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Chapter 9: Problem 67

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

Expert verified
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 sp鲁 hybridization. b. \(\mathrm{XeO}_{4}\): Tetrahedral structure with sp鲁 hybridization. c. \(\mathrm{XeOF}_{4}\): Square pyramidal structure with sp鲁d虏 hybridization. d. \(\mathrm{XeOF}_{2}\): T-shaped structure with sp鲁d hybridization. e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\): Square pyramidal structure with sp鲁d虏 hybridization.

Step by step solution

01

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 : :
02

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.
03

Describing bonding in terms of hybrid orbitals

For 4 electron pairs, Xe has to use its \(5s\) and \(5p\) orbitals to form 4 sp鲁 hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeO}_{3}\) is sp鲁. b. \(\mathrm{XeO}_{4}\)
04

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
05

Predicting the molecular structure

The Xe atom has 4 electron pairs, all bonded pairs. Hence, the molecular structure is tetrahedral.
06

Describing bonding in terms of hybrid orbitals

The hybridization for Xe in \(\mathrm{XeO}_{4}\) is also sp鲁. c. \(\mathrm{XeOF}_{4}\)
07

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
08

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.
09

Describing bonding in terms of hybrid orbitals

For 6 electron pairs, Xe has to use its \(5s\), \(5p\), and \(5d\) orbitals to form 6 sp鲁d虏 hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeOF}_{4}\) is sp鲁d虏. d. \(\mathrm{XeOF}_{2}\)
10

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 ..
11

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.
12

Describing bonding in terms of hybrid orbitals

For 5 electron pairs, Xe uses 5 sp鲁d hybrid orbitals. The hybridization for Xe in \(\mathrm{XeOF}_{2}\) is sp鲁d. e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)
13

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
14

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.
15

Describing bonding in terms of hybrid orbitals

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

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Most popular questions from this chapter

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right),\) an important industrial chemical, is produced by the following steps: $$ \begin{array}{c}{\mathrm{CaC}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{CaNCN}+\mathrm{C}} \\ {\mathrm{CaNCN} \stackrel{\mathrm{Acid}}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN}} \\\ {\mathrm{Cyanamide}}\end{array} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\) , dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the C and N atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(C-N\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\) , and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?

The oxyanion of nitrogen in which it has the highest oxidation state is the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right) .\) The corresponding oxyanion of phosphorus is \(\mathrm{PO}_{4}^{3-}\) . The \(\mathrm{NO}_{4}^{3-}\) ion is known but is not very stable. The \(\mathrm{PO}_{3}-\) ion is not known. Account for these differences in terms of the bonding in the four anions.

Draw the Lewis structures for \(\mathrm{TeCl}_{4}, \mathrm{ICl}_{5}, \mathrm{PCl}_{5}, \mathrm{KrCl}_{4},\) and \(\mathrm{XeCl}_{2}\) . Which of the compounds exhibit at least one bond angle that is approximately \(120^{\circ} ?\) Which of the compounds exhibit \(d^{2} s p^{3}\) hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

Consider the following molecular orbitals formed from the combination of two hydrogen 1s orbitals: a. Which is the bonding molecular orbital and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?
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