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. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Short answer

The Lewis structures, molecular structures, and hybrid orbitals for the central atom (Xe) of the given molecules are as follows: a. \(\mathrm{XeO}_{3}\): Trigonal pyramidal structure with sp³ hybridization. b. \(\mathrm{XeO}_{4}\): Tetrahedral structure with sp³ hybridization. c. \(\mathrm{XeOF}_{4}\): Square pyramidal structure with sp³d² hybridization. d. \(\mathrm{XeOF}_{2}\): T-shaped structure with sp³d hybridization. e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\): Square pyramidal structure with sp³d² hybridization.

Step-by-step solution

Drawing the Lewis structure

First, calculate the total number of valence electrons. Xe has 8 valence electrons and each O has 6. So, we have \(8 + 3 \times 6 = 26\) valence electrons. Arrange the atoms and distribute the electrons: .. O .. :O:Xe:O: .. O .. Now, complete the octet by adding lone pairs: .. :O--Xe--O: || : O O : :

Predicting the molecular structure

Based on VSEPR theory, the Xe atom has 4 electron pairs (3 bonded pairs and 1 lone pair). This results in a trigonal pyramidal molecular structure.

Describing bonding in terms of hybrid orbitals

For 4 electron pairs, Xe has to use its \(5s\) and \(5p\) orbitals to form 4 sp³ hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeO}_{3}\) is sp³. b. \(\mathrm{XeO}_{4}\)

Drawing the Lewis structure

We have \(8 + 4 \times 6 = 32\) valence electrons. Arrange the atoms and distribute the electrons like this: O O : : --Xe-- : : O O

Predicting the molecular structure

The Xe atom has 4 electron pairs, all bonded pairs. Hence, the molecular structure is tetrahedral.

Describing bonding in terms of hybrid orbitals

The hybridization for Xe in \(\mathrm{XeO}_{4}\) is also sp³. c. \(\mathrm{XeOF}_{4}\)

Drawing the Lewis structure

We have \(8 + 4 \times 6 + 7 = 39\) valence electrons. Arrange the atoms and distribute the electrons: F .. O : Xe --F :O: : .. F F

Predicting the molecular structure

The Xe atom has 6 electron pairs (5 bonded pairs and 1 lone pair). This results in a square pyramidal molecular structure.

Describing bonding in terms of hybrid orbitals

For 6 electron pairs, Xe has to use its \(5s\), \(5p\), and \(5d\) orbitals to form 6 sp³d² hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeOF}_{4}\) is sp³d². d. \(\mathrm{XeOF}_{2}\)

Drawing the Lewis structure

The total number of valence electrons is \(8 + 6 + 2 \times 7 = 28\). Arrange the atoms and distribute the electrons: .. O .. :O:Xe:F: .. F ..

Predicting the molecular structure

The Xe atom has 5 electron pairs (3 bonded pairs and 2 lone pairs). Thus, the molecular structure is T-shaped.

Describing bonding in terms of hybrid orbitals

For 5 electron pairs, Xe uses 5 sp³d hybrid orbitals. The hybridization for Xe in \(\mathrm{XeOF}_{2}\) is sp³d. e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Drawing the Lewis structure

We have \(8 + 3 \times 6 + 2 \times 7 = 40\) valence electrons. Arrange the atoms and distribute the electrons: F O : : --Xe-- : : O O

Predicting the molecular structure

The Xe atom has 6 electron pairs (5 bonded pairs and 1 lone pair). This gives a square pyramidal molecular structure.

Describing bonding in terms of hybrid orbitals

The hybridization for Xe in \(\mathrm{XeO}_{3} \mathrm{F}_{2}\) is sp³d² as it has 6 electron pairs.