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A metal gives two chlorides A and B. A gives black precipitate with ${NH}_{4}$ OH and B gives white. With KI, B gives a red precipitate soluble in excess of KI. A and B are respectively

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${HgCl}_{2} and {Hg}_{2} {Cl}_{2}$

${Hg}_{2} {Cl}_{2} and {HgCl}_{2}$

${HgCl}_{2} and {ZnCl}_{2}$

${ZnCl}_{2} and {HgCl}_{2}$

answer is B.

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Detailed Solution

$\underset{(A)}{H g_{2} C l_{2}} + 2 N H_{4} O H \to_{}^{} \underset{(B l a c k)}{[H_{2} N - H g - C l + H g]} ↓ + N H_{4} C l + 2 H_{2} O$

$\begin{array}{l} {HgCl}_{2} + 2 {NH}_{4} OH \to_{}^{} Hg ({NH}_{2}) Cl + {NH}_{4} Cl + H_{2} O \\ (B) (W h i t e) \end{array}$

${HgCl}_{2} + 2 KI ⟶ \underset{(Red)}{{HgI}_{2}} ↓ + 2 KCl$

${HgI}_{2} + 2 KI (excess) ⟶ K_{2} ({HgI}_{4})$

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