Question

AgCl precipitate dissolves in $N{H_3}$ due to formation of:
A. $\left[ {Ag{{\left( {N\mathop H\nolimits_4 } \right)}_2}} \right]OH$
B. $\left[ {Ag{{\left( {N\mathop H\nolimits_4 } \right)}_2}} \right]Cl$
C. $\left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]Cl$
D. $\left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]OH$

Answer 1

Hint: The IUPAC name of a complex which $AgCl$ forms with $N{H_3}$ is diamminesilver (I) chloride. The complex formed is soluble unlike $AgCl$ which is insoluble in water.

Complete step by step solution:
We know that $AgCl$ is an insoluble or sparingly soluble salt in water. This happens because the force holding the solid $AgCl$ lattice together is too strong to overcome by the forces favoring the formation of hydrate ions,
The following equilibrium exists when $AgCl$ is added in the water:
$AgCl\left( s \right) \to \mathop {Ag}\nolimits^ + \left( {aq} \right) + \mathop {Cl}\nolimits^ - \left( {aq} \right)$ ----1
But white precipitate of $AgCl$ dissolves in $N{H_3}$ and forms colorless but soluble diamminesilver(I) complex ion, ${\left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]^ + }$.

In aqueous ammonia; $N{H_3}$ act as a ligand and it forms coordination complex with $\mathop {Ag}\nolimits^+$:
$\mathop {Ag}\nolimits^ + \mathop {Cl}\nolimits^ - \left( {aq} \right) + N\mathop H\nolimits_3 \left( {aq} \right) \rightleftharpoons \left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]Cl\left( {aq} \right)$ ---- 2
We can clearly see that reaction (2) lowers the concentration of $\mathop {Ag}\nolimits^+ $ and shifts the equilibrium of reaction (1) to the left and hence, dissolving more $AgCl$ in $N{H_3}$.
Also, the coordination complex ${\left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]^ + }$ has N-H bonds which can participate in hydrogen bonding with water. so, the complex is readily soluble in water even though $\mathop {Ag}\nolimits^+$ isn’t.
Hence from above, it can be said that option C is the correct option.
Additional information:
$\left[ {Ag{{\left( {N\mathop H\nolimits_3 } \right)}_2}} \right]Cl$ is ammoniacal silver chloride. With silvernatrate it is also called Tollens reagent. It is prepared by adding $N\mathop H\nolimits_4 OH$ to $AgN\mathop O\nolimits_3$.
When an aldehyde is heated with tollen’s reagent, the aldehyde reduces silver ions to metallic silver and a bright silver mirror is produced on the inner side of the test tube and the aldehyde is oxidized to corresponding carboxylate ion.

Note: It should be remembered that $AgCl$ do not dissolve in water but readily dissolve in $N{H_3}$ due to formation of coordination complex diamminesilver(I) chloride.
The color of solid $AgCl$ precipitate is yellow, because of the presence of chloride ion.