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  4. A metal gives two chlorides A and B. A gives black precipitate with NaOH and B gives white. With KI, B gives a red precipitate which is soluble in excess of KI. A and B are respectively

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Q. A metal gives two chlorides $A$ and $B$. A gives black precipitate with $NaOH$ and $B$ gives white. With $KI$, $B$ gives a red precipitate which is soluble in excess of $KI$. $A$ and $B$ are respectively

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Solution:

$Hg_{2}Cl_{2}+2NH_{4}OH \to $
(A) image$+NH_{4}Cl+2H_{2}O$
$\underset{(B)}{{HgCl_{2}}}+2NH_{4}OH \to \underset{\text{white ppt.}}{{HgNH_{2}Cl}}+NH_{4}Cl+2H_{2}O$
$\underset{(B)}{{HgCl_{2}}}+2KI \to \underset{\text{Red ppt.}}{{HgI_{2}}}+2KCl;$
$HgI_{2}+2KI \to \underset{\text{Soluble}}{{K_{2}[HgI_{4}]}}$