JEE Chemistry 2025 Question with Solution

Given: The complexes are $[Fe(CN)_6]^{3-}, [FeF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$.

Find: The respective numbers of unpaired electrons and the correct option.

Determine the oxidation state of the metal ion in each complex, write its $d$-electron configuration, and then use the ligand strength to decide whether the complex is low spin or high spin.

For $[Fe(CN)_6]^{3-}$:

$CN^-$ is a strong field ligand and causes pairing in the lower energy orbitals. Hence this low-spin $d^5$ complex has $1$ unpaired electron.

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

$F^-$ is a weak field ligand, so the complex is high spin. Therefore all five $d$ electrons remain unpaired, giving $5$ unpaired electrons.

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

With weak field ligand $F^-$, the octahedral complex is high spin. Thus the configuration gives $4$ unpaired electrons.

For $[Mn(CN)_6]^{3-}$:

$CN^-$ is a strong field ligand, so electrons pair in the lower energy orbitals. Therefore this low-spin $d^4$ complex has $2$ unpaired electrons.

Hence, the required sequence is $1, 5, 4, 2$. Therefore, the correct option is A.

Given: The complexes are octahedral coordination species.

Find: The number of unpaired electrons in each species.

Use this principle: strong field ligands such as $CN^-$ produce large $\Delta_o$ and favor low spin configurations, whereas weak field ligands such as $F^-$ produce small $\Delta_o$ and favor high spin configurations.

1. $[Fe(CN)_6]^{3-}$

• Oxidation state of iron is $+3$.
• Neutral iron: $[Ar]3d^6 4s^2$, so $Fe^{3+}$ becomes $[Ar]3d^5$.
• Because $CN^-$ is a strong field ligand, the complex is low spin.
• A low-spin octahedral $d^5$ configuration is $t_{2g}^5 e_g^0$, which contains $1$ unpaired electron.

2. $[FeF_6]^{3-}$

• Oxidation state of iron is again $+3$.
• So the metal ion is $Fe^{3+} = [Ar]3d^5$.
• Since $F^-$ is a weak field ligand, the complex is high spin.
• A high-spin octahedral $d^5$ configuration is $t_{2g}^3 e_g^2$, which contains $5$ unpaired electrons.

3. $[CoF_6]^{3-}$

• Oxidation state of cobalt is $+3$.
• Neutral cobalt: $[Ar]3d^7 4s^2$, so $Co^{3+}$ becomes $[Ar]3d^6$.
• With weak field ligand $F^-$, the complex is high spin.
• A high-spin octahedral $d^6$ configuration is $t_{2g}^4 e_g^2$, which contains $4$ unpaired electrons.

4. $[Mn(CN)_6]^{3-}$

• Oxidation state of manganese is $+3$.
• Neutral manganese: $[Ar]3d^5 4s^2$, so $Mn^{3+}$ becomes $[Ar]3d^4$.
• Because $CN^-$ is a strong field ligand, the complex is low spin.
• A low-spin octahedral $d^4$ configuration is $t_{2g}^4 e_g^0$, which contains $2$ unpaired electrons.

Therefore, the sequence is $1, 5, 4, 2$, so the correct option is A.