When a solution of mercuric chloride treated with an excess of stannic chloride it produces, eventually

Options

A. `Hg_(2)CI_(2)`
B. `Hg_(2)SnCI_(6)`
C. Hg
D. `HgCI_(4)`

Answer

C

Explanation

Mercury (II) chloride solution is reduced by many reducing agents, e.g., formaldehyde, tin (II) chloride, sulphur diooxider, etc., precipitating with mercuryu (I) chloride first, which with excess of reducing agent turns black owing to the formation of metallic mercury.
`HgCl_(2)overset(SnCl_(2))tounderset("White")(Hg_(2)Cl_(2))overset(SnCl_(2))to underset("black")(Hg)`