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Chapter 19: Problem 93

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

Expert verified
The hybridization of the central atom for each molecule is as follows: a. \(\mathrm{SF}_{6}\): \(\mathrm{sp}^{3}\mathrm{d}^{2}\) b. \(\mathrm{ClF}_{3}\): \(\mathrm{sp}^{3}\mathrm{d}\) c. \(\mathrm{GeCl}_{4}\): \(\mathrm{sp}^{3}\) d. \(\mathrm{XeF}_{4}\): \(\mathrm{sp}^{3}\mathrm{d}^{2}\)

Step by step solution

01

a. \(\mathrm{SF}_{6}\)#

To find the central atom's hybridization in \(\mathrm{SF}_{6}\), we first need to count the electron groups around the central sulfur (S) atom. Since S is bonded to six fluorine (F) atoms, there are six electron groups. Hence, the hybridization is \(\mathrm{sp}^{3}\mathrm{d}^{2}\), since it satisfies the requirement of six orbitals (1 s, 3 p, and 2 d orbitals).
02

b. \(\mathrm{ClF}_{3}\)#

For the central atom, chlorine (Cl), in \(\mathrm{ClF}_{3}\), we count the electron groups around it. Cl is bonded to three fluorine (F) atoms. Additionally, Cl has two lone pairs, summing up to a total of five electron groups. The hybridization is \(\mathrm{sp}^{3}\mathrm{d}\), which accounts for the required five orbitals (1 s, 3 p, and 1 d orbitals).
03

c. \(\mathrm{GeCl}_{4}\)#

To find the central atom's hybridization in \(\mathrm{GeCl}_{4}\), we need to count the electron groups around the central germanium (Ge) atom. Germanium is bonded to four chlorine (Cl) atoms, so there are four electron groups. Therefore, the hybridization is \(\mathrm{sp}^{3}\), which accounts for the four orbitals needed (1 s and 3 p orbitals).
04

d. \(\mathrm{XeF}_{4}\)#

For the xenon (Xe) central atom in \(\mathrm{XeF}_{4}\), we need to count the electron groups around it. Xe is bonded to four fluorine (F) atoms and has two lone pairs, so there are a total of six electron groups. The hybridization is \(\mathrm{sp}^{3}\mathrm{d}^{2}\), which accounts for the necessary six orbitals (1 s, 3 p, and 2 d orbitals).

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Most popular questions from this chapter

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