Test Bank Answers Ch.9 Theories Of Bonding And Structure

Chemistry: Molecular Nature of Matter, 8e (Jespersen)

Chapter 9 Theories of Bonding and Structure

1) Which of the five basic geometries for molecules and ions has the largest bond angle?

A) linear

B) planar triangular

C) tetrahedral

D) square planar

E) octahedral

Diff: 1

Section: 9.1

2) Which of these examples is not one of the five basic geometries for molecules and ions?

A) planar triangular

B) octahedral

C) tetrahedral

D) square planar

E) trigonal bipyramidal

Diff: 2

Section: 9.1

3) Two different values, less than 180°, are observed for the bond angles in which of the following basic structures for molecules and ions?

A) linear

B) planar triangular

C) tetrahedral

D) trigonal bipyramidal

E) octahedral

Diff: 2

Section: 9.1

4) Which of the five basic geometries for molecules and ions has the smallest bond angle?

A) linear

B) planar triangular

C) tetrahedral

D) octahedral

E) These all have the same bond angles.

Diff: 2

Section: 9.1

5) Where is the central atom located in a trigonal bipyramidal molecule?

A) in the center of the plane shared by the two triangular pyramids, and at the point where the equatorial and axial bonds meet

B) at the end of an axial bond

C) at the end of an equatorial bond

D) on the middle edge of the plane formed by the two triangular pyramids

E) in the central face of any of the two triangular pyramids, where two bonds form a 120° angle

Diff: 2

Section: 9.1

6) The valence shell electron pair repulsion theory is based on the concept that

A) only two electrons can occupy an atomic orbital.

B) the position and speed of the electron are somewhat uncertain.

C) electron pairs tend to stay as far apart as possible to minimize repulsions between them.

D) electronegativity tends to promote the formation of polar bonds.

E) electrons are added to the lowest energy atomic orbitals first.

Diff: 1

Section: 9.2

7) Which basic arrangement would best accommodate two electron domains in the valence shell of a covalently bonded atom?

A) planar triangular

B) octahedral

C) tetrahedral

D) linear

E) trigonal bipyramidal

Diff: 1

Section: 9.2

8) Which basic arrangement would best accommodate three electron domains in the valence shell of a covalently bonded atom?

A) planar triangular

B) octahedral

C) tetrahedral

D) linear

E) trigonal bipyramidal

Diff: 1

Section: 9.2

9) Which basic arrangement would best accommodate four electron domains in the valence shell of a covalently bonded atom?

A) planar triangular

B) octahedral

C) tetrahedral

D) linear

E) trigonal bipyramidal

Diff: 1

Section: 9.2

10) Which basic arrangement would best accommodate five electron domains in the valence shell of a covalently bonded atom?

A) planar triangular

B) octahedral

C) tetrahedral

D) linear

E) trigonal bipyramidal

Diff: 1

Section: 9.2

11) Which basic arrangement would best accommodate six electron domains in the valence shell of a covalently bonded atom?

A) planar triangular

B) octahedral

C) tetrahedral

D) linear

E) trigonal bipyramidal

Diff: 1

Section: 9.2

12) Based on the Lewis structure, the number of electron domains in the valence shell of the boron atom in the BF3 molecule is

A) 1

B) 2

C) 3

D) 4

E) 5

Diff: 1

Section: 9.2

13) A molecule has a central atom surrounded by 2 lone pairs and 3 atoms. The best description for the shape of the molecule is

A) trigonal bipyramidal.

B) octahedral.

C) trigonal planar.

D) T-shaped.

E) see-saw.

Diff: 2

Section: 9.2

14) A molecule has a central atom surrounded by 1 lone pair and 3 atoms. The best description for the shape of the molecule is

A) trigonal pyramidal.

B) bent.

C) trigonal planar.

D) T-shaped.

E) see-saw.

Diff: 2

Section: 9.2

15) A molecule has a central atom surrounded by 1 lone pair and 5 atoms. The best description for the shape of the molecule is

A) octahedral.

B) square pyramidal.

C) trigonal planar.

D) square planar.

E) see-saw.

Diff: 2

Section: 9.2

16) Based on the Lewis structure, the number of electron domains in the valence shell of the arsenic atom in the AsCl3 molecule is

A) 1.

B) 2.

C) 3.

D) 4.

E) 5.

Diff: 2

Section: 9.2

17) Based on the Lewis structure, the number of electron domains in the valence shell of the carbon atom in the CO32- ion is

A) 1.

B) 2.

C) 3.

D) 4.

E) 5.

Diff: 2

Section: 9.2

18) Based on the Lewis structure, the number of electron domains in the valence shell of the xenon atom in the XeF2 molecule is

A) 1.

B) 2.

C) 3.

D) 4.

E) 5.

Diff: 2

Section: 9.2

19) Based on the Lewis structure, the number of electron domains in the valence shell of the sulfur atom in the SO2 molecule is

A) 5.

B) 4.

C) 3.

D) 2.

E) 1.

Diff: 2

Section: 9.2

20) Application of the concepts of the VSEPR theory, suggests that the shape of the SO3 molecule is

A) trigonal pyramidal.

B) square planar.

C) regular tetrahedral.

D) triangular planar.

E) distorted tetrahedron.

Diff: 2

Section: 9.2

21) Application of the concepts of VSEPR theory leads to the prediction that the shape of the PCl3 molecule is

A) bent.

B) linear.

C) regular tetrahedral.

D) triangular planar.

E) trigonal pyramidal.

Diff: 2

Section: 9.2

22) The geometry of the CS2 molecule is best described as

A) bent.

B) linear.

C) regular tetrahedral.

D) triangular planar.

E) trigonal pyramidal.

Diff: 2

Section: 9.2

23) Application of the VSEPR theory suggests that the geometric arrangement of the atoms in the carbonate ion, CO32, is

A) octahedral.

B) square planar.

C) regular tetrahedral.

D) triangular planar.

E) trigonal pyramidal.

Diff: 2

Section: 9.2

24) The bond angle in Cl2O is expected to be approximately

A) 90 degrees.

B) 109.5 degrees.

C) 120 degrees.

D) 145 degrees.

E) 180 degrees.

Diff: 2

Section: 9.2

25) The smallest F–P–F bond angle in PF6- ion is

A) 90 degrees.

B) 109.5 degrees.

C) 120 degrees.

D) 145 degrees.

E) 180 degrees.

Diff: 2

Section: 9.2

26) Application of the VSEPR model to the NCl3 molecule leads to the prediction that

A) the molecule should have trigonal pyramidal geometry.

B) the molecule should have delocalized valence electrons.

C) the molecule should be nonpolar.

D) the formal charge on the center atom is +3.

E) the molecule should have a planar triangular geometry.

Diff: 2

Section: 9.2

27) The molecular geometry of the SeF4 molecule is best described as

A) trigonal pyramidal.

B) square planar.

C) regular tetrahedral.

D) triangular planar.

E) distorted tetrahedral.

Diff: 2

Section: 9.2

28) Application of the concepts of VSEPR theory leads to the prediction that the shape of the XeOF4 molecule is

A) bent.

B) linear.

C) regular tetrahedral.

D) triangular planar.

E) square pyramidal.

Diff: 2

Section: 9.2

29) The geometry of the ClF3 molecule is best described as

A) distorted tetrahedral.

B) regular tetrahedral.

C) T-shaped.

D) trigonal pyramidal.

E) triangular planar.

Diff: 2

Section: 9.2

30) Based on conclusions from application of the VSEPR theory, which one of the following has a linear shape?

A) IF2+

B) HCN

C) H2CO

D) H2S

E) SO2

Diff: 2

Section: 9.2

31) Based on conclusions from application of the VSEPR theory, which one of the following has a linear shape?

A) IF2-

B) HCO-

C) H2CO

D) H2S

E) SO2

Diff: 2

Section: 9.2

32) Based on application of the VSEPR theory, which molecule has a trigonal bipyramidal shape?

A) BrF5

B) AsF5

C) SF4

D) XeOF4

E) SbCl52-

Diff: 2

Section: 9.2

33) Based on application of the VSEPR theory, which molecule has a bent (v-shaped) geometry?

A) CO2

B) CS2

C) KrF2

D) C2H2

E) SO2

Diff: 2

Section: 9.2

34) Based on application of the VSEPR theory, which molecule or ion has a bent (v-shaped) geometry?

A) Cl2O

B) CO2

C) HCN

D) CO

E) NO2+

Diff: 2

Section: 9.2

35) Based on application of the VSEPR theory, which molecule or ion has a regular tetrahedral geometry?

A) IF4-

B) CF4

C) H3O+

D) SF4

E) XeF4

Diff: 2

Section: 9.2

36) Based on application of the VSEPR theory, the expected bond angles in the AsF4 ion should be approximately

A) 90 degrees.

B) 109.5 degrees.

C) 120 degrees.

D) 90, 120 and 180 degrees.

E) 90 and 180 degrees.

Diff: 2

Section: 9.2

37) Which statement is true for the BrF5 molecule?

A) There are two lone pairs in the valence shell of the bromine atom.

B) There is an unpaired electron in the valence shell of the bromine atom.

C) Some of the F–Br–F bond angles are close to 90 degrees.

D) At least one of the bond angles, F–Br–F, is about 109.5 degrees.

E) Some of the F–Br–F bond angles are close to 120 degrees.

Diff: 2

Section: 9.2

38) Which best explains the polarity of the PCl3 molecule?

A) Vector addition yields a net zero dipole moment.

B) The tetragonal shape leads to no overall polarity.

C) The trigonal pyramid shape leads to a nonpolar molecule.

D) The P-Cl dipole moments sum to a net dipole moment.

E) Two nonbonding electron pairs cause an overall polarity.

Diff: 2

Section: 9.3

39) Based on observed periodic trends, arrange the following species, CH3Br, CH3Cl, CH3F, CH3I, in order of increasing dipole moment.

A) CH3F < CH3Cl < CH3Br < CH3I

B) CH3Br < CH3Cl < CH3F < CH3I

C) CH3I < CH3F < CH3Cl < CH3Br

D) CH3I < CH3Br < CH3Cl < CH3F

E) CH3Cl < CH3Br < CH3I < CH3F

Diff: 2

Section: 9.3

40) In the CSe2 molecule,

A) the bonds are all nonpolar, which makes it polar.

B) the bonds are polar, but their effect on the overall polarity is canceled by the effect of lone pairs in the valence shell of the carbon atom.

C) the bonds are polar, but their effect on the overall polarity is canceled by the fact that they are equal in magnitude and oppositely directed.

D) the bonds are polar, but the polar effect is canceled by the resonance hybrids which distribute the charge evenly.

E) the bonds are slightly polar, and so they will not affect the overall polarity of the molecule.

Diff: 2

Section: 9.3

41) Based on application of the VSEPR theory, which molecule should be nonpolar?

A) CH3Cl

B) CS2

C) H2O

D) NH3

E) OF2

Diff: 2

Section: 9.3

42) Which molecule below is polar?

A) CF4

B) BF3

C) C2H2

D) CSe2

E) IBr

Diff: 2

Section: 9.3

43) Which molecule below is most likely nonpolar?

A) NCl3

B) SO2

C) NO2

D) SO3

E) CH2Cl2

Diff: 2

Section: 9.3

44) Which of these examples is a polar molecule?

A) XeF2

B) BrF5

C) XeF4

D) CCl4

E) PBr5

Diff: 2

Section: 9.3

45) Which of these examples is a nonpolar molecule?

A) CHCl3

B) BrF5

C) CO2

D) SO2

E) NH3

Diff: 2

Section: 9.3

46) Predict the molecular geometry and polarity of the SO2 molecule by applying VSEPR theory.

A) linear, nonpolar

B) linear, polar

C) bent, 109.5° bond angle, polar

D) bent, 120° bond angle, polar

E) bent, 109.5° bond angle, nonpolar

Diff: 2

Section: 9.3

47) Predict the molecular geometry and polarity of the O–C–S molecule.

A) linear, nonpolar

B) bent, nonpolar

C) linear, polar

D) tetrahedral, nonpolar

E) bent, polar

Diff: 2

Section: 9.3

48) Compare the molecules SbCl5 and IF5, using VSEPR theory.

A) IF5 is polar, while SbF5 is nonpolar.

B) Both molecules have a trigonal bipyramidal geometry and are also nonpolar.

C) SbF5 is polar and IF5 is nonpolar.

D) Both molecules are polar and have a trigonal bipyramidal geometry.

E) Both molecules have lone pairs in the valence shell of the central atom.

Diff: 2

Section: 9.3

49) Which of the following species has the largest dipole moment?

A) CH4

B) CH3F

C) CH3Cl

D) CH3Br

E) CBr4

Diff: 2

Section: 9.3

50) Bonding in the chlorine molecule can be explained by the valence bond theory in terms of an overlap between

A) the 1s orbital of a chlorine atom and the 1s orbital of the other chlorine atom.

B) the 2s orbital of a chlorine atom and the 2s orbital of the other chlorine atom.

C) a 2p orbital of a chlorine atom and a 2p orbital of the other chlorine atom.

D) the 3s orbital of a chlorine atom and the 3s orbital of the other chlorine atom.

E) a 3p orbital of a chlorine atom and a 3p orbital of the other chlorine atom.

Diff: 2

Section: 9.4

51) Bonding in the hydrogen chloride molecule can be explained by the valence bond theory in terms of an overlap between

A) the 1s orbital of the hydrogen atom and the 1s orbital of the chlorine atom.

B) the 1s orbital of the hydrogen atom and the 2s orbital of the chlorine atom.

C) the 1s orbital of the hydrogen atom and a 2p orbital of the chlorine atom.

D) the 1s orbital of the hydrogen atom and the 3s orbital of the chlorine atom.

E) the 1s orbital of the hydrogen atom and a 3p orbital of the chlorine atom.

Diff: 2

Section: 9.4

52) Bonding in the nitrogen trifluoride molecule can be explained by the valence bond theory in terms of an overlap between

A) the 2s orbital of the nitrogen and the 2s orbital of the fluorine.

B) the 2s orbital of the nitrogen and a 2p orbital of the fluorine.

C) a 2p orbital of the nitrogen and a 2p orbital of the fluorine.

D) the 1s orbital of the nitrogen and a 2p orbital of the fluorine.

E) a 2p orbital of the nitrogen and the 3p orbital of the fluorine.

Diff: 2

Section: 9.4

53) What hybrid orbitals are used by the carbon atom in CH4 for bonding?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

54) What is the hybridization on C in C2H2?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

55) What is the hybridization of the nitrogen atom in NH3?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

56) What type of hybrid orbitals do the carbon atoms have in C2H6?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

57) What type of hybrid orbitals do the carbon atoms have in C2H4?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

58) What type of hybridization does the nitrogen atom have in H3CNO2?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

59) What hybrid orbitals are on the sulfur atom in a sulfur tetrafluoride molecule?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

60) What hybrid orbitals on an As atom in the AsF4- ion?

A) sp3d2

B) sp

C) sp3d

D) sp3

E) sp2

Diff: 2

Section: 9.5

61) Which of the following molecules or ions does not use sp3d2 hybrid orbitals on the central atom for bonding?

A) SF6

B) PF6-

C) IF4+

D) XeF4

E) BrF5

Diff: 2

Section: 9.5

62) The central atom in BrF5 uses sp3d2 hybridization. It must therefore have ________ σbonds and ________ lone pair(s) of electrons in the valence shell of the central atom.

A) 1, 5

B) 3, 3

C) 5, 0

D) 5, 1

E) 6, 0

Diff: 2

Section: 9.5

63) The central atom in SF4 has ________ σbonding pair(s) and ________ lone pair(s) of electrons in the valence shell of the central atom.

A) 3, 2

B) 4, 0

C) 4, 1

D) 5, 0

E) 5, 1

Diff: 2

Section: 9.5

64) What type of hybrid orbitals are used by the central atom for σbonding in the PCl3 molecule?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

65) What type of hybrid orbitals are used by the central atom for σ-bonding in the PCl4- ion?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

66) What kind of hybrid orbitals are used by the central atom for σbonding in the SF6 molecule?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

67) What kind of hybrid orbitals are used by the central atom for σbonding in the AsF5 molecule?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

68) What kind of hybrid orbitals are used by the central atom for σbonding in the NO3 ion?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

69) What kind of hybrid orbitals are used by the As atom for σbonding in the AsF4- ion?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

70) What kind of hybrid orbitals are used by the central atom for σbonding in the ClO3- ion?

A) sp

B) sp2

C) sp3

D) sp3d

E) sp3d2

Diff: 2

Section: 9.5

71) In the carbon monoxide molecule, there will be ________ π-bond(s) and the ________ hybrid orbitals on the carbon would be used for σ-bonding in the molecule.

A) 2, sp

B) 2, sp2

C) 3, sp

D) 3, sp2

E) 1, sp2

Diff: 2

Section: 9.6

72) How many σ-bonds and π-bond(s) are in the H3CNO2 molecule?

A) 7 σ-bonds and 0 π-bonds

B) 6 σ-bonds and 1 π-bond

C) 1 σ-bond and 5 π-bonds

D) 5 σ-bonds and 2 π-bonds

E) 2 σ-bonds and 1 π-bond

Diff: 2

Section: 9.6

73) How many σ-bonds and π-bonds are in a CO2 molecule?

A) 1 σ-bond and 2 π-bonds

B) 2 σ-bonds and 0 π-bonds

C) 2 σ-bonds and 2 π-bonds

D) 2 σ-bonds and 4 π-bonds

E) 4 σ-bonds and 0 π-bonds

Diff: 2

Section: 9.6

74) Which statement about the ammonia molecule, NH3, is true?

A) The N–H bonds are polar, but the molecule is nonpolar.

B) The molecule has both three σ- and one π-bond.

C) The nitrogen atom utilizes sp3 hybrid orbitals for σbonding.

D) The H–N–H bond angles are greater than 109.5 degrees.

E) The molecular geometry is planar triangular.

Diff: 2

Section: 9.6

75) How many π-bonds are there in the CN ion?

A) 0

B) 1

C) 1.5

D) 2

E) 3

Diff: 2

Section: 9.6

76) If formal charge considerations are taken into account, how many π-bond(s) is/are in the SO32- ion?

A) 0

B) 1

C) 2

D) 3

E) 4

Diff: 2

Section: 9.6

77) Which molecule or ion uses an sp2 set of hybrid orbitals on the central atom for σbonding?

Hint: Consider formal charge when deciding on the correct structure and remember that sulfur atoms can access d-orbitals.

A) CS2

B) N2O

C) SO2

D) NF3

E) I3-

Diff: 3

Section: 9.6

78) When formal charges are considered, how many π-bond(s) and σbonds, respectively, are in the chlorate ion, ClO3-?

Hint: Remember that chlorine atoms can access d-orbitals.

A) 1, 5

B) 2, 4

C) 2, 3

D) 3, 3

E) 3, 4

Diff: 3

Section: 9.6

79) Ozone (O3) is an allotropic form of oxygen. Based on VSEPR and valence bond theory, which statement is true for this molecule?

A) The molecule is linear, and has no lone pairs in the valence shell on the central atom.

B) The molecule is linear, and has two π-bonds and one π-bond.

C) The molecule is T-shaped, and has two π-bonds and two σ-bonds.

D) The molecule is bent, and has two lone pairs in the valence shell on the central atom.

E) The molecule is bent, and has one π-bond and two σ-bonds.

Diff: 2

Section: 9.6

80) The H2 ion, according to MO theory, has a bond order of

A) 0

B) 0.5

C) 1.0

D) 1.5

E) 2.0

Diff: 1

Section: 9.7

81) The H2+ ion, according to MO theory, has a bond order of

A) 0

B) 0.5

C) 1.0

D) 1.5

E) 2.0

Diff: 1

Section: 9.7

82) The He2 molecule, according to MO theory, has a bond order of

A) 0

B) 0.5

C) 1.0

D) 1.5

E) 2.0

Diff: 1

Section: 9.7

83) The restrictions on the placing of electrons in molecular orbitals in molecules and ions

A) must obey the Pauli exclusion principle but not Hund's rule or the Aufbau principle.

B) must obey the Pauli exclusion principle and the Aufbau principle but not Hund's rule.

C) must obey the Aufbau principle but not the Pauli exclusion principle or Hund's rule.

D) are not governed by either the Aufbau principle, Hund's rule, or the Pauli exclusion principle.

E) must obey the Aufbau principle, Hund's rule, and the Pauli exclusion principle.

Diff: 2

Section: 9.7

84) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2, lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, the bond order for the bond in the O2+ ion in its ground state is

A) 1.0

B) 1.5

C) 2.0

D) 2.5

E) 3.0

Diff: 2

Section: 9.7

85) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2 , lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, the bond order for the bond in the O2 — ion in its ground state is

A) 1.0

B) 1.5

C) 2.0

D) 2.5

E) 3.0

Diff: 2

Section: 9.7

86) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2 , lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, how many electrons are in all of the antibonding molecular orbitals of the O2 molecule in its ground state?

A) 1

B) 2

C) 3

D) 4

E) 5

Diff: 2

Section: 9.7

87) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2 , lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, how many electrons are there in all of the antibonding molecular orbitals of the F2+ ion in its ground state?

A) 1

B) 2

C) 3

D) 4

E) 5

Diff: 2

Section: 9.7

88) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2 , lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, how many unpaired electrons are in all of the antibonding molecular orbitals of the O2- ion in its ground state?

A) 1

B) 2

C) 3

D) 4

E) 5

Diff: 2

Section: 9.7

89) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2, lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, how many unpaired electrons are in all of the antibonding molecular orbitals of the O22- ion in its ground state?

A) 0

B) 1

C) 2

D) 3

E) 4

Diff: 2

Section: 9.7

90) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2, lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, which species has the largest number of unpaired electrons in its ground state?

A) Ne2

B) O2+

C) O2-

D) O2

E) F2+

Diff: 2

Section: 9.7

91) An energy level scheme for the orbitals of second row diatomic molecules O2 through Ne2, lists the molecular orbitals in the following order of increasing energy

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on this energy level scheme, which statement is true about the bonds in the species, O2, O2+, O2, O22?

A) O2- has the highest bond order.

B) O22- has the highest bond order.

C) O2 has the highest bond order.

D) O2+ has the highest bond order.

E) The bond order in all these species is the same.

Diff: 2

Section: 9.7

92) According to MO theory, the diatomic C2 molecule, has a bond order of

A) 0

B) 0.5

C) 1.0

D) 1.5

E) 2.0

Diff: 2

Section: 9.7

93) Consider the cyclobutadiene structure shown below. How many π bonds are there?

A) 1

B) 2

C) 3

D) 4

E) 8

Diff: 1

Section: 9.8

94) What is the total number of π-bonds in the benzene molecule?

A) 0

B) 1

C) 2

D) 3

E) 4

Diff: 2

Section: 9.8

95) The electrons in the delocalized molecular orbitals of benzene

A) are free to move around the six-membered ring.

B) are restricted to the space between only two carbon atoms.

C) are used to bond the hydrogens to the carbon ring.

D) only contain one electron each.

E) are responsible for bond lengths that are longer in only three bonds in the carbon ring.

Diff: 2

Section: 9.8

96) Which species from the list, NO3, NO2, O3, have delocalized molecular orbitals?

A) NO3 and NO2

B) NO2 and O3

C) NO3, NO2 and O3

D) NO3 only

E) NO3 and O3

Diff: 2

Section: 9.8

97) The energy separation between the nearest empty conduction band and the filled valence band is known as

A) the localization energy.

B) the delocalization energy.

C) the energy band.

D) the band gap.

E) the conduction energy.

Diff: 1

Section: 9.9

98) In solid sodium metal, which orbitals contribute to the formation of the localized band of energy levels?

A) 1s only

B) 1s, 2s

C) 1s, 2s, 3s

D) 1s, 2s, 2p

E) 1s, 2s, 2p, 3s

Diff: 2

Section: 9.9

99) In solid sodium metal, which orbital(s) contribute(s) to the formation of the delocalized band of energy levels?

A) 3s

B) 1s, 2s

C) 1s, 2s, 3s

D) 1s, 2s, 2p

E) 1s, 2s, 2p, 3s

Diff: 2

Section: 9.9

100) Which of the following substances is expected to have the smallest band gap?

A) carbon dioxide

B) sulfur

C) calcium

D) diamond

E) glass

Diff: 2

Section: 9.9

101) Which of the following substances is expected to have a reasonably small band gap which would allow it to show properties of a semiconductor?

A) magnesium

B) sulfur

C) calcium

D) germanium

E) sodium

Diff: 2

Section: 9.9

102) Which of the following statements is true?

A) Magnesium is a conductor of electricity, because the filled 3s valence band overlaps the vacant 3p conduction band, and this allows it to be easily populated with electrons when a voltage is applied.

B) Metallic sodium is a good conductor of electricity because the valence band is quite different from the conduction band.

C) Calcium metal is a good conductor of electricity because the band gap is fairly large.

D) Germanium is a weak conductor of electricity because the valence band is filled with electrons and these are the electrons normally used to form covalent bonds.

E) In silicon, which is a semiconductor, the band gap between the valence band and the nearest conduction band is large, but the thermal energy possessed by electrons is sufficient to promote some to the conduction band.

Diff: 2

Section: 9.9

103) Which of these elements is capable of forming the strongest π-bonds?

A) Hydrogen

B) Helium

C) Lithium

D) Carbon

E) Germanium

Diff: 2

Section: 9.10

104) Which of these elements has the greatest tendency to form multiple bonds?

A) Neon

B) Selenium

C) Oxygen

D) Magnesium

E) Germanium

Diff: 2

Section: 9.10

105) Diamond is one allotropic form of elemental carbon. Which hybrid orbitals are used by the carbon atoms in the diamond structure?

A) Hybridization does not occur in elements.

B) sp3

C) sp

D) sp2

E) sp3d

Diff: 2

Section: 9.10

106) The two vertical bonds in a trigonal bipyramidal molecule are called ________ bonds.

Diff: 1

Section: 9.1

107) The angle between two equatorial bonds in a trigonal bipyramidal molecule measures ________.

Diff: 2

Section: 9.1

108) The octahedron has exactly ______vertices, and the angles between adjacent bonds are equal to ________ degrees.

Diff: 3

Section: 9.1

109) The shape of dichlorophosphine, PHCl2, is best described as ________.

Diff: 2

Section: 9.2

110) The shape of the NH2- ion is best described as ________.

Diff: 2

Section: 9.2

111) What is the shape of the HArF molecule?

Diff: 2

Section: 9.2

112) What is the shape of the SiCl4 molecule?

Diff: 2

Section: 9.2

113) What is the shape of the boron trichloride molecule?

Diff: 2

Section: 9.2

114) What is the shape of the PH3 molecule?

Diff: 2

Section: 9.2

115) What is the shape of the SF4 molecule?

Diff: 2

Section: 9.2

116) The I3- ion has a total of ________ electron domains on the center atom and its shape is ________.

Diff: 2

Section: 9.2

117) The I3+ ion has a total of ________ electron domains on the center atom and its shape is ________.

Diff: 2

Section: 9.2

118) The F-S-F bond angles in SF6 are approximately ________.

Diff: 2

Section: 9.2

119) In XeF2, vector addition of the bond dipoles lead to a dipole moment of ________.

Diff: 2

Section: 9.3

120) The C-H bond in CHF3 is less polar than the C-F bonds, which means the overall polarity of CHF3 is ________.

Diff: 2

Section: 9.3

121) The electronegativities of the atoms in PCl3 lead to bonds that are ________. The overall polarity of the PCl3 molecule is ________.

Diff: 2

Section: 9.3

122) The electronegativities of the atoms in SO2 lead to bonds that are ________. The overall polarity of the SO2 molecule is ________.

Diff: 2

Section: 9.3

123) The electronegativities of the atoms in CS2 tell us the bonds are ________. The overall polarity of the CS2 molecule is ________.

Diff: 2

Section: 9.3

124) The overall polarity of the XeF2 is ________.

Diff: 2

Section: 9.3

125) When the portions of two ________ orbitals from different atoms share the same common space, the orbitals are said to overlap.

Diff: 2

Section: 9.4

126) Based on valence bond theory, the predicted bond angle in the H2S molecule is ________.

Diff: 2

Section: 9.4

127) According to valence bond theory, the bonding in the diatomic fluorine molecule, F2, can be explained by the overlap of two ________ orbitals.

Diff: 2

Section: 9.4

128) The C2H2 molecule has hybrid orbitals centered on the carbon atoms. What are the hybrid orbitals used by each of the carbon atoms?

Diff: 2

Section: 9.5

129) The PCl3 molecule has hybrid orbitals centered on the phosphorus atom. What are the hybrid orbitals used by the phosphorus atom?

Diff: 2

Section: 9.5

130) The C2H4 molecule has hybrid orbitals centered on the carbon atoms. What are the hybrid orbitals used by each of the carbon atoms?

Diff: 2

Section: 9.5

131) According to valence bond theory, the N2H4 molecule has hybrid orbitals centered on the nitrogen atoms. What are the hybrid orbitals used by each nitrogen atom?

Diff: 2

Section: 9.5

132) According to valence bond theory, the CCl4 molecule has hybrid orbitals centered on the carbon atom. What are the hybrid orbitals used by the carbon atom?

Diff: 2

Section: 9.5

133) According to valence bond theory, the SF6 molecule has hybrid orbitals centered on the sulfur atom. What are the hybrid orbitals used by the sulfur atom?

Diff: 2

Section: 9.5

134) The hybridization of the central atom in ClO3- is ________.

Diff: 2

Section: 9.5

135) Formation of a π-bond requires that two unhybridized ________ orbitals overlap sideways.

Diff: 2

Section: 9.6

136) The butane molecule, C4H10, has a total of ________ σ-bonds.

Diff: 2

Section: 9.6

137) The hybrid orbitals used for σ-bonding in the PO3- ion (metaphosphate) are ________ orbitals.

Diff: 2

Section: 9.6

138) In the NO+ ion, there are ________ π-bond(s) and the ________ type of hybrid orbital on the nitrogen atom is used for σ-bonding.

Diff: 2

Section: 9.6

139) If six atomic orbitals combine to form a set of molecular orbitals, how many molecular orbitals

are formed?

Diff: 2

Section: 9.7

140) If we use the same molecular orbital energy level scheme for the valence orbitals in the NO ion as is used for diatomic O2 through N2 molecules,

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

how many unpaired electrons are there in the antibonding orbitals of the system?

Diff: 2

Section: 9.7

141) If we use the same molecular orbital energy level scheme for the valence orbitals in the NO molecule as was used for diatomic O2 through N2 molecules,

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

what is the bond order of the NO molecule?

Diff: 2

Section: 9.7

142) Based on molecular orbital theory, what is the total number of π electrons in the N2 molecule?

Diff: 2

Section: 9.7

143) In molecular orbital theory, the extra stability associated with the delocalization of the s- electrons is called the ________.

Diff: 1

Section: 9.8

144) The metaphosphate ion, PO3-, has a total of ________ π-bonds, and therefore has ________ s- electrons.

Diff: 2

Section: 9.8

145) Select the species which has the largest delocalization energy: I3-, ICl2+, O3.

Hint: Draw the Lewis structures for these three species.

Diff: 3

Section: 9.8

146) The energy separation between the nearest empty conduction band and the filled valence band is known as the ________.

Diff: 1

Section: 9.9

147) In metallic solids, the band that is either vacant or partially filled with electrons, is known as the ________.

Diff: 2

Section: 9.9

148) In a typical semiconductor, the valence band is filled with electrons, and the energy gap to the nearest conduction band is ________.

Diff: 2

Section: 9.9

149) The existence of an element in more than one form due to differences in the packing of molecules in the solid is called ________.

Diff: 1

Section: 9.10

150) The square pyramid is one of the five basic geometries from which molecular shapes are determined.

Diff: 1

Section: 9.1

151) The hexagonal pyramid is one of the five basic geometries from which molecular shapes are determined.

Diff: 2

Section: 9.1

152) A tetrahedral molecule or ion is a four-sided geometric figure, shaped like a pyramid with

triangular faces. Four atoms are located at the vertices and are bonded to a fifth atom located in the center of the three-dimensional structure.

Diff: 2

Section: 9.1

153) An electron domain can be a bond, a lone pair of electrons, or even an unpaired electron.

Diff: 2

Section: 9.2

154) Electrons in a single, double, or triple bond, count as one, two, and three separate electron domains, respectively.

Diff: 2

Section: 9.2

155) Molecules like SnCl2 and NO2, which have similar empirical formulas, both contain nonbonding domains with a single unpaired valence electron on the central atom.

Diff: 2

Section: 9.2

156) All molecules with polar bonds have an overall net polarity.

Diff: 1

Section: 9.3

157) Molecules containing only nonpolar bonds are usually nonpolar.

Diff: 2

Section: 9.3

158) The PCl5 molecule has polar bonds but is a nonpolar molecule.

Diff: 2

Section: 9.3

159) The SO2 molecule is nonpolar, because the polar S–O bonds are geometrically symmetric.

Diff: 2

Section: 9.3

160) The CS2 molecule is polar because the polar C–S bond dipoles sum to a net dipole moment.

Diff: 2

Section: 9.3

161) The NH3 molecule is polar because the polar N–H bonds are not geometrically symmetric.

Diff: 2

Section: 9.3

162) The XeCl2 and XeCl4 molecules are both polar because the central atom has lone pairs of electrons.

Diff: 2

Section: 9.3

163) A covalent bond is formed when two atoms position themselves so that minimum overlap of the atomic orbitals occurs with a decrease in the potential energy of the system.

Diff: 2

Section: 9.4

164) A covalent bond is formed when two atoms position themselves so that maximum overlap of the atomic orbitals occurs with an increase in the potential energy of the system.

Diff: 2

Section: 9.4

165) Any number of electrons, with paired spins, can be shared by two overlapping atomic orbitals to yield the maximum potential energy for the entire system, and ultimately produce the strongest possible covalent bond.

Diff: 2

Section: 9.4

166) When two orbitals overlap and share electrons, the shape of the original orbitals will change in order to allow better overlap between the two orbitals.

Diff: 2

Section: 9.4

167) A molecule of CH3Cl forms a tetrahedral shape.

Diff: 2

Section: 9.5

168) According to valence bond theory, the BrF3 molecule has sp2 hybrid orbitals on the bromine atom.

Hint: Consider the valence electrons on the central atom.

Diff: 3

Section: 9.5

169) According to valence bond theory, the SF4 molecule has sp3 hybrid orbitals on the sulfur atom.

Hint: Consider the valence electrons on the central atom.

Diff: 3

Section: 9.5

170) According to valence bond theory, the H2CO molecule has sp2 hybrid orbitals on the carbon atom that form the σ-bonds.

Diff: 2

Section: 9.5

171) When two atoms interact to form a bond, an s orbital can never be employed in forming a π-bond.

Diff: 1

Section: 9.6

172) When two atoms combine to form a π-bond, it is possible that at least one of the interacting orbitals is an s orbital.

Diff: 2

Section: 9.6

173) When two atoms combine to form a σ-bond, it is possible for both of the interacting orbitals to be p orbitals.

Diff: 2

Section: 9.6

174) A pair of p orbitals can form a π-bond or an σ-bond.

Diff: 2

Section: 9.6

175) A pair of s orbitals can form a π-bond.

Diff: 2

Section: 9.6

176) When two atomic orbitals combine to form a hybrid orbital, it is possible for both of the overlapping orbitals to be p orbitals.

Diff: 2

Section: 9.6

177) When two hydrogen atoms interact so that their valence orbitals form a bonding molecular orbital and an antibonding molecular orbital, the placement of the electrons into these molecular orbitals lowers the potential energy of the system.

Diff: 2

Section: 9.7

178) The restrictions placed on the number of electrons that can occupy an orbital by the Pauli exclusion principle applies to atomic orbitals but not to molecular orbitals.

Diff: 2

Section: 9.7

179) The two atomic valence orbitals of the He2 molecule form four molecular orbitals.

Diff: 2

Section: 9.7

180) According to molecular orbital theory, the ground state electronic configuration

for He2+ is (σ1s)2 (σ*1s)1.

Diff: 2

Section: 9.7

181) According to molecular orbital theory, the bond order for the Be22- ion is 2.

Diff: 2

Section: 9.7

182) In the molecular orbital description of the benzene molecule, each carbon atom forms two C-C σ-bonds.

Diff: 2

Section: 9.8

183) All of the atoms in the nitrate ion lie in the same plane.

Diff: 2

Section: 9.8

184) The basic structure of the carbonate ion, CO32-, is determined by the σ-bonds., Each of the atoms has sp2 hybrid orbitals, and each oxygen atom also has an unhybridized p orbital, which overlaps with a p orbital on carbon to form a delocalized π-electron cloud.

Diff: 2

Section: 9.8

185) The number of energy levels in a conduction band is equal to twice the number of orbitals supplied by the entire collection of atoms.

Diff: 2

Section: 9.9

186) A substance like glass is an insulator, because its band gap is very large.

Diff: 2

Section: 9.9

187) The 3s orbital of sodium is only half-filled, so the electron in the 3s orbital of metallic sodium also belongs to the conduction band in the metal. This allows sodium to conduct electricity well.

Diff: 2

Section: 9.9

188) Elements in the third and fourth periods have a strong tendency to form multiple bonds.

Diff: 2

Section: 9.10

189) The π-bond formed by the element phosphorus is weaker than the sigma bond that it forms.

Diff: 2

Section: 9.10

190) The P-P single bond is stronger than the N-N single bond, while the N-N double bond is stronger than the P-P double bond.

Diff: 2

Section: 9.10

191) Electronic devices which are based on graphene are expected to operate at speeds which are faster than those based on silicon, as single sheets of graphene can transport electrons at a faster rate.

Diff: 1

Section: On the Cutting Edge 9.1

192) Graphene are sheet-like layers made up of carbon atoms which can be obtained from diamond which is an allotrope of carbon.

Diff: 2

Section: On the Cutting Edge 9.1

193) In graphene, carbons has sp2 hybrid orbitals.

Diff: 2

Section: On the Cutting Edge 9.1

194) One way to make graphene is to heat silicon carbide, SiC, causing the more volatile elemental carbon to undergo sublimation, after which it is collected and purified.

Diff: 2

Section: On the Cutting Edge 9.1

195) The number of nonbonding electron domains in the CH3+ ion is

A) 0.

B) 1.

C) 2.

D) 3.

E) 4.

Diff: 2

Section: 9.2

196) The total number of electron domains in the CH3 — ion is

A) 0.

B) 1.

C) 2.

D) 3.

E) 4.

Diff: 2

Section: 9.2

197) Consider a hypothetical species with an octahedral geometry shown below, with three different placements of its bonded atoms (indicated as "A") and lone pairs (indicated as blank).

Which species, 1, 2, or 3, is the preferred orientation of atoms and lone pairs?

A) 1

B) 2

C) 3

D) 2 and 3 are equally probable.

E) None of these arrangements of atoms and lone pairs is possible.

Diff: 2

Section: 9.2

198) A molecule has a measurable dipole moment. If the formula is AX3, which geometries are possible?

A) trigonal planar only

B) bent only

C) trigonal planar and trigonal pyramidal

D) T-shaped and trigonal pyramidal

E) None of these are possible.

Diff: 2

Section: 9.3

199) Kevin, the high school whiz kid, is trying to make a compound which contains neon as one of its elements. He figured that the difference in electron affinity between oxygen and neon is too small for them to form an ionic bond, so he is considering a feasibility study on the NeO molecule. He tried the same molecular orbital energy level scheme for the valence orbitals in the proposed NeO system as was used for diatomic species like O2.

σ2s < σ*2s < σ2p(z) < π2p(y), π2p(x) < π*2p(y), π*2p(x) < σ*2p(z)

Based on his proposal, a) what would be the bond order of this molecule, and b) how many unpaired electrons would there be in the antibonding orbitals of the system?

a) ________, b) ________

Diff: 2

Section: 9.7

200) Which molecular structure is most likely to contain π-bonds?

S8, P4, C60, Se8

Hint: Consider the period of the elements in these molecules.

A) S8

B) S8, P4, and C60

C) P4

D) C60

E) S8, P4, Se8

Diff: 3

Section: 9.10

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