`Fe^(2+)._((aq))+NO_(3)^(Θ)._((aq))+H_(2)SO_(4)(conc.)rarr` Brown ring .The oxidation number of iron in brown ring complex is

Explanation

To determine the oxidation number of iron in the brown ring complex formed by the reaction of Fe²⁺, NO₃⁻, and concentrated H₂SO₄, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Components:**
The reactants are:
- Fe²⁺ (iron in +2 oxidation state)
- NO₃⁻ (nitrate ion)
- H₂SO₄ (concentrated sulfuric acid)

2. **Understand the Brown Ring Complex:**
The brown ring complex formed is represented as [Fe(H₂O)₅(NO)]²⁺. In this complex:
- Fe is the central metal ion.
- H₂O is a neutral ligand.
- NO acts as a ligand and is known as nitrosyl.

3. **Assign Oxidation States:**
- The oxidation state of NO (nitrosyl) in this complex is +1.
- The oxidation state of H₂O is 0 (as it is a neutral molecule).

4. **Set Up the Equation:**
Let the oxidation state of Fe in the complex be represented as \( x \). The overall charge of the complex is +2 due to the presence of the sulfate ion (SO₄²⁻), which has a charge of -2.

The equation for the sum of oxidation states in the complex can be set up as follows:
\[ x + 5(0) + 1 = +2 \]
Simplifying this gives:
\[ x + 1 = +2 \]

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

6. **Conclusion:**
The oxidation number of iron in the brown ring complex is +1.

### Final Answer:
The oxidation number of iron in the brown ring complex is +1.