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Zinc II Oxide

Colourless so...

Question

Colourless solution of (A) gives white precipitate (B) with $A g N O_{3}$ solution, soluble in aq. $N H_{3}$ . (A) also gives white precipiatete (C) with NaOH soluble in excess of it forming (D). (D) gives white precipiate (E) with $H_{2} S$ .(A), (B), (C), (D) and (E) are as follows.
(A) : $Z n C l_{2}$ (B) : $A g C l$ (C) : $Z n (O H)_{2}$ (D) : $N a_{2} Z n O_{2}$ (E) : $Z n S$

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Solution

(A) is $Z n C l_{2}$ , (B) is $A g C l$ , (C) is $Z n (O H)_{2}$ (D) is $N a_{2} Z n O_{2}$ and (E) is $Z n S$
Zinc chloride reacts with silver nitrate to form white precipitate of silver chloride which is (B).
$Z n C l_{2} + 2 A g N O_{3} \to 2 A g C l + Z n (N O_{3})_{2}$ .
Silver chloride is soluble in aqueous ammonia as it forms $[A g (N H_{3})_{2}] C l$ complex.
$A g C l + 2 N H_{3} \to [A g (N H_{3})_{2}] C l$ .
Zinc chloride reacts with sodium hydroxide to form white precipitate of zinc hydroxide which is (C). It is soluble in excess NaOH to form sodium zincate which is (D).
$Z n C l_{2} + 2 N a O H \to Z n (O H)_{2} ↓ + 2 N a C l$
$Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O - 2 s o d i u m z i n c a t e + 2 H_{2} O$
Sodium zincate reacts with hydrogen sulphide to form white precipitate of zinc sulphide.
$N a_{2} Z n O_{2} + H_{2} S \to Z n S ↓ + 2 N a O H$

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