Brown colour of the complex `[Fe(H_(2)O)_(5)(NO)]SO_(4)` is due to C. T. spectrum which causes momentary change in oxidation state. Find out oxidation state of Fe in this complex.

Answer

To find the oxidation state of iron (Fe) in the complex \([Fe(H_2O)_5(NO)]SO_4\), we can follow these steps:

### Step 1: Identify the components of the complex
The complex consists of:
- 5 water molecules \((H_2O)\), which are neutral (oxidation state = 0).
- 1 nitroso group \((NO)\), which has an oxidation state of +1.
- 1 sulfate ion \((SO_4^{2-})\), which has a charge of -2.

### Step 2: Set up the equation for the oxidation state
Let the oxidation state of Fe be \(X\). The overall charge of the complex must balance with the charge of the sulfate ion. The equation can be set up as follows:

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

### Step 3: Simplify the equation
Since the oxidation state of water is 0, we can simplify the equation:

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

This simplifies to:

\[ X + 1 = +2 \]

### Step 4: Solve for \(X\)
Now, we can solve for \(X\):

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

### Conclusion
The oxidation state of Fe in the complex \([Fe(H_2O)_5(NO)]SO_4\) is +1.

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