JEE Chemistry 2024 Question with Solution

Given: The complex ions are $P = [FeF_6]^{3-}$, $Q = [V(H_2O)_6]^{2+}$, and $R = [Fe(H_2O)_6]^{2+}$.

Find: The correct increasing order of their spin-only magnetic moments.

The spin-only magnetic moment is given by

$$\mu_s = \sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

For $P = [FeF_6]^{3-}$: iron is in the $+3$ oxidation state, so it is $Fe^{3+} = 3d^5$. Since $F^-$ is a weak field ligand, the complex is high spin and all $5$ electrons remain unpaired.

$$\mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.}$$

For $Q = [V(H_2O)_6]^{2+}$: vanadium is in the $+2$ oxidation state, so it is $V^{2+} = 3d^3$. Water is a weak field ligand, so there are $3$ unpaired electrons.

$$\mu_s = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.}$$

For $R = [Fe(H_2O)_6]^{2+}$: iron is in the $+2$ oxidation state, so it is $Fe^{2+} = 3d^6$. Water is a weak field ligand, so the complex is high spin with $4$ unpaired electrons.

$$\mu_s = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.}$$

Therefore, the order of magnetic moments is $Q < R < P$.

The correct option is C.