A solution when diluted with water and boiled, gives a white precipitate. On addition of an excess of $N H_{4} C l / N H_{4} O H$ , the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in $N H_{4} O H / N H_{4} C l$ .

Z n {(O H)}_{2}

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A l {(O H)}_{3}

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M g {(O H)}_{2}

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C a {(O H)}_{2}

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Solution

The correct option is B $Z n {(O H)}_{2}$
A solution of $Z n^{2 +}$ ion reacts with water which on boiling gives a white precipitate of $Z n (O H)_{2}$ .

This precipitate dissolves in excess of $N H_{4} O H$ in the presence of $N H_{4} C I$ because $Z n (O H)_{2}$ on reaction with $N H_{4} O H$ in presence of $N H_{4} C I$ forms tetramine zinc (II) complex (soluble).

$Z n^{2 +} + 2 H_{2} O \to Z n (O H)_{2} ↓ W h i t e g e l a t i n o u s p p t + 2 H^{+}$

$Z n (O H)_{2} ↓ + 4 N H_{4} O H \to [Z n (N H_{3})_{4}]^{2 +} + 4 H_{2} O + 2 O H^{-}$

Hence, option A is correct.

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