Chlorine oxidises sulphur dioxide in the presence of water to give an oxyacid $A$. Chlorine also oxidises…
Chlorine oxidises sulphur dioxide in the presence of water to give an oxyacid $A$. Chlorine also oxidises iodine in the presence of water to give an oxyacid $B$. The oxidation states of $\mathrm{S}$ and I in $A$ and $B$ are respectively
+4, + 5
+6, + 3
+6, + 5
+4, + 7
Solution
(i) When chlorine oxidises sulphur dioxide in presence of water, it gives $\mathrm{H}_2 \mathrm{SO}_4$ as oxyacid $(A)$. The reaction occurs as follows :
$\mathrm{Cl}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{HCl}$
The oxidation number of sulphur in $\mathrm{H}_2 \mathrm{SO}_4$ is
$2+x+(4 x-2)=0$
or, where $x=$ oxidation state of sulphur (S).
$x+2-8=0 \Rightarrow x=+6$
Hence, oxidation state of (S) in $\mathrm{H}_2 \mathrm{SO}_4$ is (+) 6.
(ii) When chlorine $\left(\mathrm{Cl}_2\right)$ oxidises iodine $\left(\mathrm{I}_2\right)$ in presence of water, it gives $\mathrm{HIO}_3$ as oxyacid $(B)$, the reaction occurs as follows:
$5 \mathrm{Cl}_2+\mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HIO}_3+10 \mathrm{HCl} \text {. }$
The oxidation state of (I) in $\mathrm{HIO}_3$ is :
Let oxidation state of $(\mathrm{I})=x$
$\begin{aligned}
& 1+x+3 \times(-2)=0 \\
& x+1-6=0 \Rightarrow x=+5
\end{aligned}$
Hence, oxidation state of iodine (I) in $\mathrm{HIO}_3$ is $=(+) 5$
Hence, option (c) is the correct answer.