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 compound
The compound consists of a cationic part \([Fe(H_2O)_5NO]^2+\) and an anionic part \(SO_4^{2-}\). The cationic part contributes a total charge of +2.

### Step 2: Determine the charge contributions
- The water molecules \((H_2O)\) are neutral, contributing a charge of 0.
- The nitric oxide \((NO)\) in the complex has a known oxidation state of +1.

### Step 3: Set up the equation for oxidation states
Let the oxidation state of iron be \(X\). The total charge of the cationic part can be expressed as:
\[ X + 5(0) + (+1) = +2 \]
This simplifies to:
\[ X + 1 = +2 \]

### Step 4: Solve for \(X\)
Rearranging the equation gives:
\[ X = +2 - 1 \]
\[ X = +1 \]

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

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