Question

Silver chloride dissolves in excess of $$N{H}_{4}OH$$. The cation present in the solution is:
(a)- $$A{g}^{+}$$
(b)- $${\left[Ag{(N{H}_3)}_4\right]}^{+}$$
(c)- $${\left[Ag{(N{H}_3)}_2\right]}^{+}$$
(d)- $${\left[Ag{(N{H}_3)}_6\right]}^{+}$$

Answer 1

Hint: Silver chloride is a white crystalline solid with molecular formula AgCl. It is well known for its low solubility in water. Ammonium hydroxide ($$N{H}_{4}OH$$), commonly known as ammonia or ammonia water is a solution of ammonia in water.

Complete step by step solution:
Silver chloride reacts with ammonium hydroxide and yields diamine, silver chloride and water. Silver chloride is sparingly soluble in ammonia but dissolves in an excess of concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride very small solubility product value, some free silver ions are present in solution, which on addition of concentrated ammonia solution effectively consumes the free silver ion, as ammonia which is a good σ-electron pair donor ligand combines with the silver ion.
At equilibrium, ammonium hydroxide exists as per the above chemical equation. Here $$\stackrel{\bullet \bullet }{N}{H}_3$$
behaves as a ligand due to lone pairs present on nitrogen atoms.
$$N{H}_{4}OH\rightleftharpoons \stackrel{\bullet \bullet }{N}{H}_3+H^{+}+O{H}^{-}$$
Silver chloride dissociates as follows:$$AgCl\to A{g}^{+}+C{l}^{-}$$
Here, silver is a transition metal, having coordination number of 2. So, the overall reaction proceeds as follows:
$$AgCl+2N{H}_{4}OH\rightleftharpoons \left[Ag{(N{H}_3)}_2\right]Cl+2H_{2}O$$
The product $$\left[Ag{(N{H}_3)}_2\right]Cl$$ contains $${\left[Ag{(N{H}_3)}_2\right]}^{+}$$ as cation and $$C{l}^{-}$$ as anion.

So, the correct option is (c).

Note: The solubility product of silver chloride in ammonia solution is only $$2.9\times {10}^{-3}$$. Silver chloride can be used as an antidote for mercury poisoning. It assists in the elimination of mercury.