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Q.
Addition of $SnCl_{2}$ to $HgCl_{2}$ gives precipitate
AMUAMU 2004
Solution:
Stannous chloride $\left( SnCl _{2}\right)$ is a good reducing agent. It reduces mercuric chloride $\left( HgCl _{2}\right)$ to white ppt. of mercurous chloride $\left( Hg _{2} Cl _{2}\right)$ and then reduces $Hg _{2} Cl _{2}$ into grey ppt. of mercury. The following reaction takes place in two steps.
$SnCl _{2}+ HgCl _{2} \longrightarrow SnCl _{4}+\underset{\text { white ppt. }}{ Hg _{2} Cl _{2}} \downarrow$
$SnCl _{2}+ Hg _{2} Cl _{2} \longrightarrow SnCl _{4}+\underset{\text { grey ppt. }}{2 Hg } \downarrow$
Therefore, addition of $SnCl _{2}$ to $HgCl _{2}$ gives white $ppt$. turning to grey $ppt$.