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)

+6,+5

Answer 2

(i) When chlorine oxidises sulphur dioxide in presence of water, it gives $H _{2} SO _{4}$ as oxyacid $(A)$. The reaction occurs as follows:

$Cl _{2}+ SO _{2}+2 H _{2} O \longrightarrow H _{2} SO _{4}+2 HCl$

The oxidation number of sulphur in $H _{2} SO _{4}$ 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 $H _{2} SO _{4}$ is $(+) 6$.

(ii) When chlorine $\left( Cl _{2}\right)$ oxidises iodine $\left( I _{2}\right)$ in presence of water, it gives $HIO _{3}$ as oxyacid $(B)$, the reaction occurs as follows:

$SCl _{2}+ I _{2}+6 H _{2} O \longrightarrow 2 HIO _{3}+10 HCl$

The oxidation state of (I) in $HIO _{3}$ 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 $HIO _{3}$ is $=(+) 5$