JEE Advanced 2020 — Paper 2 — Question 26

(A) $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$, therefore $\mathrm{SnCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ is a Lewis acid.

(B) $\mathrm{SnO}_{2}$ is amphoteric, so it will dissolve in KOH solution forming $\mathrm{K}_{2} \mathrm{SnO}_{3}$ or $\mathrm{K}_{2}\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]$.

$$\mathrm{SnO}_{2}+2 \mathrm{KOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{~K}_{2}\left[\mathrm{Sn}(\mathrm{OH})_{6}\right]$$

(C) $\mathrm{PbCl}_{2}$ will form a complex with HCl

$\mathrm{PbCl}_{2}(\mathrm{~s})+2 \mathrm{HCl} \longrightarrow \mathrm{H}_{2}\left[\mathrm{PbCl}_{4}\right]$

So, we may assume that now this complex will dissociate as: $\mathrm{H}_{2}\left[\mathrm{PbCl}_{2}\right] \longrightarrow 2 \mathrm{H}^{+}+\left[\mathrm{PbCl}_{4}\right]^{2-}$

So, considering this possibility, option (C) should be incorrect.

(D) $\mathrm{Pb}_{3} \mathrm{O}_{4}$ reacts with conc. $\mathrm{HNO}_{3}$ to form $\mathrm{PbO}_{2}$ but the reaction is not a redox reaction.

or $\mathrm{Pb}_{3} \mathrm{O}_{4}$

So, option (D) is incorrect.