In $\mathrm{MnO}_4^{2-}$, the oxidation number of Mn is +6 . It can increase its oxidation number (to +7 ) or decrease its oxidation number (to $+4,+3,+2,0$ ). Hence, it undergoes disproportionation reaction in acidic medium.

In $\mathrm{MnO}_4^{-}, \mathrm{Mn}$ is in its highest oxidation state, i.e., +7 . It can only decrease its oxidation number. Hence, it cannot undergo disproportionation reaction.

Writing the oxidation number of each element above its symbol in the following reactions

(a) $$\mathop {2\mathop {PbO}\limits^{ + 2\,\,\,\,\, - 2} }\limits_{Basic\,oxide} + \mathop {4\mathop {HCl}\limits^{ + 1\,\,\,\,\, - 1} }\limits_{Acid} \buildrel {} \over \longrightarrow \mathop {2\mathop {PbC{l_2}}\limits^{ + 2\,\,\,\,\, - 1} }\limits_{} + \mathop {2\mathop {{H_2}O}\limits^{ + 1\,\,\,\,\, - 2} }\limits_{} $$

In this reaction, oxidation number of each element remains same hence, it is not a redox reaction. In fact, it is an example of acid-base reaction.

(b) $$\mathop {\mathop {Pb{O_2}}\limits^{ + 4\,\,\,\,\, - 2} }\limits_{} + \mathop {4\mathop {HCl}\limits^{ + 1\,\,\,\,\, - 1} }\limits_{} \buildrel {} \over \longrightarrow \mathop {\mathop {PbC{l_2}}\limits^{ + 2\,\,\,\,\, - 1} }\limits_{} + \mathop {\mathop {C{l_2}}\limits^0 + 2\mathop {{H_2}O}\limits^{ + 1\,\,\,\,\, - 2} }\limits_{} $$

In $\mathrm{PbO}_2, \mathrm{~Pb}$ is in +4 oxidation state. Due to inert pair effect Pb in +2 oxidation state is more stable. $\mathrm{So}, \mathrm{Pb}$ in +4 oxidation state $\left(\mathrm{PbO}_2\right)$ acts as an oxidising agent. It oxidises $\mathrm{Cl}^{-}$to $\mathrm{Cl}_2$ and itself gets reduced to $\mathrm{Pb}^{2+}.$

PbO is a base. It reacts with nitric acid and forms soluble lead nitrate.

$\mathrm{PbO}+2 \mathrm{HNO}_3 \longrightarrow \underset{\text { Soluble }}{\mathrm{Pb}\left(\mathrm{NO}_3\right)_2}+\mathrm{H}_2 \mathrm{O}\quad \text{(acid base reaction)}$

Nitric acid does not react with $\mathrm{PbO}_2$. Both of them are strong oxidising agents. In $\mathrm{HNO}_3$, nitrogen is in its maximum oxidation state ( +5 ) and in $\mathrm{PbO}_2$, lead is in its maximum oxidation state $(+4)$. Therefore, no reaction takes place.

(a) Ion electron method Write the skeleton equation for the given reaction. $\mathrm{MnO}_4^{-}(a q)+\mathrm{SO}_2(g) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{HSO}_4^{-}(a q)$

Find out the elements which undergo change in O.N.

Divide the given skeleton into two half equations.

Reduction half equation : $\mathrm{MnO}_4^{-}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$

Oxidation half equation: $\mathrm{SO}_2(\mathrm{~g}) \longrightarrow \mathrm{HSO}_4^{-}(a q)$

To balance reduction half equation

In acidic medium, balance H and O -atoms

$$\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{H}_2 \mathrm{O}(l)$$

To balance the complete reaction

$$\begin{aligned} & 2 \mathrm{MnO}_4(a q)+16 \mathrm{H}^{+}(a q)+10 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l) \\ & 5 \mathrm{SO}_2(g)+10 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 5 \mathrm{HSO}_4^{-}(a q)+15 \mathrm{H}^{+}(a q)+10 e^{-} \end{aligned}$$

$2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{SO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{+}(a q) \rightarrow 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{HSO}_4^{-}(a q)$

(b) Oxidation number method Write the skeleton equation for the given reaction.

$$\mathrm{N}_2 \mathrm{H}_4(l)+\mathrm{ClO}_3^{-}(a q) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}^{-}(g)$$

O.N. increases by 4 per N -atom

Multiply NO by 2 because in $\mathrm{N}_2 \mathrm{H}_4$ there are 2 N atoms

$$\mathrm{N}_2 \mathrm{H}_4(l)+\mathrm{ClO}_3^{-}(a q) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}^{-}(a q)$$

Total increase in O.N. of $\mathrm{N}=2 \times 4=8$ ( $8 \mathrm{e}^{-}$ lost) Total decrease in $\mathrm{O} . \mathrm{N}$. of $\mathrm{Cl}=1 \times 6=6\left(6 e^{-}\right.$gain $)$

Therefore, to balance increase or decrease in O.N. multiply $\mathrm{N}_2 \mathrm{H}_4$ by $3,2 \mathrm{NO}$ by 3 and $\mathrm{ClO}_3^{-}, \mathrm{Cl}^{-}$ by 4

$$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)$$

Balance O and H -atoms by adding $6 \mathrm{H}_2 \mathrm{O}$ to RHS

$$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)+6 \mathrm{H}_2 \mathrm{O}(l)$$

(c) Ion electron method Write the skeleton equation for the given reaction.

$$\mathrm{Cl}_2 \mathrm{O}_7(g)+\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{ClO}_2^{-}(\mathrm{aq})+\mathrm{O}_2(g)$$

Find out the elements which undergo a change in O.N.

Divide the given skeleton equation into two half equations.

Reduction half equation : $\mathrm{Cl}_2 \mathrm{O}_7 \longrightarrow \mathrm{ClO}_2^{-}$

Oxidation half equation : $\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{O}_2$

To balance the reduction half equation

$$\mathrm{Cl}_2 \mathrm{O}_7(g)+6 \mathrm{H}^{+}(a q)+8 \mathrm{e}^{-} \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)$$

To balance the oxidation half equation

$$\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{O}_2(g)+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

To balance the complete reaction

$$\begin{aligned} \mathrm{Cl}_2 \mathrm{O}_7(g)+6 \mathrm{H}^{+}(a q)+8 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \\ 4 \mathrm{H}_2 \mathrm{O}_2(a q) & \longrightarrow \mathrm{O}_2(g)+8 \mathrm{H}^{+}(a q)+8 e^{-} \end{aligned}$$

$$ \mathrm{Cl}_2 \mathrm{O}_7(g)+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)+4 \mathrm{O}_2(g)+2 \mathrm{H}^{+}+(a q)$$

This represents the balanced redox reaction.

(a) Suppose that the $\mathrm{O} . \mathrm{N}$. of P in $\mathrm{HPO}_3^{2-}$ be $x$.

$$\begin{aligned} \text{Then,}\quad 1+x+3(-2) & =-2 \\ \text{or,}\quad x+1-6 & =-2 \\ \text{or,}\quad x & =+3 \end{aligned}$$

(b) Suppose that the O.N. of P in $\mathrm{PO}_4^{3-}$ be $x$.

Then, $x+4(-2)=-3$

or, $x-8=-3$

or, $x=+5$