Question

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following.

$$\begin{gathered} \mathit{a}\mathbf{.}\mathit{X}\mathit{e}{\mathit{O}}_{\mathbf{3}} \\ \mathit{b}\mathbf{.}\mathit{X}\mathit{e}{\mathit{O}}_{\mathbf{4}} \\ \mathit{c}\mathbf{.}\mathit{X}\mathit{e}\mathit{O}{\mathit{F}}_{\mathbf{4}} \\ \mathit{d}\mathbf{.}\mathit{X}\mathit{e}\mathit{O}{\mathit{F}}_{\mathbf{2}} \\ \mathit{e}\mathbf{.}\mathit{X}\mathit{e}{\mathit{O}}_{\mathbf{3}}{\mathit{F}}_{\mathbf{2}} \end{gathered}$$

Short answer

Molecular Structures:

a. Trigonal Pyramidal

b. Tetrahedral

c. Square Pyramidal

d. Trigonal Bipyramidal

e. Trigonal Bipyramidal

Bonding in the compounds

a. We get the sp3 hybridization on the xenon atom in the ${\mathrm{XeO}}_3$molecule.

b. The hybridization on the xenon atom in the ${\mathrm{XeO}}_4$molecule is sp3 hybridization.

c. The hybridization state for the ${\mathrm{XeOF}}_4$molecule is sp3d2 hybridization.

d. The hybridization state for the ${\mathrm{XeOF}}_2$molecule is sp3d hybridization.

e. The hybridization of the central atom in the ${\mathrm{XeO}}_3{\mathrm{F}}_2$molecule is sp3d hybridization.

Step-by-step solution

Determination of the Lewis structures for the compounds.

A. ${\mathbf{XeO}}_{\mathbf{3}}$Molecule

a. First, count the valence electrons in the molecule.

b. Find the least electronegative molecule and place it in the center.

c. Connect the other atoms to the central atom with a single bond and then place the remaining electrons on the outer atoms to complete their octets.

d. Then, complete the octet of the central atom.

e. The required Lewis structure is formed.

B. ${\mathbf{XeO}}_{\mathbf{4}}$molecule

a. Repeat the above steps to get the required Lewis structure.

C. ${\mathbf{XeOF}}_{\mathbf{4}}$molecule

The required Lewis structure is:

D. localid="1663753793172" ${\mathbf{XeOF}}_{\mathbf{2}}$Molecule

The required Lewis structure is:

E. $\mathit{X}\mathit{e}{\mathit{O}}_{\mathbf{3}}{\mathit{F}}_{\mathbf{2}}$Molecule

The required Lewis structure is:

Step 2: To determine the bonding of the compounds

To determine the molecular geometry of the compounds, we will first find the hybridization of the compounds.

A.Number of sigma bonds on Xe atom is 3

Number of lone pairs of electrons is 1

The steric number for $\mathit{X}\mathit{e}{\mathit{O}}_{\mathbf{3}}$molecule = 3+1 = 4

Thus, hybridization of the compound is sp3.

B.Number of sigma bonds on Xe atom is 4

Number of lone pairs of electrons is 0

The steric number for ${\mathbf{XeO}}_{\mathbf{4}}$molecule = 4+0 = 4

Thus, the hybridization of the compound is sp3

C.Number of sigma bonds on Xe atom is 5

Number of lone pairs of electrons is 1

The steric number for ${\mathbf{XeOF}}_{\mathbf{4}}$molecule = 5+1 = 6

Thus, the hybridization of the compound is sp3d2

D.Number of sigma bonds on Xe atom is 3

Number of lone pairs of electrons is 2

The steric number for ${\mathbf{XeOF}}_{\mathbf{2}}$molecule = 2+3 = 5

Thus, the hybridization of the compound is sp3d

E.Number of sigma bonds on Xe atom is 5

Number of lone pairs of electrons is 0

The steric number for ${\mathbf{XeO}}_{\mathbf{3}}{\mathbf{F}}_{\mathbf{2}}$molecule = 5+0 = 5

Thus, the hybridization of the compound is sp3d.

To determine the molecular structure of the compounds

We know that when we need to find the shape of a compound, we don’t count lone pairs in it.

a.The molecular structure for ${\mathbf{XeO}}_{\mathbf{3}}$molecule is Trigonal Pyramidal.

b.The molecular structure for ${\mathbf{XeO}}_{\mathbf{4}}$molecule is Tetrahedral.

c.The molecular structure for ${\mathbf{XeOF}}_{\mathbf{4}}$molecule is Square Pyramidal.

d.The molecular structure for ${\mathbf{XeOF}}_{\mathbf{2}}$molecule is Trigonal Bipyramidal.

e.The molecular structure for ${\mathbf{XeO}}_{\mathbf{3}}{\mathbf{F}}_{\mathbf{2}}$molecule is Trigonal Bipyramidal.