We need the salt which, on treatment with $\mathrm{MnO_2}$ and $\mathrm{H_2SO_4}$, liberates a greenish-yellow gas.

The greenish-yellow gas is characteristic of chlorine gas, $\mathrm{Cl_2}$.

Now check which salt can produce $\mathrm{Cl^-}$ ions that get oxidized by $\mathrm{MnO_2}$ in acidic medium.

In acidic medium, $\mathrm{MnO_2}$ is an oxidizing agent: $\mathrm{MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O}$

Chloride ions are oxidized as: $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$

Adding the two half-reactions: $\mathrm{MnO_2 + 4H^+ + 2Cl^- \rightarrow Mn^{2+} + Cl_2 + 2H_2O}$

With sulfuric acid and ammonium chloride: $\mathrm{MnO_2 + 2NH_4Cl + H_2SO_4 \rightarrow MnSO_4 + (NH_4)_2SO_4 + Cl_2 + 2H_2O}$

So $\mathrm{NH_4Cl}$ gives chlorine gas.

Check other options:

• B: $\mathrm{CaI_2}$
  Iodide gives iodine, not greenish-yellow gas.
• C: $\mathrm{KNO_3}$
  Nitrate does not give chlorine gas here.
• D: $\mathrm{NaBr}$
  Bromide gives bromine, which is reddish-brown, not greenish-yellow.

Therefore, the salt is: $\boxed{\mathrm{NH_4Cl}}$

Comparison with stored correct answer:

Stored correct answer = A
Derived answer = A
Hence, they agree.