JEE Chemistry 2025 Question with Solution

Given: We need the complex for which $\Delta_{0}=0$ and $\mu=5.96$ B.M.

Find: Which option has the required magnetic moment and zero crystal field stabilization energy.

Use the spin-only formula:

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

where $n$ is the number of unpaired electrons.

A magnetic moment near $5.96$ B.M. corresponds to $n=5$ unpaired electrons because

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

which matches the required value closely.

Now examine the options:

1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$: Here Fe is in the $+2$ oxidation state, so it is $d^{6}$. $\mathrm{CN}^{-}$ is a strong field ligand, so the complex is low spin with no unpaired electrons. Hence it does not satisfy the condition.

2. $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$: Here Co is in the $+3$ oxidation state, so it is also $d^{6}$. $\mathrm{NH}_{3}$ gives a low-spin arrangement here, so there are no unpaired electrons. Hence it does not satisfy the condition.

3. $\left[\mathrm{FeF}_{6}\right]^{4-}$: Here Fe is $+2$, so $d^{6}$. $\mathrm{F}^{-}$ is a weak field ligand, so the complex is high spin with $4$ unpaired electrons. Its magnetic moment is around $4.90$ B.M., not $5.96$ B.M.

4. $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4-}$: Here Mn is in the $+2$ oxidation state, so it is $d^{5}$. $\mathrm{SCN}^{-}$ acts as a weak field ligand, giving a high-spin complex with $5$ unpaired electrons.

$$\mu = \sqrt{35} \text{ BM} = 5.96 \text{ BM}$$

Also for high-spin $d^{5}$, the crystal field stabilization energy is zero, so $\Delta_{0}=0$ in the sense of zero CFSE.

Therefore, the correct option is D.

Given: The required complex must have zero CFSE and magnetic moment $5.96$ B.M.

Find: Eliminate all options except the one with high-spin $d^{5}$ configuration.

• $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$: $\mathrm{Fe}^{2+} \Rightarrow 3d^{6}4s^{0}$ and $\mathrm{CN}^{-}$ is a strong field ligand, so $\mu=0$.
• $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$: $\mathrm{Co}^{3+} \Rightarrow 3d^{6}4s^{0}$ and $\mathrm{NH}_{3}$ is treated as strong field here, so $\mu=0$.
• $\left[\mathrm{FeF}_{6}\right]^{4-}$: $\mathrm{Fe}^{2+} \Rightarrow 3d^{6}4s^{0}$ and $\mathrm{F}^{-}$ is a weak field ligand, but this gives high-spin $d^{6}$, not the required $5$ unpaired electrons.
• $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4-}$: $\mathrm{Mn}^{2+} \Rightarrow 3d^{5}4s^{0}$ and $\mathrm{SCN}^{-}$ is a weak field ligand, so the complex is high spin with $5$ unpaired electrons.

Hence

$$\mu = \sqrt{35} \approx 5.96 \text{ BM}$$

and high-spin $d^{5}$ has zero CFSE.

Therefore, the correct option is D, $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4-}$.

Note: The source options omit the minus signs in some complex charges, while the solution uses the conventional charged forms.