Answer: A. Two tests narrow the identity:
1) Aqueous NaOH gives a green precipitate. This is the diagnostic test for \(\text{Fe}^{2+}\) — it forms a dirty-green \(\text{Fe(OH)}_2\) precipitate. \(\text{Fe}^{3+}\) would give a red-brown precipitate instead, eliminating B and D.
2) Heating the solid releases a gas which decolourises acidified KMnO₄ — i.e., a reducing gas. Sulfate decomposition gives \(\text{SO}_2\), which is a reducing agent. Nitrate decomposition gives \(\text{NO}_2\) and \(\text{O}_2\), neither of which decolourises KMnO₄ in the same way (KMnO₄ is already oxidising). This eliminates C.
Both tests point to \(\text{FeSO}_4\) → A.
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