Meaning of $d^2sp^3$ hybridization

$d^2sp^3$ hybridization corresponds to an octahedral arrangement using: $2d + 1s + 3p$ orbitals of the central atom.

In coordination compounds, this is typically discussed as either:

• inner orbital octahedral: $d^2sp^3$
• outer orbital octahedral: $sp^3d^2$

We now check each option.

Option A: $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$

• Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$ So metal is $\mathrm{Co}^{3+}$.

• Electronic configuration of Co: $\mathrm{Co}: [Ar]3d^7 4s^2$ $\mathrm{Co}^{3+}: [Ar]3d^6$

• Since Co is in +3 oxidation state, pairing is favored in presence of $\mathrm{NH}_3$, and the complex $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is a low-spin octahedral complex.

• Two vacant $3d$ orbitals become available, so hybridization is: $d^2sp^3$

Hence, A shows $d^2sp^3$ hybridization.

Option B: $\mathrm{SF}_6$

• $\mathrm{SF}_6$ is octahedral.
• In classical hybridization language used in JEE, sulfur is taken as: $sp^3d^2$
• This is not written as $d^2sp^3$ for sulfur in this context.

Hence, B is not the correct choice.

Option C: $\left[\mathrm{Pt}(\mathrm{Cl}_4)\right]^{2-}$

• Oxidation state of Pt: $x+4(-1)=-2 \Rightarrow x=+2$ So metal is $\mathrm{Pt}^{2+}$.

• $\mathrm{Pt}^{2+}$ is a $d^8$ system.

• Four-coordinate $d^8$ Pt(II) complexes are generally square planar with hybridization: $dsp^2$

Hence, C is not $d^2sp^3$.

Option D: $\mathrm{BrF}_5$

• $\mathrm{BrF}_5$ has 5 bond pairs and 1 lone pair on Br.
• Electron pair geometry is octahedral, molecular shape is square pyramidal.
• In classical hybridization language, this is: $sp^3d^2$

Hence, D is not $d^2sp^3$.

Conclusion

Only option A has central atom showing $d^2sp^3$ hybridization.

$\boxed{A}$

Comparison with stored correct answer

Stored correct answer: A

My derived answer is also A, so they agree.