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 ligands.
- 1 nitrosonium ion \((NO)\), which has a +1 charge.
- 1 sulfate ion \((SO_4^{2-})\), which has a -2 charge.

### Step 2: Assign oxidation states
- The oxidation state of water \((H_2O)\) is 0 since it is a neutral ligand.
- The oxidation state of the nitrosonium ion \((NO)\) is +1.
- The oxidation state of the sulfate ion \((SO_4^{2-})\) is -2.

### Step 3: Set up the equation for oxidation states
Let the oxidation state of iron (Fe) be \(x\). The overall charge of the complex is neutral (0), so we can set up the following equation:

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

### Step 4: Simplify the equation
This simplifies to:

\[ x + 1 - 2 = 0 \]

\[ x - 1 = 0 \]

### Step 5: Solve for \(x\)
Adding 1 to both sides gives:

\[ x = +1 \]

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

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