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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 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

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 sp³ hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeO}_{3}\) is sp³. 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 sp³. 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 sp³d² hybrid orbitals. Thus, the hybridization for Xe in \(\mathrm{XeOF}_{4}\) is sp³d². 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 sp³d hybrid orbitals. The hybridization for Xe in \(\mathrm{XeOF}_{2}\) is sp³d. 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 sp³d² as it has 6 electron pairs.

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Most popular questions from this chapter

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{array}{ll}{\mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g)} & {\Delta H=375 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{ONCI}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g)} & {\Delta H=158 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 \(\mathrm{pm}\) and 424.7 \(\mathrm{kJ} / \mathrm{mol}\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen (the O-H bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1\(s\) character? c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain. d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. e. Estimate the bond order for OH. f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.

Use the MO model to determine which of the following has the smallest ionization energy: $\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .$ Explain your answer.

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Which charge(s) for the \(\mathrm{N}_{2}\) molecule would give a bond order of 2.5\(?\)
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