A metal $M$ and its compound can give the observable changes as shown in the above sequence of reactions. The metal $M$ is

M g

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P b

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Z n

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S n

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Solution

The correct option is C $Z n$
Zinc reacts with dil nitric acid to form zinc nitrate.

$4 Z n + 10 H N O_{3} dil \to 4 Z n (N O_{3})_{2} + N H_{4} (N O_{3})_{2} + 3 H_{2} O$

Zinc nitrate reacts with sodium hydroxide to form sodium nitrate and white precipitate of zinc hydroxide.

$2 N a O H + Z n (N O_{3})_{2} \to 2 N a N O_{3} + Z n (O H)_{2} white precipitate$

Zinc hydroxide reacts with excess sodium hydroxide to form sodium zincate.

$Z n (O H)_{2} + 2 N a O H excess \to N a_{2} Z n O_{2} sodium zincate + 2 H_{2} O$

Sodium zincate reacts with hydrogen sulphide to form a white precipitate of zinc sulphide.

$N a_{2} Z n O_{2} sodium zincate + H_{2} S \to Z n S ↓ white precipitate + 2 N a O H + 2 H_{2} O$

Hence, the correct option is $C$

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M d i l u t e - -- \to H C l [Colorless solution] A a q u e o u s - ---- \to N a O H [White precipitate] B e x c e s s - ----- \to N a O H_{(a q)} [Colorless solution] C H_{2} S - ----- \to N a O H_{(a q)} [White precipitate] D

M

N a O H

M d i l u t e - -- \to H C l [Colorless solution] A a q u e o u s - ---- \to N a O H [White precipitate] B e x c e s s - ----- \to N a O H_{(a q)} [Colorless solution] C H_{2} S - ----- \to N a O H_{(a q)} [White precipitate] D

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P b, A u, M g, S n, F e, A l, A g, H g, N a, Z n, C u, C a

Zn {(N O_{3})}_{2}

Pb {(N O_{3})}_{2}

E_{Z n^{2 +}, Z n}^{\circ} = - 0.76 V, E_{P b^{2 +}, P b}^{\circ} = 0.13 V

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