Silver chloride dissolves in $N H_{4} O H$ , due to the formation of :

[A g (N H_{4})_{2}] C l

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[A g (N H_{3})_{4}] C l

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[A g (N H_{3})_{2}] C l

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[A g (N H_{3})_{2}] O H

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Solution

The correct options are
C $[A g (N H_{3})_{2}] O H$
D $[A g (N H_{3})_{2}] C l$
Reactions:
$A g C l + 2 N H_{4} O H \to [A g (N H_{3})_{2}] C l + 2 H_{2} O$
$A g_{2} O + 4 N H_{4} O H \to [A g (N H_{3})_{2}] O H + 3 H_{2} O$

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[A g (N H_{3})_{2}] C l

A g^{+} + 2 N H_{3} ⇌ A g (N H_{3})_{2}^{+}; K_{1} = 1.7 \times 10^{7}

A g^{+} + C l^{-} ⇌ A g C l; K_{2} = 5.4 \times 10^{9}

A g C l + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} + C l^{-}

A g^{+} + 2 N H_{3} ⇌ {[A g (N H_{3})_{2}]}_{3}^{+}

K_{1} = 1.7 \times 10^{7}

A g^{+} + C l^{-} ⇌ A g C l

K_{2} = 5.4 \times 10^{9}

A g C l + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} C l^{-}

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