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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $[A g {(N H_{3})}_{2}] C l$
Option (C) is correct.
$A g C l$ precipitate dissolves in $N H_{3}$ due to the formation of $[A g {(N H_{3})}_{2}] C l$
$A g C l$ is soluble in ammonia because it form complexes $[A g {(N H_{3})}_{2}] C l$ . The formation of the complex $[A g (N H_{3})_{2}]^{+}$ ion makes it behave like ionic therefore making it soluble.If we add nitric acid it will break the coordinate bond between silver and ammonia molecule, the ammonia will react with nitric acid to form ammonium nitrate and the white precipitate of silver chloride will reappear.

Suggest Corrections

Similar questions

{N H}_{3}

1

0.1 M

A g {N O}_{3}

{A g}^{+}

2 \times 10^{- 7} M

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

K_{d i s s .} = \frac{[{A g}^{+}] {[{N H}_{3}]}^{2}}{[A g {({N H}_{3})}_{2}^{+}]} = 6.8 \times 10^{- 8}

AgCl dissolved in excess $N H_{3}$ , $K C N$ and $N a_{2} S_{2} O_{3}$ solutions the complex produce ions.

A g C l (K_{s p} = 1.8 \times 10^{- 10})

K C N, N H_{3}

N a_{2} S_{2} O_{3}

K [A g {(C N)}_{2}]

K_{4} [F e {(C N)}_{6}]

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

[C r {(N H_{3})}_{6}] {C l}_{3}

Join BYJU'S Learning Program