The brown ring complex compound of iron is formulated as `[Fe (H_(2)O)_(5) (NO)] SO_(4)`. The oxidation state of iron is

Explanation

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

### Step 1: Identify the components of the complex
The complex consists of:
- Iron (Fe)
- Five water molecules (H₂O), which are neutral ligands
- One nitric oxide (NO) ligand

### Step 2: Assign oxidation states to the ligands
- The oxidation state of water (H₂O) is 0 since it is a neutral molecule.
- The oxidation state of nitric oxide (NO) is +1.

### Step 3: Write the equation for the oxidation state
Let the oxidation state of iron (Fe) be represented as \( X \). The overall charge of the complex must equal the charge of the sulfate ion (SO₄²⁻), which is -2.

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

### Step 4: Simplify the equation
This simplifies to:
\[ X + 1 = -2 \]

### Step 5: Solve for \( X \)
Now, we can solve for \( X \):
\[ X = -2 - 1 \]
\[ X = -3 \]

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