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Page 379
Chapter Summary
Section 9.1
  • According to the valence-shell electron-pair repulsion (VSEPR) model, electron pairs in the valence shell of an atom repel one another. An electron domain is a lone pair or a bond. Any bond (single, double, or triple) constitutes one electron domain.

  • The arrangement of electron domains about a central atom, determined using the VSEPR model, is called the electron-domain geometry. The arrangement of atoms in a molecule is called the molecular geometry. The basic molecular geometries are linear, bent, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw-shaped, T-shaped, octahedral, square pyramidal, and square planar.

  • The bond angle is the angle between two adjacent bonds in a molecule or polyatomic ion. A trigonal bipyramid contains two types of bonds: axial and equatorial.

Section 9.2
  • The polarity of a molecule depends on the polarity of its individual bonds and on its molecular geometry. Even a molecule containing polar bonds may be nonpolar overall if the bonds are distributed symmetrically.

  • Structural isomers are molecules with the same chemical formula, but different structural arrangements.

Section 9.3
  • According to valence bond theory, bonds form between atoms when atomic orbitals overlap, thus allowing the atoms to share valence electrons. A bond forms, furthermore, when the resulting molecule is lower in energy than the original, isolated atoms.

Section 9.4
  • To explain the bonding in some molecules, we need to employ the concept of hybridization, in which atomic orbitals mix to form hybrid orbitals.

  • To use hybrid-orbital analysis, we must already know the molecular geometry and bond angles in a molecule. Hybrid orbitals are not used to predict molecular geometries.

Section 9.5
  • Sigma (σ) bonds form when the region of orbital overlap lies directly between the two atoms. Pi (π) bonds form when parallel, unhybridized p orbitals interact. A double bond consists of one sigma bond and one pi bond. A triple bond consists of one sigma bond and two pi bonds.

Section 9.6
  • A paramagnetic species is one that contains unpaired electrons.

    A diamagnetic species is one in which there are no unpaired electrons. Paramagnetic species are weakly attracted by a magnetic field, whereas diamagnetic species are weakly repelled by a magnetic field.

  • According to molecular orbital theory, atomic orbitals combine to form new molecular orbitals that are associated with the molecule, rather than with individual atoms. Molecular orbitals may be sigma, if the orbital lies directly along the internuclear axis, or pi, if the orbital does not lie directly along the internuclear axis.

  • Molecular orbitals may be bonding or antibonding. A bonding molecular orbital is lower in energy than the isolated atomic orbitals that combined to form it. The corresponding antibonding molecular orbital is higher in energy than the isolated atomic orbitals. Bond order is a measure of the strength of a bond and can be determined using a molecular orbital diagram.

Section 9.7
  • It is generally best to use the bonding theory that most easily describes the bonding in a particular molecule or polyatomic ion. In species that can be represented by two or more resonance structures, the pi bonds are delocalized, meaning that they are spread out over the molecule and not constrained to just two atoms. Localized bonds are those constrained to two atoms. Many species are best described using a combination of valence bond theory and molecular orbital theory.

Key Words
Key Equation
Page 380
Questions and Problems
Section 9.1: Molecular Geometry
Review Questions
  • 9.1

    How is the geometry of a molecule defined, and why is the study of molecular geometry important?

  • 9.2

    Sketch the shape of a linear triatomic molecule, a trigonal planar molecule containing four atoms, a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule. Give the bond angles in each case.

  • 9.3

    How many atoms are directly bonded to the central atom in a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule?

  • 9.4

    Discuss the basic features of the VSEPR model. Explain why the magnitude of repulsion decreases in the following order: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair.

  • 9.5

    In the trigonal bipyramidal arrangement, why does a lone pair occupy an equatorial position rather than an axial position?

  • 9.6

    Explain why the CH4 molecule is not square planar, although its Lewis structure makes it look as though it could be.

Problems
  • 9.7

    Predict the geometries of the following species using the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, (d) TeCl4.

    Answer

  • 9.8

    Predict the geometries of the following species: (a) AlCl3, (b) ZnCl2, (c) HgBr2, (d) N2O (arrangement of atoms is NNO).

  • 9.9

    Predict the geometry of the following molecules and ion using the VSEPR model: (a) CBr4, (b) BCl3, (c) NF3, (d) H2Se, (e) .

    Answer

  • 9.10

    Predict the geometry of the following molecules and ion using the VSEPR model: (a) CH3I, (b) CIF3, (c) H2S, (d) SO3, (e) .

  • 9.11

    Predict the geometry of the following ions using the VSEPR method: (a) SCN (arrangement of atoms is SCN), (b) , (c) , (d) H3O+, (e) .

    Answer

  • 9.12

    Predict the geometries of the following ions: (a) , (b) , (c) , (d) , (e) .

  • 9.13

    Describe the geometry around each of the three central atoms in the CH3COOH molecule.

    Answer

  • 9.14

    Which of the following species are tetrahedral: SiCl4, SeF4, XeF4, CI4, ?

Section 9.2: Molecular Geometry and Polarity
Review Questions
  • 9.15

    Explain why an atom cannot have a permanent dipole moment.

  • 9.16

    The bonds in beryllium hydride (BeH2) molecules are polar, and yet the dipole moment of the molecule is zero. Explain.

Problems
  • 9.17

    Determine whether (a) BrF5 and (b) BCl3 are polar.

    Answer

  • 9.18

    Determine whether (a) OCS and (b) XeF4 are polar.

  • 9.19

    Which of the molecules shown is polar?

    Answer

  • 9.20

    Which of the molecules shown is polar?

Section 9.3: Valence Bond Theory
Review Questions
  • 9.21

    What is valence bond theory? How does it differ from the Lewis concept of chemical bonding?

  • 9.22

    Use valence bond theory to explain the bonding in Cl2 and HCl. Show how the atomic orbitals overlap when a bond is formed.

  • 9.23

    According to valence bond theory, how many bonds would you expect each of the following atoms (in the ground state) to form: Be, C?

  • 9.24

    According to valence bond theory, how many bonds would you expect each of the following atoms (in the ground state) to form: P, S?

Section 9.4: Hybridization of Atomic Orbitals
Review Questions
  • 9.25

    What is the hybridization of atomic orbitals? Why is it impossible for an isolated atom to exist in the hybridized state?

  • 9.26

    How does a hybrid orbital differ from a pure atomic orbital? Can two 2p orbitals of an atom hybridize to give two hybridized orbitals?

  • 9.27

    What is the angle between the following two hybrid orbitals on the same atom: (a) sp and sp hybrid orbitals, (b) sp2 and sp2 hybrid orbitals, (c) sp3 and sp3 hybrid orbitals?

Problems
  • 9.28

    Describe the bonding scheme of the AsH3 molecule in terms of hybridization.

  • 9.29

    What is the hybridization state of Si (a) in SiH4 and (b) in H3Si—SiH3?

    Answer

  • 9.30

    Describe the change in hybridization (if any) of the Al atom in the following reaction:

  • 9.31

    Consider the reaction

    Describe the changes in hybridization (if any) of the B and N atoms as a result of this reaction.

    Answer

  • 9.32

    What hybrid orbitals are used by nitrogen atoms in the following species: (a) NH3, (b) H2N—NH2, (c) ?

  • 9.33

    Describe the hybridization of phosphorus in PF5.

    Answer

Page 381
Section 9.5: Hybridization in Molecules Containing Multiple Bonds
Visualizing Chemistry

Figure 9.10

VC 9.1 How is a sigma bond different from a pi bond?

  1. A sigma bond is a bonding molecular orbital; a pi bond is an antibonding molecular orbital.

  2. A sigma bond is a single bond, whereas a pi bond is a double bond.

  3. The electron density in a sigma bond lies along the internuclear axis; that of a pi bond does not.

VC 9.2 Pi bonds form when atomic orbitals on atom(s) overlap.

  1. perpendicular, adjacent

  2. parallel, adjacent

  3. parallel, the same

VC 9.3 Formation of two pi bonds requires the combination of atomic orbitals.

  1. two

  2. four

  3. six

VC 9.4 Why are there no pi bonds in ethane (C2H6)?

  1. The remaining unhybridized p orbitals do not contain any electrons.

  2. There are no unhybridized p orbitals remaining on either C atom.

  3. The remaining unhybridized p orbitals are not parallel to each other.

Review Questions
  • 9.34

    How would you distinguish between a sigma bond and a pi bond?

  • 9.35

    Which of the following pairs of atomic orbitals of adjacent nuclei can overlap to form a sigma bond? Which overlap to form a pi bond? Which cannot overlap (no bond)? Consider the x axis to be the internuclear axis, that is, the line joining the nuclei of the two atoms. (a) 1s and 1s, (b) 1s and 2p x, (c) 2px and 2py, (d) 3py and 3py, (e) 2px and 2px, (f) 1s and 2s.

Problems
  • 9.36

    What are the hybrid orbitals of the carbon atoms in the following molecules?

    1. H3C—CH3

    2. H3C—CH==CH2

    3. CH3—C==CH2OH

    4. CH3CH==O

    5. CH3COOH

  • 9.37

    Specify which hybrid orbitals are used by carbon atoms in the following species: (a) CO, (b) CO2, (c) CN.

    Answer

  • 9.38

    The allene molecule (H2C==C==CH2) is linear (the three C atoms lie on a straight line). What are the hybridization states of the carbon atoms? Draw diagrams to show the formation of sigma bonds and pi bonds in allene.

  • 9.39

    What is the hybridization of the central N atom in the azide ion ()? (The arrangement of atoms is NNN.)

    Answer

  • 9.40

    How many sigma bonds and pi bonds are there in each of the following molecules?

  • 9.41

    How many pi bonds and sigma bonds are there in the tetracyanoethylene molecule?

    Answer

  • 9.42

    Tryptophan is one of the 20 amino acids in the human body. Describe the hybridization state of the C and N atoms, and determine the number of sigma and pi bonds in the molecule.

  • 9.43

    Benzo(a)pyrene is a potent carcinogen found in coal and cigarette smoke. Determine the number of sigma and pi bonds in the molecule.

    Answer

Section 9.6: Molecular Orbital Theory
Review Questions
  • 9.44

    What is molecular orbital theory? How does it differ from valence bond theory?

  • 9.45

    Define the following terms: bonding molecular orbital, antibonding molecular orbital, pi molecular orbital, sigma molecular orbital.

  • 9.46

    Sketch the shapes of the following molecular orbitals: σ1s, , π2p, . How do their energies compare?

  • 9.47

    Explain the significance of bond order. Can bond order be used for quantitative comparisons of the strengths of chemical bonds?

Problems
  • 9.48

    Explain in molecular orbital terms the changes in H—H internuclear distance that occur as the molecular H2 is ionized first to and then to .

  • 9.49

    The formation of H2 from two H atoms is an energetically favorable process. Yet statistically there is less than a 100 percent chance that any two H atoms will undergo the reaction. Apart from energy considerations, how would you account for this observation based on the electron spins in the two H atoms?

    Answer

  • 9.50

    Draw a molecular orbital energy level diagram for each of the following species: He2, HHe, . Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.)

  • Page 382
  • 9.51

    Arrange the following species in order of increasing stability: , . Justify your choice with a molecular orbital energy level diagram.

    Answer

  • 9.52

    Use molecular orbital theory to explain why the Be2 molecule does not exist.

  • 9.53

    Which of these species has a longer bond, B2 or ? Explain in terms of molecular orbital theory.

    Answer

  • 9.54

    Acetylene (C2H2) has a tendency to lose two protons (H+) and form the carbide ion (), which is present in a number of ionic compounds, such as CaC2 and MgC2. Describe the bonding scheme in the ion in terms of molecular orbital theory. Compare the bond order in with that in C2.

  • 9.55

    Compare the Lewis and molecular orbital treatments of the oxygen molecule.

    Answer

  • 9.56

    Explain why the bond order of N2 is greater than that of , but the bond order of O2 is less than that of .

  • 9.57

    Compare the relative bond orders of the following species and indicate their magnetic properties (that is, diamagnetic or paramagnetic): O2, , (superoxide ion), (peroxide ion).

    Answer

  • 9.58

    Use molecular orbital theory to compare the relative stabilities of F2 and .

  • 9.59

    A single bond is almost always a sigma bond, and a double bond is almost always made up of a sigma bond and a pi bond. There are very few exceptions to this rule. Show that the B2 and C2 molecules are examples of the exceptions.

    Answer

Section 9.7: Bonding Theories and Descriptions of Molecules with Delocalized Bonding
Review Questions
  • 9.60

    How does a delocalized molecular orbital differ from a molecular orbital such as that found in H2 or C2H4? What do you think are the minimum conditions (for example, number of atoms and types of orbitals) for forming a delocalized molecular orbital?

  • 9.61

    In Chapter 8 we saw that the resonance concept is useful for dealing with species such as the benzene molecule and the carbonate ion. How does molecular orbital theory deal with these species?

Problems
  • 9.62

    Both ethylene (C2H4) and benzene (C6H6) contain the C==C bond. The reactivity of ethylene is greater than that of benzene. For example, ethylene readily reacts with molecular bromine, whereas benzene is normally quite inert toward molecular bromine and many other compounds. Explain this difference in reactivity.

  • 9.63

    Explain why the symbol on the left is a better representation of benzene molecules than that on the right.

    Answer

  • 9.64

    Determine which of these molecules has a more delocalized orbital, and justify your choice. (Hint: Both molecules contain two benzene rings. In naphthalene, the two rings are fused together. In biphenyl, the two rings are joined by a single bond, around which the two rings can rotate.)

  • 9.65

    Nitryl fluoride (FNO2) is very reactive chemically. The fluorine and oxygen atoms are bonded to the nitrogen atom. (a) Write a Lewis structure for FNO2. (b) Indicate the hybridization of the nitrogen atom. (c) Describe the bonding in terms of molecular orbital theory. Where would you expect delocalized molecular orbitals to form?

    Answer

  • 9.66

    Describe the bonding in the nitrate ion in terms of delocalized molecular orbitals.

  • 9.67

    What is the state of hybridization of the central O atom in O3? Describe the bonding in O3 in terms of delocalized molecular orbitals.

    Answer

Additional Problems
  • 9.68

    Which of the following species is not likely to have a tetrahedral shape: (a) SiBr4, (b) , (c) SF4, (d) , (e) , (f) ?

  • 9.69

    Draw the Lewis structure of mercury(II) bromide. Is this molecule linear or bent? How would you establish its geometry?

    Answer

  • 9.70

    Although both carbon and silicon are in Group 4A, very few Si==Si bonds are known. Account for the instability of silicon-tosilicon double bonds in general. (Hint: Compare the atomic radii of C and Si in Figure 7.6. What effect would the larger size have on pi bond formation?)

  • 9.71

    Predict the geometry of sulfur dichloride (SCl2) and the hybridization of the sulfur atom.

    Answer

  • 9.72

    Antimony pentafluoride (SbF5) reacts with XeF4 and XeF6 to form ionic compounds, and . Describe the geometries of the cations and anions in these two compounds.

  • 9.73

    The molecular model of vitamin C is shown here. (a) Write the molecular formula of the compound. (b) What is the hybridization of each C and O atom? (c) Describe the geometry about each C and O atom.

    Answer

  • 9.74

    The molecular model of nicotine (a stimulant) is shown here. (a) Write the molecular formula of the compound. (b) What is the hybridization of each C and N atom? (c) Describe the geometry about each C and N atom.

  • 9.75

    Predict the bond angles for the following molecules: (a) BeCl2, (b) BCl3, (c) CCl4, (d) CH3Cl, (e) Hg2Cl2 (arrangement of atoms: ClHgHgCl), (f) SnCl2, (g) H2O2, (h) SnH4.

    Answer

  • Page 383
  • 9.76

    Briefly compare the VSEPR and hybridization approaches to the study of molecular geometry.

  • 9.77

    Draw Lewis structures and give the other information requested for the following molecules: (a) BF3. Shape: planar or nonplanar? (b) . Shape: planar or nonplanar? (c) HCN. Polar or nonpolar? (d) OF2. Polar or nonpolar? (e) NO2. Estimate the ONO bond angle.

    Answer

  • 9.78

    Describe the hybridization state of arsenic in arsenic pentafluoride (AsF5).

  • 9.79

    The disulfide bond, —S—S—, plays an important role in determining the three-dimensional structure of proteins. Describe the nature of the bond and the hybridization state of the S atoms.

    Answer

  • 9.80

    Draw Lewis structures and give the other information requested for the following: (a) SO3. Polar or nonpolar molecule? (b) PF3. Polar or nonpolar? (c) F3SiH. Polar or nonpolar? (d) . Shape: planar or pyramidal? (e) Br2CH2. Polar or nonpolar molecule?

  • 9.81

    Which of the following molecules are linear: , , OF2, SnI2, CdBr2?

    Answer

  • 9.82

    Draw the Lewis structure for the ion. Predict its geometry, and describe the hybridization state of the Be atom.

  • 9.83

    The N2F2 molecule can exist in either of the following two forms:

    Answer

  • 9.84

    Cyclopropane (C3H6) has the shape of a triangle in which a C atom is bonded to two H atoms and two other C atoms at each corner. Cubane (C8H8) has the shape of a cube in which a C atom is bonded to one H atom and three other C atoms at each corner. (a) Draw Lewis structures of these molecules. (b) Compare the CCC angles in these molecules with those predicted for an sp3-hybridized C atom. (c) Would you expect these molecules to be easy to make?

  • 9.85

    The compound 1,2-dichloroethane (C2H4Cl2) is nonpolar, while cis-dichloroethylene (C2H2Cl2) has a dipole moment: The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in 1,2-dichloroethane but not in cis-dichloroethylene.

    Answer

  • 9.86

    Does the following molecule have a dipole moment? Explain.

  • 9.87

    So-called greenhouse gases, which contribute to global warming, have a dipole moment or can be bent or distorted during molecular vibration into shapes that have a temporary dipole moment. Which of the following gases are greenhouse gases: N2, O2, O3, CO, CO2, NO2, N2O, CH4, CFCl3?

    Answer

  • 9.88

    (a) From what group must the terminal atoms come in an ABx molecule where the central atom is from Group 5A, for both the electron-domain geometry and the molecular geometry to be trigonal bipyramidal? (b) From what group must the terminal atoms come in an ABx molecule where the central atom is from Group 6A, for the electron-domain geometry to be tetrahedral and the molecular geometry to be bent?

  • 9.89

    The compound 3′-azido-3′-deoxythymidine, commonly known as AZT, is one of the drugs used to treat AIDS. What are the hybridization states of the C and N atoms in this molecule?

    Answer

  • 9.90

    The following molecules (AX4Y2) all have an octahedral geometry. Group the molecules that are equivalent to each other.

  • 9.91

    The compounds carbon tetrachloride (CCl4) and silicon tetrachloride (SiCl4) are similar in geometry and hybridization. However, CCl4 does not react with water but SiCl4 does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the Si atom in SiCl4.)

    Answer

  • 9.92

    Write the ground-state electron configuration for B2. Is the molecule diamagnetic or paramagnetic?

  • 9.93

    What is the hybridization of C and of N in this molecule?

    Answer

  • 9.94

    The stable allotropic form of phosphorus is P4, in which each P atom is bonded to three other P atoms. Draw a Lewis structure of this molecule and describe its geometry. At high temperatures, P4 dissociates to form P2 molecules containing a P==P bond. Explain why P4 is more stable than P2.

  • Page 384
  • 9.95

    Use molecular orbital theory to explain the difference between the bond enthalpies of F2 and . (See Problem 8.121.)

    Answer

  • 9.96

    Use molecular orbital theory to explain the bonding in the azide ion (The arrangement of atoms is NNN.)

  • 9.97

    The ionic character of the bond in a diatomic molecule can be estimated by the formula

    where μ is the experimentally measured dipole moment (in C · m), e is the electronic charge, and d is the bond length (in meters). (The quantity ed is the hypothetical dipole moment for the case in which the transfer of an electron from the less electronegative to the more electronegative atom is complete.) Given that the dipole moment and bond length of HF are 1.92 D and 91.7 pm, respectively, calculate the percent ionic character of the molecule. (1D = 3.336 × 10−30 Cm)

    Answer

  • 9.98

    Draw three Lewis structures for compounds with the formula C2H2F2. Indicate which of the compounds are polar.

  • 9.99

    Greenhouse gases absorb (and trap) outgoing infrared radiation (heat) from Earth and contribute to global warming. A molecule of a greenhouse gas either possesses a permanent dipole moment or has a changing dipole moment during its vibrational motions. Consider three of the vibrational modes of carbon dioxide

    where the arrows indicate the movement of the atoms. (During a complete cycle of vibration, the atoms move toward one extreme position and then reverse their direction to the other extreme position.) Which of the preceding vibrations are responsible for CO2 behaving as a greenhouse gas?

    Answer

  • 9.100

    Aluminum trichloride (AlCl3) is an electron-deficient molecule. It has a tendency to form a dimer (a molecule made up of two AlCl3 units):

    (a) Draw a Lewis structure for the dimer. (b) Describe the hybridization state of Al in AlCl3 and Al2Cl6. (c) Sketch the geometry of the dimer. (d) Do these molecules possess a dipole moment?

  • 9.101

    Progesterone is a hormone responsible for female sex characteristics. In the usual shorthand structure, each point where lines meet represents a C atom, and most H atoms are not shown. Draw the complete structure of the molecule, showing all C and H atoms. Indicate which C atoms are sp 2- and sp3-hybridized.

    Answer

  • 9.102

    The molecule benzyne (C6H4) is a very reactive species. It resembles benzene in that it has a six-membered ring of carbon atoms. Draw a Lewis structure of the molecule and account for the molecule's high reactivity.

  • 9.103

    Assume that the third-period element phosphorus forms a diatomic molecule, P2, in an analogous way as nitrogen does to form N2. (a) Write the electronic configuration for P2. Use [Ne2] to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

    Answer

  • 9.104

    Consider an N2 molecule in its first excited electronic state, that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular orbital. (a) Identify the molecular orbitals involved, and sketch a diagram to show the transition. (b) Compare the bond order and bond length of with N2, where the asterisk denotes the excited molecule. (c) Is diamagnetic or paramagnetic? (d) When loses its excess energy and converts to the ground state N2, it emits a photon of wavelength 470 nm, which makes up part of the auroras lights. Calculate the energy difference between these levels.

  • 9.105

    The Lewis structure for O2 is

    Use molecular orbital theory to show that the structure actually corresponds to an excited state of the oxygen molecule.

    Answer

  • 9.106

    Draw the Lewis structure of ketene (C2H2O) and describe the hybridization states of the C atoms. The molecule does not contain O—H bonds. On separate diagrams, sketch the formation of the sigma and pi bonds.

  • 9.107

    The compound TCDD, or 2,3,7,8-tetrachlorodibenzo-p-dioxin, is highly toxic:

    It gained considerable notoriety in 2004 when it was implicated in the attempted murder of a Ukrainian politician. (a) Describe its geometry, and state whether the molecule has a dipole moment. (b) How many pi bonds and sigma bonds are there in the molecule?

    Answer

  • 9.108

    Write the electron configuration of the cyanide ion (CN). Name a stable molecule that is isoelectronic with the ion.

  • 9.109

    Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to Fe2+ in the hemoglobin molecule. The molecular orbitals of CO have the same energy order as those of the N2 molecule. (a) Draw a Lewis structure of CO and assign formal charges. Explain why CO has a rather small dipole moment of 0.12 D. (b) Compare the bond order of CO with that from molecular orbital theory. (c) Which of the atoms (C or O) is more likely to form bonds with the Fe2+ ion in hemoglobin?

    Answer

  • 9.110

    Which of the following ions possess a dipole moment: (a) , (b) , (c) , (d) ?

  • 9.111

    Which of the following geometries has a greater stability for tin(IV) hydride (SnH4)?

    Answer

  • 9.112

    Carbon dioxide has a linear geometry and is nonpolar. Yet we know that the molecule exhibits bending and stretching motions that create a dipole moment. How would you reconcile these seemingly conflicting descriptions of CO2?

  • 9.113

    Carbon suboxide (C3O2) is a colorless pungent-smelling gas. Does this molecule possess a dipole moment? Explain.

    Answer

Page 385
Pre-Professional Practice Exam Questions: Physical Sciences

These questions are not based on a descriptive passage.

  1. What is the shape of the ICl3 molecule?

    1. Trigonal planar

    2. Trigonal pyramidal

    3. T-shaped

    4. Tetrahedral

  2. Which of the following molecules is nonpolar?

    1. NCl3

    2. BCl3

    3. PCl3

    4. BrCl3

  3. Which of the following has a bond order of 2?

    1. I only

    2. II only

    3. III only

    4. I and III

  4. Which of the following are paramagnetic?

    1. I and II

    2. II and III

    3. III and IV

    4. I and III

Answers to In-Chapter Materials
Answers to Practice Problems
  • 9.1A
    1. linear,

    2. bent.

  • 9.1B
    1. Group 6A,

    2. Group 7A.

  • 9.2A

    Bent about O, tetrahedral about each C, trigonal pyramidal about N. All bond angles are ∽109.5°. Angles labeled in blue are <109.5°.

  • 9.2B

    SO2 contains double bonds, which are not pushed together by the central atom's lone pair as easily as the single bonds in NH3.

  • 9.3A
    1. polar,

    2. nonpolar.

  • 9.3B
    1. Trigonal bipyramidal, square planar, octahedral.

    2. Octahedral.

  • 9.4A

    Singly occupied 3p orbitals from the P atom overlap with s orbitals from H atoms.

  • 9.4B

    We cannot use valence bond theory to explain the bonding in SO2 or CH4. In the case of SO2, although the central atom has two unpaired electrons and can form two bonds, the unpaired electrons on S are in 3p orbitals. Formation of two bonds by the overlap of two 3p orbitals on S would be expected to result in a bond angle of approximately 90°. In the case of CH4, the central atom does not have enough unpaired electrons to form four bonds.

  • 9.5A

    Two of the 4p electrons in Br are promoted to empty d orbitals. The s orbital, all three porbitals, and two of the d orbitals hybridize to form six sp3d2 hybrid orbitals. One of the hybrid orbitals contains the lone pair. Each of the remaining hybrid orbitals contains one electron and overlaps with a singly occupied 2p orbital on an F atom. The arrangement of hybrid orbitals is octahedral and the bond angles are ∽90°.

  • 9.5B

    One of the 2s electrons in Be is promoted to an empty p orbital. The s orbital and one p orbital hydridize to form two sp hybrid orbitals. Each hybrid orbital contains one electron and overlaps with a singly occupied 2p orbital on an F atom. The arrangement is linear with a bond angle of ∽180°.

  • 9.6A

    16 σ bonds and 4 π bonds.

  • 9.6B

    17 σ bonds and 5 π bonds.

  • 9.7A

    C and N atoms are sp-hybridized. The triple bond between C and N is composed of one sigma bond (from overlap of hybrid orbitals) and two pi bonds (from interaction of remaining p orbitals). The single bond between H and C is the result of an sp orbital from C overlapping with an s orbital from H.

  • 9.7B

    Each N atom is sp-hybridized. One sp orbital on each is singly occupied and one contains a lone pair. The singly occupied sp orbitals overlap to form a sigma bond between the N atoms. The remaining unhybridized p orbitals interact to form two pi bonds.

  • 9.8A

    paramagnetic; bond order = 2.

  • 9.8B

    paramagnetic; bond order = 2.

  • 9.9A

    Two different resonance structures are possible; therefore we consider all three atoms to be sp2-hybridized. One of the hybrid orbitals on the central O atom contains the lone pair, the other two form sigma bonds to the terminal O atoms. Each atom has one remaining unhybridized p orbital. The p orbitals combine to form p molecular orbitals.

  • 9.9B

    Two different resonance structures are possible; therefore we consider all three atoms to be sp2-hybridized. One of the hybrid orbitals on the central N atom contains the lone pair, the other two form sigma bonds to the terminal O atoms. Each atom has one remaining unhybridized p orbital. The p orbitals combine to form π molecular orbitals.

Answers to Checkpoints
  • 9.1.1

    d.

  • 9.1.2

    b.

  • 9.1.3

    e.

  • 9.1.4

    c.

  • 9.1.5

    d.

  • 9.2.1

    a.

  • 9.2.2

    c.

  • 9.3.1

    b, e

  • 9.3.2

    b.

  • 9.4.1

    b.

  • 9.4.2

    d.

  • 9.5.1

    a, c.

  • 9.5.2

    c.

  • 9.5.3

    a, b, e.

  • 9.5.4

    c.

  • 9.6.1

    b.

  • 9.6.2

    c, e.

  • 9.6.3

    b.

  • 9.6.4

    a, c, e.

  • 9.7.1

    c.

  • 9.7.2

    b.

  • 9.7.3

    a, b.

  • 9.7.4

    e.

Answers to Applying What You've Learned
    1. There are 14 carbon-carbon sigma bonds in musk ketone and four pi bonds.

    2. Each of the top two indicated C atoms has three electron domains around it and is therefore sp2-hybridized. The bottom indicated C atom has four electron domains around it and is therefore sp3-hybridized. The geometry about the sp2-hybridized C atoms is trigonal planar, whereas the geometry about the sp3-hybridized C atom is tetrahedral.

    3. Of the pi bonds, only the carbon-carbon and nitrogen-oxygen pi bonds are delocalized. Only the six-membered ring and the nitro groups can be represented with two different resonance structures.

    4. The bonding in the ring can be described as follows: each of the C atoms is sp2-hybridized. The sigma bonds between the C atoms result from the overlap of sp2 hybrid orbitals. Each C atom has one leftover, singly occupied p orbital. As in the case for benzene, the unhybridized p orbitals on all six C atoms combine to form six pi molecular orbitals (three bonding and three antibonding) that lie above and below the plane of the molecule. All six electrons in these pi orbitals reside in the bonding molecular orbitals.