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Chapter 9: Problem 65

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

Short Answer

Expert verified
The Lewis structures, molecular geometries, and hybridizations of the given Xenon compounds are: a. \(\mathrm{XeO}_3\): trigonal pyramidal, sp3d hybridization b. \(\mathrm{XeO}_4\): tetrahedral, sp3 hybridization c. \(\mathrm{XeOF}_4\): square pyramid, sp3d hybridization d. \(\mathrm{XeOF}_2\): T-shaped, sp3d hybridization e. \(\mathrm{XeO}_3 \mathrm{F}_2\): trigonal bipyramid, sp3d hybridization

Step by step solution

01

(a. \(\mathrm{XeO}_{3}\))

Step 1: Draw the Lewis structure Valence electrons in Xe: 8, and in O: 6. Total valence electrons in \(\mathrm{XeO}_{3}\): \(8 + 3 \times 6 = 26 \) Xe is the central atom, surrounded by O (oxygen) atoms. Each O atom forms a double bond with Xe, consuming 2 electrons per bond, while the remaining 3 valence electrons on each O complete the octet. Xe is left with two lone pairs. Step 2: Predict the molecular structure (geometry) The terminal atoms are arranged in a trigonal pyramidal configuration because Xe has three bonding domains and two lone pairs. Step 3: Describe the bonding in terms of hybrid orbitals The Xe atom has 5 electron domains (three bonding domains and two lone pairs of electrons), requiring hybridization of five orbitals. Thus, Xe exhibits sp3d hybridization.
02

(b. \(\mathrm{XeO}_{4}\))

Step 1: Draw the Lewis structure Total valence electrons in \(\mathrm{XeO}_{4}\): \(8 + 4 \times 6 = 32 \) Xe forms double bonds with each of the 4 O atoms, consuming 2 electrons per bond, and satisfying the octet rule for all atoms. Step 2: Predict the molecular structure (geometry) The terminal atoms are arranged in a tetrahedral configuration. Step 3: Describe the bonding in terms of hybrid orbitals Xe has 4 electron domains (all bonding domains), requiring hybridization of four orbitals. Thus, Xe exhibits sp3 hybridization.
03

(c. \(\mathrm{XeOF}_{4}\))

Step 1: Draw the Lewis structure Total valence electrons in \(\mathrm{XeOF}_{4}\): \( 8 + 6 + 7 \times 4 = 42 \) Xe forms double bonds with the O atom and single bonds with the 4 F (fluorine) atoms. Step 2: Predict the molecular structure (geometry) The terminal atoms are in a square pyramid configuration. Step 3: Describe the bonding in terms of hybrid orbitals Xe has 5 electron domains (all bonding domains), requiring hybridization of five orbitals. Thus, Xe exhibits sp3d hybridization.
04

(d. \(\mathrm{XeOF}_{2}\))

Step 1: Draw the Lewis structure Total valence electrons in \(\mathrm{XeOF}_{2}\): \( 8 + 6 + 7 \times 2 = 36 \) Xe forms double bonds with the O atom and single bonds with the 2 F atoms. Two lone pairs of electrons remain on Xe. Step 2: Predict the molecular structure (geometry) The terminal atoms are arranged in a T-shaped configuration, with 2 lone electron pairs on Xe. Step 3: Describe the bonding in terms of hybrid orbitals Xe has 5 electron domains (three bonding domains and two lone pairs of electrons), requiring hybridization of five orbitals. Thus, Xe exhibits sp3d hybridization.
05

(e. \(\mathrm{XeO}_{3} \mathrm{~F}_{2}\))

Step 1: Draw the Lewis structure Total valence electrons in \(\mathrm{XeO}_{3} \mathrm{~F}_{2}\): \( 8 + 3 \times 6 + 2 \times 7 = 48 \) Xe forms double bonds with each of the 3 O atoms and single bonds with the 2 F atoms. Step 2: Predict the molecular structure (geometry) The terminal atoms have a trigonal bipyramid configuration. Step 3: Describe the bonding in terms of hybrid orbitals Xe has 5 electron domains (all bonding domains), requiring hybridization of five orbitals. Thus, Xe exhibits sp3d hybridization.

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Most popular questions from this chapter

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}{ }^{+}\), \(\mathrm{F}_{2}\), and \(\mathrm{F}_{2}^{-}\), predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

\(\mathrm{FClO}_{2}\) and \(\mathrm{F}_{3} \mathrm{ClO}\) can both gain a fluoride ion to form stable anions. \(\mathrm{F}_{3} \mathrm{ClO}\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\) will both lose a fluoride ion to form stable cations. Draw the Lewis structures and describe the hybrid orbitals used by chlorine in these ions.

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with \(25-\mathrm{nm}\) light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{aligned} \mathrm{N}_{2}(g) & \longrightarrow 2 \mathrm{~N}(g) & \Delta H &=941 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) & \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1501 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}(g) & \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1402 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{~kJ} / \mathrm{mol})\) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{~kJ} / \mathrm{mol})\).

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

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?
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