JEE Chemistry 2024 Question with Solution

Given: We need to identify the species in which the central atom shows $d^2sp^3$ hybridization.

Find: The correct option among the given species.

From the solution text:

• $[\text{Co}(\text{NH}_3)_6]^{3+}$ is octahedral and shows $d^2sp^3$ hybridization.
• $\text{BrF}_5$ shows $sp^3d^2$ hybridization.
• $[\text{PtCl}_4]^{2-}$ shows $dsp^2$ hybridization.
• $\text{SF}_6$ shows $sp^3d^2$ hybridization.

For $[\text{Co}(\text{NH}_3)_6]^{3+}$, cobalt is in the $+3$ oxidation state:

As stated in the solution, it uses two $d$ orbitals, one $s$ orbital, and three $p$ orbitals to form an octahedral complex, so the hybridization is $d^2sp^3$.

Therefore, the species showing $d^2sp^3$ hybridization is $[\text{Co}(\text{NH}_3)_6]^{3+}$. The correct option is A.

The second provided approach mentions option label C for the same species, but this conflicts with the listed options; the species itself matches option A, so A is the defensible answer.

Given: Four species are provided.

Find: Which central atom shows $d^2sp^3$ hybridization.

1. $[\text{Co}(\text{NH}_3)_6]^{3+}$

• Central atom: cobalt in oxidation state $+3$
• As given in the solution, $\text{Co}^{3+}$ has configuration $[\text{Ar}]\,3d^6$
• This complex is octahedral and shows $d^2sp^3$ hybridization

2. $\text{BrF}_5$

• The solution states bromine has $sp^3d^2$ hybridization
• Hence it is not $d^2sp^3$

3. $[\text{PtCl}_4]^{2-}$

• The solution states platinum shows $dsp^2$ hybridization
• Hence it is not $d^2sp^3$

4. $\text{SF}_6$

• The solution states sulfur shows $sp^3d^2$ hybridization
• Hence it is not $d^2sp^3$

Only $[\text{Co}(\text{NH}_3)_6]^{3+}$ satisfies the required hybridization. Therefore, the correct option is A.