The brown ring complex compound `[Fe(H_2O)_5NO]SO_4` the oxidation state of iron is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

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

### Step 1: Identify the components of the compound
The compound consists of a complex ion \([Fe(H_2O)_5NO]^+\) and a sulfate ion \((SO_4)^{2-}\). The sulfate ion has a charge of -2.

### Step 2: Assign oxidation states to the ligands
- Water \((H_2O)\) is a neutral ligand, so its oxidation state is 0.
- The nitric oxide \((NO)\) ligand has an oxidation state of +1.

### Step 3: Set up the equation for the oxidation state of iron
Let the oxidation state of iron be \(X\). The total charge of the complex ion \([Fe(H_2O)_5NO]^+\) is +1. Therefore, we can write the equation as follows:

\[ X + 5(0) + 1 = +1 \]

### Step 4: Simplify the equation
Since the oxidation state of water is 0, the equation simplifies to:

\[ X + 1 = +1 \]

### Step 5: Solve for \(X\)
To find \(X\), we can rearrange the equation:

\[ X = +1 - 1 \]
\[ X = 0 \]

### Step 6: Conclusion
The oxidation state of iron in the brown ring complex \([Fe(H_2O)_5NO]SO_4\) is +1.

### Final Answer
The oxidation state of iron is +1 (Option A).

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