1. In the brown ring test for nitrate ion, the characteristic brown complex formed is:

$[\mathrm{Fe(H_2O)_5(NO)}]SO_4$

So, the complex cation is:

$[\mathrm{Fe(H_2O)_5(NO)}]^{2+}$

2. Let the oxidation number of iron be $x$.

• Water is a neutral ligand, so $5\,\mathrm{H_2O}$ contributes $0$ charge.
• In this complex, the nitrosyl ligand is treated as $\mathrm{NO}^+$.
• Total charge on the complex ion is $+2$.

Therefore,

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

$x = +1$

3. Hence, the oxidation number of iron in the brown ring complex is:

$\boxed{+1}$

4. Comparison with stored answer:

Stored correct answer = $1$

This matches the derived answer.