Question

The valency factor of $I_2$ when, (i) it is formed by the reaction of potassium iodide and potassium iodate in acid medium and (ii) when it reacts with hypo, are respectively :

• A. $2,2$
• B. $\frac{5}{3},2$
• C. $\frac{3}{5},2$
• D. $5,2$

Answer 1

Answer: (B)

$(i)I{O}_3^{-}+I^{-}+H^{+}\to I_2+H_{2}O$
$I_2$ can be said to have undergone disproportionation.
$\therefore v.f.of{I}_2$ for oxidation, $n_1=2\left(5-0\right)=10$
and $v.f.of{I}_2$ for reduction, $n_2=2\left(0-(-1)\right)=2$
$\therefore v.f.of{I}_2$ (overall) $\frac{n_1{n}_2}{n_1+n_2}=\frac{10\times 2}{10\times 2}=\frac{5}{3}$
$\left(ii\right)I_2+2N{a}_2{S}_2{O}_3\to N{a}_2{S}_4{O}_6+2NaI$
$I_2+2e^{-}\to 21^{-}\Rightarrow v.f=2$