JEE Chemistry 2025 Question with Solution

Given: The species are $[Co(NH_3)_6]^{3+}, SF_6, [CrF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$ and $[MnCl_6]^{3-}$.

Find: The number of species showing $sp^3d^2$ hybridization.

Use the ligand field idea:

• Strong field ligands such as $NH_3$ and $CN^-$ generally lead to inner orbital octahedral complexes with $d^2sp^3$ hybridization.
• Weak field ligands such as $F^-$ and $Cl^-$ generally lead to outer orbital octahedral complexes with $sp^3d^2$ hybridization.
• $SF_6$ is a standard example of $sp^3d^2$ hybridization.

Now check each species:

1. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $d^6$ and $NH_3$ is a strong field ligand, so this is $d^2sp^3$, not $sp^3d^2$.

2. $SF_6$: octahedral with $sp^3d^2$ hybridization.

3. $[CrF_6]^{3-}$: $Cr^{3+}$ is $d^3$ and $F^-$ is a weak field ligand, so it forms an outer orbital octahedral complex, i.e. $sp^3d^2$.

4. $[CoF_6]^{3-}$: $Co^{3+}$ with weak field ligand $F^-$ gives outer orbital octahedral hybridization, i.e. $sp^3d^2$.

5. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ with strong field ligand $CN^-$ gives $d^2sp^3$, not $sp^3d^2$.

6. $[MnCl_6]^{3-}$: $Cl^-$ is a weak field ligand, so the complex shows $sp^3d^2$ hybridization.

Therefore, the species showing $sp^3d^2$ hybridization are

Hence the number of such species is $4$. Therefore, the correct option is B.

The first solution block on the solution's lists only three species in one intermediate sentence, but it finally concludes the count as $4$. The second solution explicitly lists all four species, which resolves the discrepancy.