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 hybridization, evaluate the number of electron pairs or regions of electron density surrounding the central atom.
Step 1: Evaluate the central atom's coordination and oxidation state in each complex.
1.
\( [NiCl_4]^{2-} \):
Nickel exhibits a \( +2 \) oxidation state and a coordination number of 4.
A coordination number of 4 implies \( sp^3 \) hybridization and a tetrahedral geometry, not \( sp^3d^2 \).2.
\( [Ni(CO)_4] \):
Nickel is in the \( 0 \) oxidation state with a coordination number of 4.
Due to CO being a strong field ligand, this complex also exhibits \( sp^3 \) hybridization, resulting in a tetrahedral geometry.
3.
\( SF_6 \):
The central sulfur atom has a coordination number of 6, as it is bonded to six fluorine atoms.
A coordination number of 6 corresponds to \( sp^3d^2 \) hybridization, which results in an octahedral geometry.4.
\( [Ni(CN)_4]^{2-} \):
Nickel shows a \( +2 \) oxidation state and a coordination number of 4.
Similar to \( [NiCl_4]^{2-} \), this complex has \( sp^3 \) hybridization.
Step 2: Conclusion.
The complex exhibiting \( sp^3d^2 \) hybridization is \( SF_6 \) (option 3), characterized by a coordination number of 6 and an octahedral geometry.