White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of:

🧠 The Tollens / AgCl Recipe

Insoluble $\mathrm{AgCl}$ dissolves in aqueous $\mathrm{NH_3}$ because $\mathrm{Ag^+}$ is greedy for nitrogen donors. With ammonia present, $\mathrm{Ag^+}$ forms the diammine-silver(I) cation, $[\mathrm{Ag(NH_3)_2}]^+$ — linear, two-coordinate, very stable.

The released $\mathrm{Cl^-}$ stays in solution as the counter-ion. Net product: $[\mathrm{Ag(NH_3)_2}]\mathrm{Cl}$.

🗺️ Eliminate the Wrong Salts

• (1) $[\mathrm{Ag(NH_3)_4}]\mathrm{Cl_2}$ — would need $\mathrm{Ag^{2+}}$ (silver(I) is the only common Ag oxidation state in aqueous chemistry). ✗
• (2) $[\mathrm{Ag(Cl)_2(NH_3)_2}]$ — invokes a 4-coordinate neutral Ag(I) species; not the ammonia-dissolution product. ✗
• (3) $[\mathrm{Ag(NH_3)_2}]\mathrm{Cl}$ — the textbook diammine complex. ✓
• (4) $[\mathrm{Ag(NH_3)Cl}]\mathrm{Cl}$ — would need $\mathrm{Ag^{2+}}$. ✗

⚡ The "Ag(I) = 2-Coordinate Linear" Anchor

$\mathrm{Ag^+}$ is $\mathrm{d^{10}}$ and almost always forms 2-coordinate linear complexes: $[\mathrm{Ag(NH_3)_2}]^+$, $[\mathrm{Ag(CN)_2}]^-$, $[\mathrm{Ag(S_2O_3)_2}]^{3-}$. If you see $\mathrm{Ag^+}$ + ligand → expect two ligands, linear.

⚠️ Don't Promote Ag

Aqueous Ag(I) does not become Ag(II) just to fit 4 ligands. The only common oxidation state in solution is $+1$. Reject any option that demands Ag(+2) or higher.

$\boxed{\text{Answer: (3) } [\mathrm{Ag(NH_3)_2}]\mathrm{Cl}}$