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Chapter 9: Problem 19

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape, assuming they are all in the gaseous state? (a) \(\mathrm{SiH}_{4}\) (b) \(\mathrm{PF}_{3},\) (c) \(\mathrm{HBr}\), (d) \(\mathrm{HCN},\) (e) \(\mathrm{SO}_{2}\)

Short Answer

Expert verified
In the gaseous state, the presence of nonbonding electron pairs produce an effect on molecular shape for (b) PF3 and (e) SO2. PF3 has a trigonal pyramidal shape and SO2 has a bent (V-shaped) shape, both affected by their nonbonding electron pairs.

Step by step solution

01

Identify the central atom of each molecule/ion

For each molecule/ion, we will identify the central atom which is usually the least electronegative atom (except for hydrogen). (a) SiH4: The central atom is Si. (b) PF3: The central atom is P. (c) HBr: The central atom is Br. (d) HCN: The central atom is C. (e) SO2: The central atom is S.
02

Determine the number of nonbonding electron pairs on each central atom

We will now find the number of nonbonding electron pairs on each central atom: (a) SiH4: Si has 4 valence electrons, which are all bonded to four H atoms; thus, there are 0 nonbonding electron pairs. (b) PF3: P has 5 valence electrons, three of which are bonded to three F atoms. Thus, there remains 1 lone pair (nonbonding electron pair). (c) HBr: Br has 7 valence electrons, one of which is bonded to the H atom. Thus, there are 3 nonbonding electron pairs. (d) HCN: C has 4 valence electrons, two of which are bonded to the H atom and two others forming a triple bond with the N atom. Thus, there are 0 nonbonding electron pairs. (e) SO2: S has 6 valence electrons, four of which are bonded to two O atoms (with double bonds). Thus, there remains 1 nonbonding electron pair.
03

Use VSEPR theory to predict the molecular shape and the effect of nonbonding electron pairs

We will now use VSEPR theory to predict the molecular shape of each species and determine if the nonbonding electron pairs affect the shape: (a) SiH4: 0 nonbonding electron pairs; the shape is tetrahedral and unaffected by nonbonding electron pairs. (b) PF3: 1 nonbonding electron pair; the shape is trigonal pyramidal and affected by the nonbonding electron pair. (c) HBr: 3 nonbonding electron pairs; the shape is linear and unaffected by nonbonding electron pairs due to Br and H forming single bond with no other atoms present. (d) HCN: 0 nonbonding electron pairs; the shape is linear and unaffected by nonbonding electron pairs. (e) SO2: 1 nonbonding electron pair; the shape is bent (V-shaped) and affected by the nonbonding electron pair.
04

Conclusion

Among the given molecules/ions, the presence of nonbonding electron pairs produce an effect on the molecular shape for (b) PF3 and (e) SO2.

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Most popular questions from this chapter

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(C\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

Methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO},\) was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3,800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate. (b) Draw a ball-and-stick model of the structure, including estimates of all the bond angles in the compound. (c) Predict all the bond distances in the molecule. (d) Do you predict that the molecule will have a dipole moment? Explain.

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the \(3 d\) orbitals (Figure 6.23\()\) can be used to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases of orbitals, explain why a node is generated in the \(\sigma_{3 d}^{\star}\) MO. (d) Sketch the energy-level diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2}\) ?

An \(\mathrm{AB}_{2}\) molecule is described as linear, and the \(\mathrm{A}-\mathrm{B}\) bond length is known. (a) Does this information completely describe the geometry of the molecule? (b) Can you tell how many nonbonding pairs of electrons are around the A atom from this information?

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)
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