JEE Chemistry 2023 Question with Solution

Given: The complexes are $[Fe(H_2O)_6]^{3+}$ and $[Fe(CN)_6]^{3-}$.

Find: The spin only magnetic moments of Fe in both complexes.

For $Fe^{3+}$, the electronic configuration is $[Ar]3d^5$.

In $[Fe(H_2O)_6]^{3+}$, $H_2O$ is a weak field ligand, so no pairing occurs. Hence the number of unpaired electrons is $n=5$.

The spin-only magnetic moment is

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

Substituting $n=5$,

$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{BM}$$

In $[Fe(CN)_6]^{3-}$, $CN^-$ is a strong field ligand, so pairing occurs. Hence the number of unpaired electrons is $n=1$.

Again using

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

Substituting $n=1$,

$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \, \text{BM}$$

Therefore, the spin-only magnetic moments are $5.92 \, \text{BM}$ and $1.732 \, \text{BM}$ respectively. The correct option is C.

Given: Both complexes contain $Fe^{3+}$.

Find: How ligand strength changes the number of unpaired electrons and hence the magnetic moment.

Neutral Fe has configuration $[Ar]3d^64s^2$. Removing three electrons gives

$$Fe^{3+} : [Ar]3d^5$$

For $[Fe(H_2O)_6]^{3+}$, $H_2O$ is a weak field ligand, so the complex is high spin. Thus all five $3d$ electrons remain unpaired.

For $[Fe(CN)_6]^{3-}$, $CN^-$ is a strong field ligand, so the complex is low spin. The electrons pair as much as possible, leaving only one unpaired electron.

Now apply

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

For the first complex, $n=5$ gives $\mu \approx 5.92 \, \text{BM}$. For the second complex, $n=1$ gives $\mu \approx 1.732 \, \text{BM}$.

Hence the correct choice is C. The solution says option A, but its worked values clearly match option C. Therefore the worked solution determines the answer.