Question

Which of the following has $ sp^3d^2 $ hybridization?

• A. \( [NiCl_4]^{2-} \)
• B. \( [Ni(CO)_4] \)
• C. \( SF_6 \)
• D. \( [Ni(CN)_4]^{2-} \)

Hint:

For a central atom with a coordination number of 6, the hybridization is typically \( sp^3d^2 \), corresponding to an octahedral geometry.

Answer 1

The Correct Option is C

To determine the hybridization, we need to consider the number of electron pairs or regions of electron density around the central atom.
Step 1: Analyze the coordination and oxidation state of the central atom in each complex.
1.
\( [NiCl_4]^{2-} \):
The oxidation state of \( Ni \) is \( +2 \), and the coordination number is 4.
With a coordination number of 4, the hybridization is \( sp^3 \), corresponding to a tetrahedral geometry, not \( sp^3d^2 \). 2.
\( [Ni(CO)_4] \):
The oxidation state of \( Ni \) is \( 0 \), and the coordination number is 4.
Since CO is a strong field ligand, this complex also follows \( sp^3 \) hybridization, leading to a tetrahedral geometry.
3.
\( SF_6 \):
The central atom \( S \) has a coordination number of 6 (as there are 6 fluorine atoms attached to sulfur).
A coordination number of 6 corresponds to \( sp^3d^2 \) hybridization, which leads to an octahedral geometry. 4.
\( [Ni(CN)_4]^{2-} \):
The oxidation state of \( Ni \) is \( +2 \), and the coordination number is 4.
As with \( [NiCl_4]^{2-} \), this complex has \( sp^3 \) hybridization.
Step 2: Conclusion.
The complex with \( sp^3d^2 \) hybridization is \( SF_6 \) (option 3), because it has a coordination number of 6 and an octahedral geometry.