JEE Chemistry 2025 Question with Solution

Given: The complexes are $[FeF_6]^{3-}$, $[CoF_6]^{3-}$, $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$.

Find: The correct order based on the number of unpaired electrons present.

We analyze the oxidation state, electronic configuration and ligand field strength for each complex.

1. For $[FeF_6]^{3-}$:

• Iron is in the $+3$ oxidation state, so it is $Fe^{3+}$.
• Its electronic configuration is $[Ar] \, 3d^5$.
• $F^-$ is a weak field ligand, so the complex is high spin.
• Therefore, the number of unpaired electrons is $5$.

2. For $[CoF_6]^{3-}$:

• Cobalt is in the $+3$ oxidation state, so it is $Co^{3+}$.
• Its electronic configuration is $[Ar] \, 3d^6$.
• $F^-$ is a weak field ligand, so the complex is high spin.
• Therefore, the number of unpaired electrons is $4$.

3. For $[Ni(CO)_4]$:

• Nickel is in the $0$ oxidation state, so it is $Ni^0$.
• The configuration is $[Ar] \, 3d^8 \, 4s^2$.
• $CO$ is a strong field ligand and causes pairing of electrons.
• Therefore, the number of unpaired electrons is $0$.

4. For $[Ni(CN)_4]^{2-}$:

• Nickel is in the $+2$ oxidation state, so it is $Ni^{2+}$.
• Its electronic configuration is $[Ar] \, 3d^8$.
• $CN^-$ is a strong field ligand and causes pairing of electrons.
• Therefore, the number of unpaired electrons is $0$.

Hence, the required order is

$$[FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4]$$

Therefore, the correct option is D.

Given: The same four complexes are to be compared by unpaired electrons.

Find: Their decreasing order of unpaired electrons.

Use ligand strength directly:

• $F^-$ is a weak field ligand, so $[FeF_6]^{3-}$ and $[CoF_6]^{3-}$ are high spin.
• $CO$ and $CN^-$ are strong field ligands, so $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ have paired electrons.

Now compare the metal ions:

• $Fe^{3+} : 3d^5$ gives $5$ unpaired electrons in high spin state.
• $Co^{3+} : 3d^6$ gives $4$ unpaired electrons in high spin state.
• Both nickel complexes here have $0$ unpaired electrons.

So the order is

$$[FeF_6]^{3-} > [CoF_6]^{3-} > [Ni(CN)_4]^{2-} = [Ni(CO)_4]$$

Therefore, the correct option is D.