BiO₃ + 6H⁺ + 2e — Bi³⁺ + 3H₂O…………………….(2)

Multiplying equation (1) by 2 and equation (2) by 5 and then adding them we get—

2Mn²⁺ + 8H₂O + 5BiO₃ ^– +30H⁺→ 2MnO₄^– + 16H⁺ + 5Bi³⁺ +15H₂O

2Mn²⁺ + 5BiO₃ + 14H⁺→ 5Bi³⁺+ + 2MnO₄^–  +7HO

This is the balanced ionic equation of the reaction. The equation in the molecular form stands as—

2Mn(NO₃)₂ + 5NaBiO₃+ 14HNO₃ →  5Bi(NO₃)₃ + 2NaMnO₄ + 7H₂O

Ionic reaction: IO₃^– + I^–+ H⁺→ I₂ + H₂O

Oxidation half-reaction: 2I^– → I₂ + 2e ……………………….(1)

Reduction half-reaction: 2 IO₃^– + 12H⁺+ 10 e → I₂ + 6H₂O ……………………….(2)

Multiplying equation (1) by 5 and then adding to equation (2) we get,

10I^– + 2IO₃^–+ 12H⁺ → 6I₂+ 6H₂O or, 5I^– + 1O₃^–+ 6H⁺ — 3I₂ + 3H₂O

This is the balanced ionic equation of the reaction.

Oxidation number method

In any redox reaction, the increase in the oxidation number of some of the atoms is balanced by the decrease in the oxidation number of some other atoms.

The steps which are to be followed while balancing the oxidation-reduction equation by this method are given below— After identifying the oxidant and reductant, the skeleton equation for the reaction is written.

• The elements of the reactants and the products changing oxidation number are identified and the oxidation number of the concerned atoms is mentioned.
• The reactant in which the element undergoes a decrease in oxidation number is the oxidant, while the reactant in which the element undergoes an increase in oxidation number is the reductant.
• As oxidation and reduction are complementary to each other, die increase and decrease in oxidation numbers should necessarily be equal, For this reason, the respective formulae of the oxidants and reductants are multiplied by a possible suitable integer so that the changes in oxidation numbers arc equalised.
• For balancing the equation, it may sometimes be necessary to multiply the formula of other substances participating in the reaction by a suitable integer.
• If the reactions are carried out in an acidic medium, then, to balance the number of O -atoms, one molecule of O^–is added for each O -atom to the side of the equation deficient in oxygen.

To balance the number of -atoms, H+ ions are added to the side deficient in hydrogen.

In case of a reaction occurring in an alkaline medium, to balance the number of O -atoms on both sides of the equations, for each O -atom one molecule of water is added to the side deficient in O -atoms and to the opposite side two OH- ions for each water molecule are added.

Again for balancing the number of FI^– atoms on both sides of the equation, for each 2 -atom one OH^– ion is added to the side which contains excess 2 -atoms and the same number of FI₂O molecules are added to the other side.

Example

1. Copper dissolves in concentrated HNO₃ to form Cu(NO₃)₂, NO₂ and H₂O

Reaction:

In the given reaction, the increase in oxidation number of Cu -atom =(+2)-0 = 2 unit (oxidation) and the decrease in oxidation number of N -atom =(+5)-(+4) = 1 unit (reduction).

To nullify the effect of increase and decrease in the oxidation numbers, the ratio of the number of Cu -atoms and Cu(NO₃)₂  molecules in the reaction should be 1:2. So the equation may be written as

Cu + 2HNO₃ → Cu(NO₃)₂ + 2NO₂ + H₂O

Now, to produce one molecule of Cu(NO₃)₂ two NO₂ radicals i.e. two molecules of UNO₂ are required. Hence in the reaction further addition of two molecules of HNO₃ is necessary. So the balanced equation is expressed as

Cu + 4HNO₃→ Cu(NO₃)₂ + 2NO₂ + 2H₂O

Now, to produce one molecule of Cu(NO₃)₂, two NO₃ radicals i.e. two molecules of UNO₂ are required.

Hence in the reaction further addition of two molecules of HNO₂ is necessary. So the balanced equation is expressed as

Cu+4HNO₃ Cu(NO₃)^{→ 2+}2NO²⁺2H₂O

When H₂S gas is passed through chlorine water H₂SO₄ is produced.

Reaction:

In this reaction, an increase in the oxidation number of S = (+6) — (— 2) = 2 units (oxidation) and a decrease In the oxidation number of Cl = 0 – (— 1 ) I unit (reduction). So decrease In oxidation number for two (‘,1 -atoms or I molecule of Cl₂ -2 unit.

To neutralise the effect of Increase and decrease In oxidation number in the given equation, the number of molecules of H₂S and Cl₂ should be in the ratio of 2: i.e., 1: <1.

Therefore, the equation becomes

H₂S+4CI₂+HCl+H₂SO₄

Balancing the number of 11 and O -atoms on both sides gives the balanced equation —

3. NH₃ gas when passed over heated Cut) produces Cu, N₂ and H₂O.

In this reaction, increase In oxidation number of N=0-(-3) = 3 unit (oxidation) and decrease In oxidation number of Cu =(+2)-0 = 2 unit (reduction). As, in a redox reaction, the total increase in oxidation number is equal to the total decrease In oxidation number, the number of molecules of CuO and NH₃ in the reaction should be in the ratio of 3:2. Hence, the balanced equation will be—

3CuO+2NH₃→3Cu+N₂+3H₂O

4. In the reaction between KMnO₄ and H₂O₂, the products obtained were K₂SO₂ MnSO₂, H₂O And O₂.

Reaction:

In this reaction increases in oxdination number of O = 0-(-1)=1 (oxidaxtion) and dexrease in oxidation number of MN= (+7)-(+20)

= 5 unit (reduction).

The total increase in the oxidation number of two 0 -atoms presents one molecule of H₂O₂

To balance the decrease and increase in oxidation numbers, the ratio of the number of KMnO₂ and H₂O, molecules in the equation for the reaction will be 2:5.

Again from 2 molecules of KMnO₄ and 5 molecules of H₂O₂, 2 molecules of MnSO₂ and 5 molecules of O₂ are produced respectively. Thus the equation becomes—

2KMnO₄ + 5H₂O₂+ H₂SO₄→ K₂SO₄+ 2MnSO₄ + 5O₂ + H₂O

Again, for the formation of 1 molecule of K₂SO₄ and 2 molecules of MnSO₄, three SO₄– radicals are required and hence three H₂SO₄ molecules are necessary on the left-hand side. Besides this, the total number of H-atoms in 5 molecules of H₂O₂ and 3 molecules of H₂SO₄ = 16.

These H-atoms produce water molecules. Therefore, 8 molecules of H₂O are to be placed on the right-hand side. So the balanced equation will be—

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5O₂+ 8H₂O

5. White phosphorus and concentrated NaOH react together to yield NaH₂PO₂ and PH₃. Reaction

The increase in oxidation number of P [P to NaH₂PO₂ ] = +1- 0 = 1 unit (oxidation). The decrease in oxidation number of P [P to PH₃] = 0-(-3) = 3 unit (reduction). To balance the increase and decrease in oxidation number, three P atoms for oxidation and one P -atom for reduction are required. Thus four P -atoms are necessary.

Now, in the oxidation of P, NaH₂PO₂ and its reduction, PH₃ are formed. So the oxidation of three P atoms forms 3 molecules of NaH₂PO₉ and for this, three NaOH molecules are required. Again 1 atom of P reduction produces 1 molecule of PH₃. So the equation will be

P₄+3NaOH + H₂O→ 3NaH₂PO₂ + PH₃

On the right side of the equation, there are 6 oxygen atoms, out of which 3 atoms will come from 3 molecules of NaOH and for the rest three atoms, 3 molecules of H₂O will be necessary. Hence, the balanced equation will be

P₄+ 3NaOH + H₂O → 3NaH₂PO₂ + PH₃

6. In NaOH solution, Zn reacts with NaNO₃ to yield Na₂ZnO₂, NH₃ and H₂O.

Reaction:

In the reaction, an increase in the oxidation number of Zn =(+2) -0 = 2 unit (oxidation) and a decrease in the oxidation number of N =(+5)-(-3) = 8 unit (reduction). As the increase and decrease in oxidation number in the reaction must be equal, the number of Zn -atoms and the number of molecules of NaNO₃ should be in the ratio of 4:1. Now, 1 molecule of NaNO₃ and 4 atoms of Zn produce 1 molecule of NH3 and 4 molecules of Na₂ZnO₂ respectively. Therefore the reaction is—

4Zn + NaNO₃ + NaOH →4Na₂ZnO₂ + NH₃ + H₂O

Again formation of 4 molecules of Na₂ZnO₂ requires 8 Na -atoms, out of which 1 atom is supplied by 1 molecule of NaNO₃. Additional 7 Na -atoms come from NaOH on the left side. To balance H -atoms on both sides, 1 H₂O molecule is to be placed on the right side. Thus the balanced equation will be —

4Zn + NaNO₃ + 7NaOH →4Na₂ZnO₂ + NH₃ + 2H₂O

7. In the reaction between Cr₂O₃ and Na₂O₂, Na₂CrO₄ and NaOH are produced.

Reaction:

In this reaction, an increase in the oxidation number of Cr = (+6)- (+3) = 3 unit (oxidation) and a decrease in the oxidation number of O =(- 1 )-(- 2) = 1 unit (reduction). Thus a total increase in the oxidation number of two Cr -atoms = 3 × 2 = 6 units and the total decrease in the oxidation number of two O -atoms = 1× 2

= 2 units.

To balance the increase and decrease in oxidation number, the ratio of Cr₂O₃ and Na₂O₂ should be =1:3. Now 2 molecules of Na₂CrO₄ are produced from 1 molecule of Cr₂O₃. Hence the equation will be as follows—

Cr₂O₃+3Na₂O₂+H₂O→ 2Na₂CrO₄+NaOH

If Na, H and O- atoms are balanced on both sides, the balanced equation will stand as

Cr₂O3+ 3Na₂O₂ + H₂O→ 2Na₂CrO₄ + 2NaOH

8. White phosphorus reacts with copper sulphate solution to produce Cu, H₃PO₄ and H₂SO₄.

Reaction:

In this reaction, an increase in the oxidation number of P = (+5)-0 = 5 unit (oxidation) and a decrease in the oxidation number of Cu =(+2)-0 = 2 unit (reduction).

Since the increase and decrease in oxidation number must be equal, in the given reaction, the ratio of the number of atoms of P and the number of CuSO₄ molecules should be in the ratio of 2: 5. Again 2 molecules of H₃PO₄ and five Cu -atoms will be produced respectively from two P atoms and five CuSO₄ molecules. As a result, the equation becomes—

2P + 5CuSO₄ + H₂O→ 5Cu + 2H₃PO₄ + H₂SO₄

To balance the number of SO²‾₄ radicals on both sides of the equation, 5 molecules of H₂SO₄ are to be added to the right-hand side ofthe equation.

2P + 5CuSO₄ + H₂O→5Cu + 2H₃PO₄ + 5H₂SO₄

Now, the total number of H-atoms present in 2 molecules of H₃PO₄ and 5 molecules of H₂SO₄ =16. So, for balancing the number of H-atoms, 8 water molecules are to be placed on the left-hand side. So, the balanced equation will be

2P + 5CuSO₄ + 8H₂O→5Cu + 2H₃PO₄ + 5H₂SO₄

9. Aluminium powder when boiled with caustic soda solution yields sodium aluminate and hydrogen gas.

Reaction:

Aluminium is oxidised in this reaction to produce sodium aluminate. On the other hand, the H -atoms of NaOH and H₂O are reduced to produce H₂. Therefore, the change in oxidation number in the reaction may be shown as follows—

The increase in oxidation number of A1 = (+3) -0 = 3 unit (oxidation), the decrease in oxidation number of 1 H atom of NaOH molecule \(=(+1)-\left(\frac{1}{2} \times 0\right)\) (reduction) and decrease in oxidation number of 2 H -atoms
of water molecule = 2 x (+1) -2×0 = 2 unit(reduction).

Hence, the total decrease in oxidation number for the Hatoms in 1 molecule of NaOH and 1 molecule of H₂O =3 unit.

Since, in a chemical reaction, the increase and decrease in oxidation number are the same, the ratio of the number of A1 atoms, NaOH molecule and water molecules in the given reaction should be =1: 1: 1.

⇒ \(\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NaAlO}_2+\frac{1}{2} \mathrm{H}_2+\mathrm{H}_2\)

Hence, the given reaction may be represented as:

Now, to express the number of molecules of reactants and products in terms of whole numbers, both sides of the equation should be multiplied by 2.

So, the balanced equation will be as follows:

⇒ \(\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NaAlO}_2+\frac{1}{2} \mathrm{H}_2+\mathrm{H}_2\)

Determination of equivalent mass of an element or compound in disproportionation reaction:

If in oxidation and reduction reaction, the change in oxidation number of any element or an element of any compound participating in a disproportionation reaction be n1 and n2 respectively and M be the molecular mass of that element or compound, then the equivalent mass of that element or compound \(=\frac{M}{n_1}+\frac{M}{n_2}\)

In oxidation reaction (P4— change in oxidation number of each P -atom = 1 unit. So the total change in oxidation number of four P-atoms = 4 × 1 =4 units.

In the reduction reaction, (P₄→PH₃), the change in oxidation number of each P-atom is 3 units. So the total change in oxidation number of four P-atoms = 4×3 = 12 units.

Thus in this reaction, the equivalent mass of P₄.

⇒ \(\frac{M}{4}+\frac{M}{12}=\frac{4 \times 31}{4}+\frac{4 \times 31}{12}=31+10.33 \text {= } 41.33\)

∴ The atomic mass of p = 31

Redox Titration

A process by which a standard solution of an oxidant (or a standard solution of a reductant) is completely reacted with a solution of an unknown concentration of a reductant (or with a solution of an unknown concentration of an oxidant) in the presence of a suitable indicator is called redox titration.

In a redox titration, an oxidant (or a reductant) reacts completely with an equivalent amount of a reductant (or an oxidant). Therefore, in a redox titration, the number of grams equivalent of oxidant = number of grams equivalent of reductant.

Types of redox titrations

1. Permanganometry titration

A titration in which KMnO₄ solution is used as the standard solution. In this titration, no indicators are needed.

Example:

The amount of iron present in an acidic ferrous ion (Fe²⁺) solution can be estimated by titrating the solution with a standard solution of KMnO₄.

Reaction:

MnO₄ + 5Fe²⁺ + 8H+ → Mn²⁺ + 5Fe3+ + 4H₂O …………………….(1)

Oxidation reaction: Fe²⁺→ Fe3+ + e

Reduction reaction: MnO₄ ^–+ 8H⁺ + 5e → Mn²⁺ + 4H₂O

So, in this reaction, the equivalent mass of Fe²⁺ — an atomic mass of Fe and the equivalent mass of KMnO₄ \(=\frac{1}{5} \times\) Molecular or formula mass of KMnO₄

According to the reaction (1), 1 mol of KMnO₄ = 5 mol of Fe²⁺ ions or, 1000 mLof 1 mol of KMnO₄ solution = 5 × 55.85g of Fe²⁺ ions or, 1 mLof l(M) KMnO₄ solution = 0.2792g of Fe²⁺ ions mL of 5(N) KMnO₄ solution = 0.2792g of Fe²⁺ ions. [In the given reaction, the normality of KMnO₄ solution is five times its molarity.]

lmL of (N) KMnO₄ solution = 0.05585g of Fe²⁺ ions

2.  Dichromatometry titration

A ptration in which a standard solution of potassium dichromate (K₂Cr₂O₇) is used.

In this titration, sodium or barium diphenylamine sulphonate or diphenylamine is used as an indicator.

Example:

The amount of iron present in an acidic ferrous ion (Fe²⁺) solution can be calculated by titrating the solution with a standard solution of K₂Cr₂O₇.

Reaction: 

Cr₂O₇+ 14H+ + 6Fe²⁺→ 2Cr³⁺+ 6Fe²⁺ + H₂O …………….(1)

In reaction (1), the equivalent mass of Fe²⁺ is equal to the atomic mass of Fe, & the equivalent mass of K₂Cr₂O₇ is equal to one-sixth of its molecular or formula mass.

According to the reaction (1), 1 mol of K₂Cr₂O₇ H 6mol of Fe²⁺ ions

Or, 1000mL of 1M K₂Cr₂O₇ = 6 × 55.85 g of Fe²⁺ ions

Or, lmLof1M K₂Cr₂O₇ solution s 0.3351g of Fe²⁺ ions

Or, lmL of 6N K₂Cr₂O₇ solution s 0.3351g of Fe²⁺ ions

[In the given reaction, the normality of K₂Cr₂O₇ solution is six times its molarity.]

lmL of IN K₂Cr₂O₇ solution = 0.05585g of Fe²⁺ ions

3. Iodometry titration

In this titration, KI in excess is added to a neutral or an acidic solution of an oxidant. Consequently, the oxidant quantitatively oxidises I^– ions (reductant), to form I₂. The liberated I₂ is then titrated with a standard Na₂S₂O₃ solution using starch as an indicator.

The amount of liberated iodine is calculated from the volume of standard Na₂S₂O₃ solution consumed in one titration. After the amount of liberated iodine is known, one can calculate the amount of oxidant by using the balanced chemical equation for the reaction of oxidant with iodine.

Example:

Iodometric titration is often used for quantitative estimation of Cu²⁺ ions. The addition of excess KI to a neutral or an acidic solution of Cu²⁺ ions results in oxidation of 1 to I2 and reduction of Cu²⁺ to Cu⁺.

So, in the reaction(l), the equivalent mass of \(\mathrm{Cu}^{2+}=\frac{2 \times \text { atomic mass of } \mathrm{Cu}}{2}=\text { atomic mass of } \mathrm{Cu}\)

The reaction of I₂ with Na₂S₂O₃ is:

⇒  \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

In this reaction, Oxidation reaction:

⇒  \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 e\)

Reduction reaction: I₂ + 2e → 2I^–

Therefore, the equivalent mass of Na₂S₂O₃

⇒ \(\frac{2 \times \text { molecular or formula mass of } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}{2}\)

= molecular or formula mass of Na₂S₂O₃

Equivalent mass of I₂ \(=\frac{\text { molecular mass of } \mathrm{I}_2}{2}=\text { atomic mass of } \mathrm{I}\)

According to the reactions (1) and (2), 2mol of Cu²⁺ = 1 mol of I2 and 2 mol of Na₂S₂O₃ s 1 mol of I₂

Therefore, 2mol of Na₂S₂O₃= 2 mol of Cu²⁺ or, 1 mol of Na₂S₂O₃ = l mol of Cu²⁺ = 63.5g of Cu²⁺ or, l mol of lM Na₂S₂O₃ solution = 63.5 g of Cu²⁺ or, 1 mol of IN Na₂S₂O₃ solution = 63.5gof Cu²⁺ [As in the reaction of Na₂S₂O₃ with I₂, the equivalent mass of Na₂S₂O₃ is equal to its molecular mass].

Question 1. The reaction Fe₂O₃(s) + 3CO(gH→2Fe(s) + 3CO₂(g) is carried out separately in a closed vessel and an open vessel. In which case do you expect a higher yield of CO₂(g)?
Answer:

If the reaction Fe₂O₃(s) + 3CO(g)→2Fe(s) + 3CO₂(g) is carried out in a closed vessel, an equilibrium is established between the reactants and the products. As a result, the reaction vessel always contains a mixture of reactants and products.

On the other hand, if the reaction is carried out in an open vessel, CO2(g) formed in the reaction diffuses out from the vessel and mixes with the air and does not have the opportunity to react with Fe(s).

As a result, the reaction becomes irreversible and goes to completion. At the end of the reaction, the vessel only contains Fe(s). Since the reaction reaches completion in an open vessel, the yield of CO₂(g) will be higher in this condition.

Question 2. If a reversible reaction is carried out in a closed vessel, the reactant(s) is/are never used up completely. Explain the reaction with an example.
Answer:

When a reversible reaction is carried out in a closed vessel, an equilibrium is established between the reactants and the products.

As a result, the reaction system always contains a mixture of reactants and products. In other words, a reversible reaction occurring in a closed vessel never gets completed. Consequently, the reactants are never used up.

Question 3. 2BrP₂(g) ⇌ Br₂(g) + 5F₂(g); At constant temperature, how the increase in pressure will affect the following at equilibrium— Equilibrium constant, Position of the equilibrium, and yield of the product?
Answer:

Pressure does not affect the magnitude of the equilibrium constant.

The given reversible reaction involves a decrease in several gas molecules in the backward direction. So, increasing pressure at the equilibrium of the reaction will favor the backward reaction Consequently, the yield of the product will decrease.

Question 4. Write the conjugate bases of the following acids and give the reason: HN₃, CH₃OH, [Al(H₂O)₆]³⁺, NH₄⁺, HPO₄^2-, H₂O₂, OH^–
Answer:

According to the Bronsted-Lowry concept, an acid is a substance that can donate protons. The species formed when an acid donates a proton is called the conjugate base of the acid. The conjugate base of an acid has one fewer H-atom than the acid.

Therefore, the conjugate bases of HN₃, CH₃OH, [Al(H₂O)₆]³⁺, NH₄⁺, HPO₄^2-, H₂O₂ and OH^– are N^–₃, CH₃O^–, [Al(H₂O)₅OH]²⁺, NH₃, PO^3-₄, HO^–₂ and O^-2, respectively.

Question 5. Write the conjugate acids of the following bases and give the reason: OH^–, H₂PO^–₄, O^2-, HS^–, SO^2-₃, H₂O, HCO₃^–, NH^–₂, NH₃, H^–, C₆H₅NH₂, S₂O₈^2- CO₃^2-
Answer

According to the Bronsted-Lowry concept, a base is a proton acceptor. When a base accepts a proton, the conjugate acid of the die base is formed. The conjugate acid of a base contains one more H-atom than the base.

Therefore, the conjugate acids of OH^–, H₂PO₄ , O^2-, HS^–, SO^2–₃, H₂O, HCO^–₃, NH^–₂ , NH₃, H^–, C₆H₅NH₂ , S₂O₈^2- and CO₃^2-– are H₂O , H₃PO₄ , OH^– , H₂S, HSO^–₃, H₃O⁺, H₂CO₃, NH₃, NH₄⁺+, H₂, C₆H₅NH⁺₃, HS₂O₈^– and HCO^–₃ respectively.

Question 6. To find out the equilibrium constant (K) of a reaction, it is compulsory to mention the balanced equation of the reaction—why?
Answer:

The value of the equilibrium constant (K) depends on how the balanced equation of the reaction is written, example the formation of H₂ (g) from H₂ (g) and I₂ (g) can be written in two different ways

⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \text { (2) } \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \rightleftharpoons \mathrm{HI}(g)\)

Both 1 and 2 express the same reaction but the coefficients of reactants and products are different. As a result, the value of the equilibrium constant (K) for the reaction will not be the same in the above two cases.

Question 7. Find Out K_p/K_c for the solutions \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g)\)
Answer:

We know that \(K_p=K_c(R T)^{\Delta n}.\)

For the given reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)

Thus \(K_p=K_c(R T)^{-\frac{1}{2}}\)

⇒ \(\frac{K_p}{K_c}=\frac{1}{\sqrt{R T}}\)

Question 8. By what factor will the concentration of H₃O⁺ ions in an aqueous solution be increased or decreased if its pH is increased by one unit?
Answer:

If the pH of an aqueous solution be x, then [H₃O⁺] in the solution = 10^-pH

= 10-x mol.L^-1

Increasing the pH of this solution by one unit makes

pH =  1 + x.So, (H₃O⁺) In the solution

= 10^-pH = 10^-(1+x)

= \(\frac{10^{-x}}{10} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

Question 9. Why is the ionic product of water at 50 greater than that at 25°C?
Answer:

Ionisation of water \(\left[2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right]\)

Is an endothermic process. The equilibrium constant referring to the ionization equilibrium of water is called the ionic product of water (K_w). As the process is endothermic, Kw increases with the temperature rise. Thus, Kw of water at 50 °C is greater than that at 25 °C.

Question 10. The pH of solution A is twice that of solution B. If the concentrations of H₃O⁺ ions in A and B are x (M) and y (M), respectively, then what is the relation between x and y?
Answer:

Given:

pH of the solution A = 2 × pH of the solution B,

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=x \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B=y \mathrm{~mol} \cdot \mathrm{L}^{-1} \text {. }\)

∴ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=2 \times-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B\)

Or, \(x=y^2 \text { or, } y=\sqrt{x}\)

 

• 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

Read and Learn More CBSE Class 11 Chemistry Notes

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).