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)

$(i) IO^{-}_3+ I^{-} + H^{+} \to I_2+ H_2O$
$I_2$ 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\times2}{10\times2}=\frac{5}{3} $
$\left(ii\right)I_{2}+2Na_{2}S_{2}O_{3}\to Na_{2}S_{4}O_{6}+2NaI$
$I_{2}+2e^{-}\to21^{-} \Rightarrow v.f =2$