Question

(A) (black) + dil.$HCl\underset{⟶}{△}\left(B\right)\left(l\right)$ + (C)(g). gas (C) turns lead acetate paper black. (B) gives orange ppt (D) soluble in excess of KI forming (E).

State True or False.

(A) is $HgS$.

• A. True
• B. False

Answer 1

The correct option is A True

When black coloured HgS is heated with dil HCl, mercuric chloride (B ) and hydrogen sulphide gas (C) is obtained.
$HgS+2HCl\to HgC{l}_2+2H_{2}S$
Hydrogen sulphide gas turns lead acetate paper black.
Mercury(II) ions react with KI to form orange precipitate of $Hg{I}_2$ (D). It is soluble in excess of KI to form (E) $K_{2}Hg{I}_4$.