JEE Main 2022 — Coordination Compounds Question with Solution

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 ($H_{2}O$): 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 ($H_{2}O$): 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.