Question

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 $S$ and $I$ in $A$ and $B$ are respectively

• A. +4 , + 5
• B. +6,+3
• C. +6,+5
• D. +4,+7

Answer 1

Answer: (C)

(i) When chlorine oxidises sulphur dioxide in presence of water, it gives $H_{2}S{O}_4$ as oxyacid $(A)$. The reaction occurs as follows:

$C{l}_2+S{O}_2+2H_{2}O⟶H_{2}S{O}_4+2HCl$

The oxidation number of sulphur in $H_{2}S{O}_4$ is

$2+x+(4\times -2)=0$

or, where $x=$ oxidation state of sulphur $(S)$.

$x+2-8=0\Rightarrow x=+6$

Hence, oxidation state of (S) in $H_{2}S{O}_4$ is $(+)6$.

(ii) When chlorine $\left(C{l}_2\right)$ oxidises iodine $\left(I_2\right)$ in presence of water, it gives $HI{O}_3$ as oxyacid $(B)$, the reaction occurs as follows:

$SC{l}_2+I_2+6H_{2}O⟶2HI{O}_3+10HCl$

The oxidation state of (I) in $HI{O}_3$ is :

Let oxidation state of $(I)=x$

$1+x+3\times (-2)=0$

$x+1-6=0\Rightarrow x=+5$

Hence, oxidation state of iodine (I) in $HI{O}_3$ is $=(+)5$