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Chapter 8: Problem 109

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 105 and 106.) a. \(\mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{3}\) c. \(\mathrm{TeF}_{4}\) d. \(\mathrm{PCl}_{5}\)

Short Answer

Expert verified
The molecular structures and bond angles for the given molecules are as follows: a. \(\mathrm{XeCl}_{2}\): Bent molecular geometry with a bond angle of approximately 120°. b. \(\mathrm{ICl}_{3}\): T-shaped molecular geometry with bond angles of approximately 90°. c. \(\mathrm{TeF}_{4}\): Tetrahedral molecular geometry with bond angles of approximately 109.5°. d. \(\mathrm{PCl}_{5}\): Trigonal bipyramidal molecular geometry with equatorial bond angles of approximately 120° and axial bond angles of approximately 90°.

Step by step solution

01

Identify the central atom

In this molecule, the central atom is Xenon (Xe) because it is less electronegative than chlorine.
02

Determine the number of valence electrons of the central atom

Xenon is a noble gas and has 8 valence electrons.
03

Count the number of bonded atoms and nonbonded electron pairs

There are 2 chlorine atoms bonded to the central atom, and with 2 single bonds, 4 electrons are used. This leaves 4 more electrons as two nonbonded (lone) electron pairs.
04

Use VSEPR theory to predict geometry and bond angles

According to VSEPR theory, Xenon has 2 bonded atoms and 2 lone pairs, resulting in a bent molecular geometry. The bond angle between the chlorine atoms is approximately 120° due to the repulsion between the electron pairs. #b. ICl3#
05

Identify the central atom

In this molecule, the central atom is iodine (I) because it is less electronegative than chlorine.
06

Determine the number of valence electrons of the central atom

Iodine has 7 valence electrons.
07

Count the number of bonded atoms and nonbonded electron pairs

There are 3 chlorine atoms bonded to the central atom, and with 3 single bonds, 6 electrons are used. This leaves 1 more electron as a nonbonded (lone) electron pair.
08

Use VSEPR theory to predict geometry and bond angles

According to VSEPR theory, iodine has 3 bonded atoms and 1 lone pair, resulting in a T-shaped molecular geometry. The bond angles between I and Cl are approximately 90° due to the repulsion between the electron pairs. #c. TeF4#
09

Identify the central atom

In this molecule, the central atom is tellurium (Te) as it is less electronegative than fluorine.
10

Determine the number of valence electrons of the central atom

Tellurium has 6 valence electrons.
11

Count the number of bonded atoms and nonbonded electron pairs

There are 4 fluorine atoms bonded to the central atom, and with 4 single bonds, 8 electrons are used. This leaves no electrons as lone pairs.
12

Use VSEPR theory to predict geometry and bond angles

According to VSEPR theory, tellurium has 4 bonded atoms and no lone pairs, resulting in a tetrahedral molecular geometry. The bond angles between Te and F are approximately 109.5° due to the tetrahedral shape. #d. PCl5#
13

Identify the central atom

In this molecule, the central atom is phosphorus (P) because it is less electronegative than chlorine.
14

Determine the number of valence electrons of the central atom

Phosphorus has 5 valence electrons.
15

Count the number of bonded atoms and nonbonded electron pairs

There are 5 chlorine atoms bonded to the central atom, and with 5 single bonds, 10 electrons are used. This leaves no electrons as lone pairs.
16

Use VSEPR theory to predict geometry and bond angles

According to VSEPR theory, phosphorus has 5 bonded atoms and no lone pairs, resulting in a trigonal bipyramidal molecular geometry. There are two types of bond angles in this configuration: the equatorial angles, between the three equatorial chlorine atoms, are approximately 120°, while the axial angles, between the two axial chlorine atoms and the central atom, are approximately 90° due to the shape.

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Most popular questions from this chapter

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

What is meant by a chemical bond? Why do atoms form bonds with each other? Why do some elements exist as molecules in nature instead of as free atoms?

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3}\) Give a possible explanation for why these two reactions occur.

Borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\right)\) has often been called "inorganic" benzene. Write Lewis structures for borazine. Borazine contains a sixmembered ring of alternating boron and nitrogen atoms with one hydrogen bonded to each boron and nitrogen.
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