Chlorine oxidises sulphur dioxide in the pres | vedclass

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(C) $(i)$ When chlorine oxidises sulphur dioxide in the presence of water,it gives $H_2SO_4$ as oxyacid $A$. The reaction is: $Cl_2 + SO_2 + 2H_2O \lo

Vedclass · Accepted Answer

$+6, +5$

Vedclass · Answer

(C) $(i)$ When chlorine oxidises sulphur dioxide in the presence of water,it gives $H_2SO_4$ as oxyacid $A$. The reaction is: $Cl_2 + SO_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl$ In $H_2SO_4$,the oxidation state of $S$ is calculated as: $2(+1) + x + 4(-2) = 0$ $\Rightarrow x - 6 = 0$ $\Rightarrow x = +6$. $(ii)$ When chlorine oxidises iodine in the presence of water,it gives $HIO_3$ as oxyacid $B$. The reaction is: $5Cl_2 + I_2 + 6H_2O \longrightarrow 2HIO_3 + 10HCl$ In $HIO_3$,the oxidation state of $I$ is calculated as: $1 + x + 3(-2) = 0$ $\Rightarrow x - 5 = 0$ $\Rightarrow x = +5$. Thus,the oxidation states of $S$ and $I$ are $+6$ and $+5$ respectively. Therefore,option $C$ is correct.

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

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