`Fe^(2+) (aq)+NO_(3)^(-) (aq)+H_(2)SO_(4) (conc.) to` Brown ring
The brown ring is due to the formation of complex, `[Fe(H_(2)O)_(5)NO]SO_(4)`. What is the oxidation state of iron in the complex ?

Explanation

To determine the oxidation state of iron in the complex `[Fe(H₂O)₅NO]SO₄`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Complex**: The complex given is `[Fe(H₂O)₅NO]SO₄`. Here, we need to focus on the iron ion (Fe) and its coordination with water (H₂O) and nitric oxide (NO).

2. **Assign Charges to Ligands**:
- **Water (H₂O)**: This is a neutral ligand, so it has a charge of 0.
- **Nitric Oxide (NO)**: This is a neutral ligand as well, but it can have a +1 charge when it acts as a ligand in complexes. Here, we will consider it as +1 for the purpose of calculating the oxidation state.

3. **Identify the Charge of the Sulfate Ion**: The sulfate ion (SO₄²⁻) has a charge of -2.

4. **Set Up the Equation for Oxidation State**:
- Let the oxidation state of iron be \( x \).
- The total charge of the complex ion `[Fe(H₂O)₅NO]` must balance with the charge of the sulfate ion.
- The equation can be set up as follows:
\[ x + (5 \times 0) + 1 = 2 \]
- Here, \( 5 \times 0 \) accounts for the five water molecules, which contribute no charge.

5. **Solve for \( x \)**:
- Simplifying the equation gives:
\[ x + 1 = 2 \]
- Rearranging this gives:
\[ x = 2 - 1 \]
\[ x = 1 \]

6. **Conclusion**: The oxidation state of iron (Fe) in the complex `[Fe(H₂O)₅NO]SO₄` is +1.

### Final Answer:
The oxidation state of iron in the complex is +1.