QJEEM22Checeb: JEE Main 2022 Chemistry - Class 12 JEEM22Checeb Solution: In the following brown complex, the oxidation state of iron is…

To determine the oxidation state of iron in the brown complex ${[Fe({H_2}O)_5(NO)]^{2+}}$ , we need to analyze the ligands and their charges, as well as the overall charge of the complex.

Step 1: Understanding the Complex

The complex is:

${[Fe({H_2}O)_5(NO)]^{2+}}$

Iron (Fe): The central metal atom whose oxidation state we need to find.

Ligands:

5 Water Molecules ( $H_2O$ ): Neutral ligands (charge = 0).

1 Nitrosyl Group (NO): Can have different charges depending on its mode of bonding.

Step 2: Assigning Charges to Ligands

Water ( $H_2O$ ): Neutral ligand, so it contributes 0 to the overall charge.

Nitrosyl (NO): Can act as:

NO $^+$ (nitrosonium ion) with a +1 charge.

NO (neutral molecule) with 0 charge.

NO $^-$ (nitroxide ion) with a –1 charge.

Step 3: Setting Up the Oxidation State Equation

Let $x$ be the oxidation state of Fe.

The sum of the oxidation states of all components equals the overall charge of the complex:

$x + (5 \times 0) + (\text{Charge of NO}) = +2$

Simplifying:

$x + (\text{Charge of NO}) = +2$

Step 4: Determining the Charge of NO in the Complex

In the brown ring complex, experimental evidence shows that the nitrosyl ligand acts as NO $^+$ . This is because:

The NO ligand forms a linear bond with Fe, characteristic of NO $^+$ .

The complex is known to involve a reduction of the oxidation state of Fe to an unusual value.

Thus, Charge of NO = +1.

Step 5: Calculating the Oxidation State of Fe

Substitute the charge of NO into the equation:

$x + (+1) = +2$

Solving for $x$ :

$x = +2 - (+1) = +1$

Therefore, the oxidation state of Fe in the brown complex is $+1$ .

Conclusion

The oxidation state of iron in ${[Fe({H_2}O)_5(NO)]^{2+}}$ is +1.