$A g C l$ precipitate dissolves in $N H_{3}$ due to the formation of:

[A g (N H_{4})_{2}] O H

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[A g (N H_{4})_{2}] 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 option is C $[A g (N H_{3})_{2}] C l$
$A g C l + 2 N H_{3} ⟶ [A g (N H_{3})_{2}] C l$
White ppt. of $A g C l$ dissolves in $N H_{3}$ due to the formation of complex diammine silver chloride.

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Similar questions

N H_{4} O H

A g C l

N H_{3}

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

A g C l

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

A g 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^{-}

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