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Question

The hybridisation of sulphur in $$SF_{4}$$ is:


A
sp3
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B
sp2
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C
sp3d
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D
sp3d2
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Solution

The correct option is C $$sp^{3}d$$
When the central metal atom is surrounded with five pairs of electrons the electronic geometry is trigonal bi pyramidal. If all the electron pairs are shared, the molecular geometry will also be trigonal bi pyramidal. An example for such a molecule is $$PF_5$$. If one of the pairs of electrons is not shared, then the molecular shape is called distorted tetrahedron which sometimes is also called as seesaw. An example of a molecule with trigonal bi pyramidal electronic geometry and the molecular shape as distorted tetrahedron is $$SF_4$$.  In this molecule $$SF_4$$, there are five electron pairs which should lead to trigonal bi pyramidal structure. However, out of the five electron pairs, one is a lone pair. This lone pair could occupy an axial position or an equatorial position. When the lone pair is in axial position, then there will be three Lone pair - Bond pair repulsion but when lone pair is in equatorial position, then there will be two Lone pair - Bond pair repulsion. One is the new hybridization using one 3d orbital of the sulfur atom with the normal 3 $$sp^3$$, resulting in the formation of the five coordinate bonds $$sp^3d$$ required for the decet

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Chemistry

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