Question

What is the shape of ${\mathrm{SF}}_4$ (including bond angles)?

Answer 1

Hybridization:

• It is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shapes.
• The hybridized orbitals are always equivalent in energy and shape.

The formula used to calculate the hybridization of a molecule is as follows:

• $\mathrm{Number}\mathrm{of}\mathrm{electron}\mathrm{pair}=\frac{1}{2}[\mathrm{V}+\mathrm{N}-\mathrm{C}+\mathrm{A}]$

where,

$\mathrm{V}$ is the number of valence electrons present in the central atom

$\mathrm{N}$ is the number of monovalent atoms bonded to the central atom

$\mathrm{C}$ is the charge of cation

$\mathrm{A}$ is the charge of anion

• The given molecule is, ${\mathrm{SF}}_4$

$\mathrm{Number}\mathrm{of}\mathrm{electron}\mathrm{pair}=\frac{1}{2}[6+4-0+0]=\frac{10}{2}=5$

• The number of electron pairs is $5$ which means the hybridization will be ${\mathrm{sp}}^3\mathrm{d}$ and the electronic geometry of the molecule will be trigonal bipyramidal.
• But as there are $4$ atoms around the central sulphur atom, the $5^{\mathrm{th}}$ position will be occupied by lone pair of electrons.
• Hence the molecular geometry will be a see-saw.

Therefore, the shape of ${\mathrm{SF}}_4$ is see-saw and bond angle is ${90}^{\mathrm{o}}\mathrm{and}{120}^{\mathrm{o}}$.