The oxidation state of Fe in brown complex `[Fe(H_(2)O)_(5)NO]SO_(4)` is

Answer

To determine the oxidation state of Fe in the complex \([Fe(H_2O)_5NO]SO_4\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the components of the complex**: The complex consists of five water molecules (H2O), one nitric oxide (NO), and a sulfate ion (SO4).

2. **Determine the oxidation states of the ligands**:
- Water (H2O) is a neutral ligand, which means its oxidation state is 0.
- Nitric oxide (NO) is a neutral ligand but can also be considered as a positive ligand in some contexts. For this problem, we will consider its oxidation state as +1.
- The sulfate ion (SO4) has an oxidation state of -2.

3. **Set up the equation**: Let the oxidation state of Fe be \(x\). The overall charge of the complex \([Fe(H_2O)_5NO]SO_4\) is neutral, which means the sum of the oxidation states must equal 0.

The equation can be set up as follows:
\[ x + (5 \times 0) + (+1) + (-2) = 0 \]

4. **Simplify the equation**:
\[ x + 0 + 1 - 2 = 0 \]
\[ x - 1 = 0 \]

5. **Solve for \(x\)**:
\[ x = +1 \]

6. **Conclusion**: The oxidation state of Fe in the complex \([Fe(H_2O)_5NO]SO_4\) is +1.