1. Identify the curdy white precipitate formed with $\mathrm{AgNO_3}$

A chloride salt solution, when acidified with dilute $\mathrm{HNO_3}$ and treated with $\mathrm{AgNO_3}$, gives the characteristic white curdy precipitate of silver chloride:

$\mathrm{Ag^+ + Cl^- \rightarrow AgCl\,(s)}$

So,

$A = \mathrm{AgCl}$

2. Action of $\mathrm{NH_4OH}$ on $\mathrm{AgCl}$

Silver chloride dissolves in aqueous ammonia due to formation of the diamminesilver(I) complex:

$\mathrm{AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-}$

Thus the clear solution contains:

$\mathrm{[Ag(NH_3)_2]Cl}$

So,

$B = \mathrm{[Ag(NH_3)_2]Cl}$

3. Match with options

The pair is:

• $A = \mathrm{AgCl}$
• $B = \mathrm{[Ag(NH_3)_2]Cl}$

This corresponds to Option B.

4. Why other options are incorrect

• A: $\mathrm{H[AgCl_3]}$ is not the precipitate formed in this test.
• C: First species is incorrect.
• D: $\mathrm{(NH_4)[Ag(OH)_2]}$ is not the expected product with ammonia; the actual dissolution is due to the ammine complex $\mathrm{[Ag(NH_3)_2]^+}$.

Therefore, the correct answer is B.