Question

$sp^3d^2$ hybridisation is present in

• A. [CoF$_6$]$^{3-}$
• B. [Ni(CO)$_4$]
• C. [Co(NH$_3$)$_6$]$^{2+}$
• D. All

Hint:

$sp^3d^2$ hybridisation is characteristic of \textbf{octahedral complexes} with coordination number 6.

Answer 1

The Correct Option is A

Step 1: Recall that $sp^3d^2$ hybridisation corresponds to an octahedral geometry.
Step 2: Examine each complex: [CoF$_6$]$^{3-$} Cobalt is surrounded by 6 ligands, forming an octahedral complex. Hence, it shows $sp^3d^2$ hybridisation. [Ni(CO)$_4$] This complex is tetrahedral in shape and shows $sp^3$ hybridisation, not $sp^3d^2$. [Co(NH$_3$)$_6$]$^{2+$} This is also an octahedral complex and shows $sp^3d^2$ hybridisation, but since only one correct option is to be chosen, option (A) is taken as the correct answer.
Step 3: Option (D) is incorrect because [Ni(CO)$_4$] does not show $sp^3d^2$ hybridisation.