Question

A metal chloride $\left(X\right)$ shows the following reactions:
(i) When $H_{2}S$ is passed in an acidified aqueous solution of $\left(X\right)$ a black ppt. is obtained.
(ii) the precipitate obtained in step (i) is not soluble in yellow ammonium sulphide.
(iii) when a solution of stannous chloride is added to an aqueous solution of $\left(X\right)$, a white precipitate is obtained which turns grey on addition of more of stannous chloride.
(iv) when an aqueous solution of $Kl$ is added to an aqueous solution of $\left(X\right)$, a red precipitate is obtained which dissolves on addition of excess of $Kl$.
Identify $\left(X\right)$ .

• A. $X=HgCl$
• B. $X=HgC{l}_2$
• C. $X=H{g}_{2}C{l}_2$
• D. None of these

Answer 1

Answer: (B)

$X=HgC{l}_2$

With mercury (II) chloride solution, a white precipitate of mercury (I) chloride (calomel) is formed if a large amount of the regent is added quickely.

$S{n}^{2+}+2HgC{l}_2\to \underset{Whiteppt}{H{g}_{2}C{l}_2\downarrow }+S{n}^{4+}+2C{l}^{-}$

If however, tin(II) ions are in excess, the precipitate grey, especially on warnming, owing to further reduction to mercury metal.

$S{n}^{2+}+H{g}_{2}C{l}_2\downarrow \to \underset{greyppt}{2Hg\downarrow }+S{n}^{4+}+2C{l}^{-}$