The brown ring complex compound is formulated as `[Fe(H_(2)O_(5))No]SO_(4)`. The oxidation state of `Fe` is

Options

A. `1`
B. `2`
C. `3`
D. `0`

Explanation

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

### Step 1: Identify the components of the complex
The complex can be broken down into its components:
- The cation part: \([Fe(H_2O)_5NO]^{2+}\)
- The anion part: \(SO_4^{2-}\)

### Step 2: Write the oxidation states of known components
- Water (\(H_2O\)) is a neutral molecule, so its oxidation state is 0.
- The nitric oxide (\(NO\)) in this complex is known to have an oxidation state of +1.

### Step 3: Set up the equation for the oxidation state of Fe
Let the oxidation state of Fe be \(X\). The total charge of the complex cation \([Fe(H_2O)_5NO]^{2+}\) is +2. We can express this relationship mathematically:

\[ X + (5 \times 0) + 1 = +2 \]

### Step 4: Solve for X
Now, we can simplify the equation:

\[ X + 0 + 1 = 2 \]
\[ X + 1 = 2 \]
\[ X = 2 - 1 \]
\[ X = +1 \]

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