Question

Which of the following has $\text{sp}^3\text{d}^2$ hybridisation?

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

Hint:

When determining hybridization, consider the number of ligands around the central atom and the geometry of the complex.

Answer 1

The Correct Option is A

To determine the hybridization of a complex, we need to consider the geometry of the complex and the number of electron pairs around the central atom.
- In the complex \([NiCl_4]^{2-}\), nickel is in the +2 oxidation state, and there are four chloride ions surrounding it. This suggests that the geometry of the complex is tetrahedral, and to form a tetrahedral structure, nickel will use sp\textsuperscript{3} hybridization.
- For \([Ni(CO)_4]\), nickel in the zero oxidation state has a square planar geometry, which indicates sp\textsuperscript{2} hybridization.
- \(SF_6\) has a central sulfur atom surrounded by six fluorine atoms, indicating sp\textsuperscript{3}d\textsuperscript{2} hybridization.
- \([Ni(CN)_4]^{2-}\) also has a square planar geometry, which corresponds to sp\textsuperscript{2} hybridization.
Therefore, the correct answer is (1) \([NiCl_4]^{2-}\) because it has sp\textsuperscript{3} hybridization.