Among the following, the brown ring is due to the formation of

When a freshly prepared solution of $$FeSO_4$$ is mixed with $$NaNO_3$$ and then concentrated $$H_2SO_4$$ is allowed to flow down the side of the test-tube, the classic “brown-ring” test for nitrate ions is performed.
The brown ring appears exactly at the junction of the two layers because a brown coloured nitrosyl-iron(II) complex is formed.

Step-1 Nitrate is reduced by $$Fe^{2+}$$ present in $$FeSO_4$$ to nitric oxide (NO):
$$\ce{NO_3^- + 3\,Fe^{2+} + 4\,H^+ \rightarrow 3\,Fe^{3+} + NO + 2\,H_2O}$$

Step-2 The freshly liberated NO immediately coordinates with the excess $$Fe^{2+}$$ still present in the lower layer (which is rich in $$FeSO_4$$) along with five water ligands to give the brown coloured complex:
$$\ce{[Fe(H_2O)_6]^{2+} + NO \;\;\longrightarrow\;\; [Fe(H_2O)_5(NO)]^{2+} + H_2O}$$

The coordination entity responsible for the colour is therefore
$$[Fe(NO)(H_2O)_5]^{2+}$$

Checking the given options:

• Option A $$[Fe(NO)_2(SO_4)_2]^{2-}$$ contains two NO ligands and sulphate ligands; such a species is not produced in the test.
• Option B $$[Fe(NO)_2(H_2O)_4]^{3+}$$ again has two NO ligands and the wrong charge.
• Option C $$[Fe(NO)_4(SO_4)_2]$$ is an unrealistically crowded tetranitrosyl species.
• Option D $$[Fe(NO)(H_2O)_5]^{2+}$$ matches the experimentally established composition of the brown-ring complex.

Hence, the brown ring is due to the formation of
Option D which is: $$[Fe(NO)(H_2O)_5]^{2+}$$.