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Covalent Bond

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Question

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(A) (black) + dil. $H C l △ ⟶ (B) (l)$ + (C)(g). gas (C) turns lead acetate paper black. (B) gives orange ppt (D) soluble in excess of KI forming (E). Above mentioned unknown compounds are (A) : HgS, (B) : $H g C l_{2}$ , (C) : $H_{2} S$ , (D) : $H g I_{2}$ , (E) : $K_{2} H g I_{4}$ .

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Solution

(A) is HgS, (B) is $H g C l_{2}$ , (C) is $H_{2} S$ , (D) is $H g I_{2}$ , and (E) is $K_{2} H g I_{4}$
When black coloured HgS is heated with dil HCl, mercuric chloride (B ) and hydrogen sulphide gas (C) is obtained.
$H g S + 2 H C l \to H g C l_{2} + 2 H_{2} S$
Hydrogen sulphide gas turns lead acetate paper black.
Mercury(II) ions react with KI to form orange precipitate of $H g I_{2}$ which is D. It is soluble in excess of KI to form E which is $K_{2} H g I_{4}$ .
$H g C l_{2} + 2 K I \to H g I_{2} ↓ o r a n g e p r e c i p i t a t e + 2 K C l$
$H g I_{2} + 2 K I (e x c e s s) \to K_{2} [H g I_{4}]$

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