JEE Chemistry 2025 Question with Solution

Given: The complexes are $[FeF_6]^{3-}, [Fe(CN)_6]^{3-}, [Mn(CN)_6]^{3-}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}$ and $[CoF_6]^{3-}$.

Find: The number of complexes that are both paramagnetic and involve $d^2sp^3$ hybridization.

A complex shows $d^2sp^3$ hybridization only when it is an inner orbital octahedral complex, which generally requires a strong field ligand to cause pairing.

Now check each complex:

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

• $Fe^{3+} : 3d^5$
• $F^-$ is a weak field ligand.
• Weak field gives $sp^3d^2$, not $d^2sp^3$.
• So this complex is not counted.

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

• $Fe^{3+} : 3d^5$
• $CN^-$ is a strong field ligand.
• Pairing occurs, so the complex is $d^2sp^3$.
• One unpaired electron remains, so it is paramagnetic.
• This complex is counted.

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

• $Mn^{3+} : 3d^4$
• $CN^-$ is a strong field ligand.
• Pairing occurs, so the complex is $d^2sp^3$.
• Two unpaired electrons remain, so it is paramagnetic.
• This complex is counted.

4. $[Co(C_2O_4)_3]^{3-}$

• $Co^{3+} : 3d^6$
• From the solution, this is treated as giving $d^2sp^3$ hybridization.
• All electrons are paired, so it is diamagnetic.
• Hence this is not counted.

5. $[MnCl_6]^{3-}$

• $Mn^{3+} : 3d^4$
• $Cl^-$ is a weak field ligand.
• Weak field gives $sp^3d^2$, not $d^2sp^3$.
• So this complex is not counted.

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

• $Co^{3+} : 3d^6$
• $F^-$ is a weak field ligand.
• Weak field gives $sp^3d^2$, not $d^2sp^3$.
• So this complex is not counted.

Therefore, the complexes that are both paramagnetic and involve $d^2sp^3$ hybridization are $[Fe(CN)_6]^{3-}$ and $[Mn(CN)_6]^{3-}$.

Thus, the required number is $2$.

Given: We must count only those complexes which satisfy both conditions:

• they are paramagnetic, and
• they involve $d^2sp^3$ hybridization.

Find: Total number of such complexes.

The key idea is:

• strong field ligands can produce inner orbital octahedral complexes, that is $d^2sp^3$,
• weak field ligands usually produce outer orbital octahedral complexes, that is $sp^3d^2$.

Now apply this to the given set.

So exactly two complexes satisfy both requirements.

Therefore, the final answer is $2$.