The brown ring complex compound is given as [Fe(H₂O)₅(NO)⁺]SO₄. To determine the oxidation state of iron (Fe) in this complex, we need to analyze the components and their charges:

1. Let the oxidation state of Fe be $$x$$.

2. The water molecules (H₂O) are neutral ligands; hence, they do not contribute to the overall charge.

3. The nitrosyl ligand (NO) in this complex is positively charged, $$NO^+$$, contributing a +1 charge.

4. The sulfate ion (SO₄) is doubly charged, with a -2 charge.

The overall charge of the complex compound, [Fe(H₂O)₅(NO)⁺], must balance the charge of the sulfate ion (SO₄):

Since the sulfate ion $$SO_4^{2-}$$ has a charge of -2, the complex ion must have a charge of +2 to balance it out:

Therefore:

$$ x + 5(0) + 1 = +2 $$

$$ x + 1 = +2 $$

$$ x = +1 $$

Thus, the oxidation state of iron (Fe) in the brown ring complex compound is +1.

The correct answer is:

Option A

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