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Chapter 8: Problem 121

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 115 and 116.) 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. XeCl2: The molecular geometry is bent, with bond angles of approximately 120 degrees. b. ICl3: The molecular geometry is T-shaped, with bond angles of approximately 90 degrees. c. TeF4: The molecular geometry is see-saw, with bond angles of approximately 120 degrees and 90 degrees. d. PCl5: The molecular geometry is trigonal bipyramidal, with bond angles of approximately 120 degrees in the equatorial plane and 90 degrees between the axial and equatorial positions.

Step by step solution

01

Molecule XeCl2

Step 1: Identify the central atom and count the number of electron domains around it Xe is the central atom, and it has two Cl atoms bonded to it. Xe has 8 valence electrons, and 2 of these electrons form covalent bonds with the 2 Cl atoms. So, the central atom has 3 electron domains: 2 bonding domains (one for each Cl atom) and 1 nonbonding domain (3 lone pairs). Step 2: Determine the molecular geometry using the VSEPR table With 3 electron domains (2 bonding and 1 nonbonding), the molecular geometry of XeCl2 is bent, and the bond angles are approximately 120 degrees.
02

Molecule ICl3

Step 1: Identify the central atom and count the number of electron domains around it I is the central atom, and it has three Cl atoms bonded to it. I has 7 valence electrons, and 3 of these electrons form covalent bonds with the 3 Cl atoms, leaving 2 lone pairs on the I atom. So, there are 5 electron domains: 3 bonding domains (one for each Cl atom) and 2 nonbonding domains (2 lone pairs). Step 2: Determine the molecular geometry using the VSEPR table With 5 electron domains (3 bonding and 2 nonbonding), the molecular geometry of ICl3 is T-shaped, and the bond angles are approximately 90 degrees.
03

Molecule TeF4

Step 1: Identify the central atom and count the number of electron domains around it Te is the central atom, and it has four F atoms bonded to it. Te has 6 valence electrons, and 4 of these electrons form covalent bonds with the 4 F atoms, leaving 1 lone pair on the Te atom. So, there are 5 electron domains: 4 bonding domains (one for each F atom) and 1 nonbonding domain (1 lone pair). Step 2: Determine the molecular geometry using the VSEPR table With 5 electron domains (4 bonding and 1 nonbonding), the molecular geometry of TeF4 is see-saw, and the bond angles are approximately 120 degrees and 90 degrees.
04

Molecule PCl5

Step 1: Identify the central atom and count the number of electron domains around it P is the central atom, and it has five Cl atoms bonded to it. P has 5 valence electrons, and all of these electrons form covalent bonds with the 5 Cl atoms, leaving no lone pairs on the P atom. So, there are 5 electron domains, which are all bonding domains (one for each Cl atom). Step 2: Determine the molecular geometry using the VSEPR table With 5 electron domains (all bonding), the molecular geometry of PCl5 is trigonal bipyramidal, and the bond angles are approximately 120 degrees in the equatorial plane and 90 degrees between the axial and equatorial positions.

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Most popular questions from this chapter

Oxidation of the cyanide ion produces the stable cyanate ion, \(\mathrm{OCN}^{-}\) . The fulminate ion, \(\mathrm{CNO}^{-}\), on the other hand, is very unstable. Fulminate salts explode when struck; \(\mathrm{Hg}(\mathrm{CNO})_{2}\) is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{N}\) is the central atom in \(\mathrm{CNO}^{-}\) )

Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb} \quad\) c. $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga} \qquad\) d. $\mathrm{S}, \mathrm{O}, \mathrm{F}$

List the bonds $\mathrm{P}-\mathrm{Cl}, \mathrm{P}-\mathrm{F}, \mathrm{O}-\mathrm{F},\( and \)\mathrm{Si}-\mathrm{F}$ from least polar to most polar.

Two different compounds exist having the formula $\mathrm{N}_{2} \mathrm{F}_{2}$ . One compound is polar whereas the other is nonpolar. Draw Lewis structures for \(\mathrm{N}_{2} \mathrm{F}_{2}\) consistent with these observations.

The molecules $\mathrm{BF}_{3}, \mathrm{CF}_{4}, \mathrm{CO}_{2}, \mathrm{PF}_{5},\( and \)\mathrm{SF}_{6}$ are all nonpolar, even though they contain polar bonds. Why?
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