Question

What is the hybridization of the central atom in each of the following molecules? a. \(\mathrm{SF}_{6}\) b. \(\mathrm{ClF}_{3}\) c. \(\mathrm{GeCl}_{4}\) d. \(\mathrm{XeF}_{4}\)

Short answer

The hybridizations of the central atoms in the given molecules are: a. \(\mathrm{SF}_{6}\): \(\mathrm{Sp^{3}d^{2}}\) b. \(\mathrm{ClF}_{3}\): \(\mathrm{Sp^{3}d}\) c. \(\mathrm{GeCl}_{4}\): \(\mathrm{Sp^{3}}\) d. \(\mathrm{XeF}_{4}\): \(\mathrm{Sp^{3}d^{2}}\)

Step-by-step solution

Identify the central atom for each molecule

First, it is critical to identify the central atom for each molecule. It is the atom that is single in its type within each molecule. Hence, in \(\mathrm{SF}_{6}\), the central atom is S (Sulphur), in \(\mathrm{ClF}_{3}\) is Cl (Chlorine), in \(\mathrm{GeCl}_{4}\) is Ge (Germanium) and in \(\mathrm{XeF}_{4}\) is Xe (Xenon).

Count the number of charge centres around each central atom

Next, count the number of charge centres around each central atom. Remember that charge centres include atoms and lone pairs of electrons on the central atom. - In \(\mathrm{SF}_{6}\), there are 6 fluorine atoms connected to the central S atom. - In \(\mathrm{ClF}_{3}\), there are 3 Fluorine atoms connected to the central Cl atom and it also has 2 lone pairs of electrons. - In \(\mathrm{GeCl}_{4}\), there are 4 Chlorine atoms connected to the central Ge atom. - In \(\mathrm{XeF}_{4}\), there are 4 Fluorine atoms connected to the central Xe atom and it also has 2 lone pairs of electrons.

Determine the type of hybridization

Then, use the number of charge centres from step 2 to determine the type of hybridization. - In \(\mathrm{SF}_{6}\), there are 6 charge centres meaning the hybridization is \(\mathrm{Sp^{3}d^{2}}\). - In \(\mathrm{ClF}_{3}\), there are 5 charge centres meaning the hybridization is \(\mathrm{Sp^{3}d}\). - In \(\mathrm{GeCl}_{4}\), there are 4 charge centres meaning the hybridization is \(\mathrm{Sp^{3}}\). - In \(\mathrm{XeF}_{4}\), there are 6 charge centres meaning the hybridization is \(\mathrm{Sp^{3}d^{2}}\).