Princeton 2012/Barron 4th ed.
AP Practice Problems
Unit 4 – Bonding
Multiple Choice (no calculator)
For questions 1-4, one or more of the
following responses will apply; each
response may be used more than once or not
at all in these questions.
A.
B.
C.
D.
E.
Metallic bonding
Network covalent bonding
Hydrogen bonding
Ionic bonding
London dispersion forces
1. Solids exhibiting this kind of
bonding are excellent conductors of
heat. (P4.1)
2. This kind of bonding is the reason
that water is more dense than ice.
(P4.2)
6. This molecule contains two pi (π)
bonds. (P4.6)
7. This substance undergoes hydrogen
bonding. (P4.7)
For questions 8-10, one or more of the
following responses will apply; each
response may be used more than once or not
at all in these questions.
(A) BF3
(B) CO2
(C) H2O
(D) CF4
(E) PH3
8. The central atom in this molecule
forms sp2 hybrid orbitals. (P4.8)
3. This kind of bonding exists between
atoms with very different
electronegativities. (P4.3)
9. This molecule has a tetrahedral
structure. (P4.9)
4. The stability exhibited by diamonds
is due to this type of bonding. (P4.4)
10. This molecule has a linear structure.
(P4.10)
For questions 5-7, one or more of the
following responses will apply; each
response may be used more than once or not
at all in these questions.
11. A liquid that is held together by
which of the following forces would
be expected to have the lowest
boiling point? (P4.11)
(A) CH4
(B) NH3
(C) NaCl
(D) N2
(E) H2
5. This substance undergoes ionic
bonding. (P4.5)
a.
b.
c.
d.
e.
Ionic bonds
London dispersion forces
Hydrogen bonds
Metallic bonds
Network bonds
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12. Hydrogen bonding would be seen in
a sample of which of the following
substances? (P4.12)
a.
b.
c.
d.
e.
CH4
H2
H2O
HI
All of the above
13. Which of the following species does
NOT have a tetrahedral structure?
(P4.13)
a.
b.
c.
d.
e.
CH4
NH4+
SF4
AlCl4CBr4
14. Which form of orbital hybridization
can form molecules with shapes that
are either trigonal pyramidal or
tetrahedral? (P4.14)
a.
b.
c.
d.
e.
sp
sp2
sp3
d2sp
dsp3
15. The six carbon atoms in a benzene
molecule are shown in different
resonance forms as three single
bonds and three double bonds. If the
length of a single carbon-carbon
bond is 154 pm and the length of a
double carbon-carbon bond is 133
pm, what length would be expected
for the carbon-carbon bonds in
benzene? (P4.15)
a.
b.
c.
d.
e.
126 pm
133 pm
140 pm
154 pm
169 pm
16. Which of the following structures
could be the Lewis structure for
sulfur trioxide? (P4.16)
a.
b.
c.
d.
e.
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17. In which of the following species
does the central atom NOT form sp2
hybrid orbitals? (P4.17)
a.
b.
c.
d.
e.
SO2
BF3
NO3SO3
PCl3
18. A molecule whose central atom has
d2sp3 hybridization can have which
of the following shapes? (P4.18)
I. Tetrahedral
II. Square pyramidal
III. Square planar
a.
b.
c.
d.
e.
I only
III only
I and II only
II and III only
I, II, and III
19. Which of the following molecules
will have a Lewis dot structure with
exactly one unshared electron pair on
the central atom? (P4.19)
a.
b.
c.
d.
e.
H2O
PH3
PCl5
CH2Cl2
BeCl2
20. Which of the following lists of
species is in order of increasing
boiling points? (P4.20)
a.
b.
c.
d.
e.
H2, N2, NH3
N2, NH3, H2
NH3, H2, N2
NH3, N2, H2
H2, NH3, N2
21. Solid NaCl melts at a temperature of
800°C, while solid NaBr melts at
750°C. Which of the following is an
explanation for the higher melting
point of NaCl? (P4.21)
a. A chlorine ion has less mass
than a bromine ion.
b. A chlorine ion has a greater
negative charge than a
bromine ion.
c. A chlorine ion has a lesser
negative charge than a
bromine ion.
d. A chlorine ion is smaller than
a bromine ion.
e. A chlorine ion is larger than a
bromine ion.
22. Which of the compounds listed
below would require the greatest
energy to separate it into ions in the
gaseous state? (P4.22)
a.
b.
c.
d.
e.
NaCl
NaI
MgO
Na2O
MgCl2
23. Which sample of the following
compounds contains both ionic and
covalent bonds? (P4.23)
a.
b.
c.
d.
e.
H2O2
CH3Cl
C2H3OH
NaNO3
NH2OH
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24. Which of the molecules listed below
has the largest dipole moment?
(P4.24)
a.
b.
c.
d.
e.
Cl2
HCl
SO3
NO
N2
a.
b.
c.
d.
e.
25. Which of the following statements
about boiling points is (are) correct?
(P4.25)
I.
II.
III.
a.
b.
c.
d.
e.
26. Which of these molecules has a
shape related to a tetrahedron (or has
sp3 bonding)? (B5.1)
H2O boils at a higher
temperature than CO2.
Ar boils at a higher
temperature than He.
Rb boils at a higher
temperature than Na.
I only
I and II only
I and III only
II and III only
I, II, and III
For questions 26-30, one or more of the
following responses will apply; each
response may be used more than once or not
at all in these questions.
I.
II.
III.
IV.
V.
CBr4
PF5
NH3
SO3
HCN
I and III
II
III and V
IV
V
27. Which of these molecules is linear?
(B5.2)
a.
b.
c.
d.
e.
I
II
III
IV
V
28. Which of these molecules has the
most pi bonds? (B5.3)
a.
b.
c.
d.
e.
I
II
III
IV
V
29. Which of these molecules has all
atoms lying in the same plane?
(B5.4)
a.
b.
c.
d.
e.
I and III
II
III and V
IV
IV and V
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30. Which of these molecules uses more
than an octet of electrons in its Lewis
structure? (B5.5)
a.
b.
c.
d.
e.
I and III
II
III and V
IV
V
31. In which of the following are the
elements listed in order of increasing
electronegativity? (B5.6)
a.
b.
c.
d.
e.
Ba, Zn, C, Cl
N, O, S, Cl
N, P, As, Sb
K, Ba, Si, Ga
Li, K, Na, Ca
32. Which of the following bonds is
expected to be the most polar?
(B5.7)
a.
b.
c.
d.
e.
C—Si
C—N
O—C
S—C
H—C
33. For which of the following may we
draw both polar and nonpolar Lewis
structures? (B5.8)
a.
b.
c.
d.
e.
CHCl3
NH3
BF3
SF2Cl2
PCl5
34. Which of the following has the
fewest pi bonds and is nonpolar?
(B5.9)
a.
b.
c.
d.
e.
HCCH
CO2
CO3-2
N2
SO2
35. The SF5- ion has a square pyramid
structure. The hybridization of the s
orbitals in sulfur is (B5.10)
a.
b.
c.
d.
e.
sp3d
sp
sp3d2
sp3
sp2
36. Which of the following is NOT a
linear structure? (B5.11)
a.
b.
c.
d.
e.
I2
I3CO2
H2S
H—C≡C—H
37. The Lewis structure of the cyanide
ion most closely resembles (B5.12)
a.
b.
c.
d.
e.
N2
O2
CO2
NO
C2H2
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38. In which of the following pairs are
the two items NOT properly related?
(B5.13)
a.
b.
c.
d.
e.
sp3 and 109.5°
trigonal planar and 120°
octahedral and sp3d
sp and 180°
square planar and sp3d2
39. How many resonance structures are
possible for the SO3 molecule?
(B5.14)
a.
b.
c.
d.
e.
None
2
3
4
4/3
40. Which of the following has a
nonbonding pair of electrons on the
central atom? (B5.15)
a.
b.
c.
d.
e.
BCl3
NH3
CCl2Br2
PF5
SO4-2
41. Which of the following is true when
C=C and C≡C bonds are compared?
(B5.16)
a. The triple bond is shorter
than the double bond.
b. The double bond vibrates at a
lower frequency than the
triple bond.
c. The double-bond energy is
lower than the triple-bond
energy.
d. Both are composed of sigma
and pi bonds.
e. All of the above are true.
42. Which angle is NOT expected in any
simple molecule? (B5.19)
a.
b.
c.
d.
e.
60°
90°
109.5°
120°
All of these are reasonable
angles
43. Sulfur forms the following
compounds: SO2, SF6, SCl4, SCl2.
Which form of hybridization is NOT
represented by these molecules?
(B5.20)
a.
b.
c.
d.
e.
sp
sp2
sp3
sp3d
sp3d2
Princeton 2012/Barron 4th ed.
Essays
1. Use the principles of bonding and molecular structure to explain the following
statements. (P4.1)
a. The boiling point of argon is -186°C, whereas the boiling point of neon is -246°C.
b. Solid sodium melts at 98°C, but solid potassium melts as 64°C.
c. More energy is required to break up a CaO(s) crystal into ions than to break up a
KF(s) crystal into ions.
d. Molten KF conducts electricity, but solid KF does not.
2. The carbonate ion CO3-2 is formed when carbon dioxide, CO2, reacts with slightly basic
cold water. (P4.2)
a. (i) Draw the Lewis electron dot structure for the carbonate ion. Include resonance
forms if they apply.
(ii) Draw the Lewis electron dot structure for carbon dioxide.
b. Describe the hybridization of carbon in the carbonate ion.
c. (i) Describe the relative lengths of the three C-O bonds in the carbonate ion.
(ii) Compare the average bond length of the C-O bonds in the carbonate ion to the
average length of the C-O bonds in carbon dioxide.
3. (P4.3)
Substance
H2
N2
O2
Cl2
Boiling Point (°C)
-253°
-196°
-182°
-34°
Bond Length (Ǻ)
0.75
1.10
1.21
1.99
Bond Strength (kcal/mol)
104.2
226.8
118.9
58.0
a. Explain the differences in the properties given in the table above for each of the
following pairs.
i. The bond strengths of N2 and O2
ii. The bond strengths of H2 and Cl2
iii. The boiling points of O2 and Cl2
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b. Use the principles of molecular bonding to explain why H2 and O2 are gases at
room temperature, while H2O is a liquid at room temperature.
4. H2S, SO4-2, XeF2, ICl4- (P4.4)
a. Draw a Lewis electron dot diagram for each of the dot molecules listed above.
b. Use the valence shell electron-pair repulsion (VSEPR) model to predict the
geometry of each of the molecules.
5. Use the principles of bonding and molecular structure to explain the following
statements. (P4.5)
a. The angle between the N-F bonds in NF3 is smaller than the angle between the BF bonds in BF3.
b. Diamond is one of the hardest substances on Earth.
c. HCl has a lower boiling point than either HF or HBr.
6. Answer the following questions regarding the concepts and properties concerning the
structure and geometry of molecular compounds. (B5b-e)
a. Draw the Lewis structure for NCl3. Explain the geometric shape of this molecule
based on the VSEPR theory. Explain what hybrid orbitals are needed to produce
this structure. Why is the molecule NCl5 impossible?
b. Use the results for part (a) to answer this question. Based on this structure, are the
bonds in this molecule polar? If so, indicated the positive and negative poles in
an appropriate manner. Explain your reasoning.
c. Use the results for part (a) to answer this question. Finally, based on this
structure, is this molecule polar? If so, indicate the positive and negative poles in
an appropriate manner. Explain your reasoning.
d. Explain the difference between sigma and pi bonds using the appropriate theories
and examples.
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Answer Key – Unit 4: Bonding
Multiple Choice
1. A
2. C
3. D
4. B
5. C
6. D
7. B
8. A
44.
9. D
10. B
11. B
12. C
13. C
14. C
15. C
16. A
17. E
18. D
19. B
20. A
21. D
22. C
23. D
24. B
25. B
26. A
27. E
28. E
29. E
30. B
31. A
32. C
33. D
34. C
35. C
36. D
37. A
38. C
39. C
40. B
41. E
42. A
43. A
Essays
1. (a) Molecules of noble gases in the liquid phase are held together by London dispersion
forces, which are weak interactions brought about by instantaneous polarities in nonpolar
atoms and molecules. Atoms with more electrons are more easily polarized and
experience stronger London dispersion forces. Argon has more electrons than neon, so it
experiences stronger London dispersion forces and boils at a higher temperature.
(b) Sodium and potassium are held together by metallic bonds and positively charged
ions in a delocalized sea of electrons. Potassium is larger than sodium, so the
electrostatic attractions that hold the atoms together act at a greater distance, reducing the
attractive force and resulting in its lower melting point.
(c) Both CaO(s) and KF(s) are held together by ionic bonds in crystal lattices. CaO is
more highly charged, with Ca+2 bonded to O-2. KF is not as highly charged, with K+
bonded to F-. CaO is held together by stronger forces and is more difficult to break apart.
(d) KF is composed of K+ and F- ions. In the liquid (molten) state, these ions are free to
move and can thus conduct electricity. In the solid state, the K+ and F- ions are fixed in a
crystal lattice and their electrons are localized around them, so there is no charge that is
free to move and thus no conduction of electricity.
2. (a) (i)
(ii)
(b) The central carbon atom forms three sigma bonds with oxygen atoms and has no free
electron pairs, so its hybridization must be sp2.
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(c) (i) All three bonds will be the same length because no particular resonance form is
preferred over the others. The actual structure is an average of the resonance structures.
(ii) The C-O bonds in the carbonate ion will be shorter than single bonds and longer than
double bonds, while the C-O bonds in carbon dioxide are both double bonds. The bonds
in the carbonate ion will be shorter than single bonds and longer than double bonds, so
the carbonate bonds will be longer than the carbon dioxide bonds.
3. (a) (i) The bond strength of N2 is larger than the bond strength of O2 because N2
molecules have triple bonds and O2 molecules have double bonds. Triple bonds are
stronger and shorter than double bonds.
(ii) The bond length of H2 is smaller than the bond length of Cl2 because hydrogen is a
smaller atom than chlorine, allowing the hydrogen nuclei to be closer together.
(iii) Liquid oxygen and liquid chlorine are both nonpolar substances that experience only
London dispersion forces of attraction. These forces are greater for Cl2 because it has
more electrons (which makes it more polarizable), so Cl2 has a higher boiling point than
O2.
(b) H2 and O2 are both nonpolar molecules that experience only London dispersion
forces, which are too weak to form the bonds required for a substance to be liquid at
room temperature. H2O is a polar substance whose molecules form hydrogen bonds with
each other. Hydrogen bonds are strong enough to form the bonds required in a liquid at
room temperature.
4. (a)
(b) H2S has two bonds and two free electron pairs on the central S atom. The greatest
distance between the electron pairs is achieved by tetrahedral arrangement. The electron
pairs at two of the four corners will cause the molecule to have a bent shape, like water.
SO4-2 has four bonds around the central atom and no free electron pairs. The four bonded
pairs will be farthest apart when they are arranged in a tetrahedral shape, so the molecule
is tetrahedral.
XeF2 has two bonds and three free electron pairs on the central Xe atom. The greatest
distance between the electron pairs can be achieved by a trigonal bipyramidal
arrangement. The three free electron pairs will occupy the equatorial positions, which are
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120 degrees apart, to minimize repulsion. The two F atoms are at the poles, so the
molecule is linear.
ICl4- has four bonds and two free electron pairs on the central I atom. The greatest
distance between the electron pairs can be achieved by an octahedral arrangement. The
two free electron pairs will be opposite each other to minimize repulsion. The four Cl
atoms are in the equatorial positions, so the molecule is square planar.
5. (a) BF3 has three bonds on the central B atom and no free electron pairs, so the structure
of BF3 is trigonal planar, with each of the bonds 120 degrees apart. NF3 has three bonds
and one free electron pair on the central N atom. The four electron pairs are pointed
toward the corners of a tetrahedron, 109.5 degrees apart. The added repulsion from the
free electron pair causes the N-F bonds to be even closer to together, and the angle
between them is more like 107.
(b) The carbon atoms in diamond are bonded together in a tetrahedral network, with each
carbon atom bonded to three other carbon atoms. The tetrahedral structure of the
network bonds does not leave any seams along which the diamond can be broken, so a
diamond behaves as one big molecule with no weaknesses.
(c) HBr and HCl are polar molecules. In liquid form, both substances are held together
by dipole-dipole interactions. These interactions are stronger for molecules with more
electrons, so HBr has stronger intermolecular bonds and a higher boiling point. HF has a
higher boiling point than HCl because HF undergoes hydrogen bonding, while HCl does
not; this causes HF to remain a liquid at higher temperatures than HCl, although HF is a
polar molecule with fewer electrons than HCl.
6. (a) This molecule is an AX3E shape, where three atoms and a nonbonding electron pair
are on the central atom, N. The VSEPR theory states that these four will repel each other
and form an arrangement where the atoms and electron pairs are as far apart as possible.
This results in a tetrahedral shape. If the electrons are not “seen,” then the shape of the
atoms alone is a triangular pyramid. To obtain the tetrahedral shape, a set of four sp3
hybrid orbitals must form. This results when the 2s (one orbital) and the 2p (three
orbitals) energy levels combine to form one energy level with four equivalent bonds.
The NCl5 would require the formation of an sp3d2 hybrid. However, there are no
available d-orbitals to use to make these hybrids, and therefore NCl5 is impossible. Only
elements in the 3rd period and below are able to have more than 8 valence electrons
surrounding the central atom.
(b) The N—Cl bonds in this molecule are polar because a significant difference in
electronegativity is expected between Cl and N. This polarity of the bond can be
indicated by the following
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(c) The structure and direction of the polar bonds in a molecule are arranged to see if the
bond polarity cancels. Generally, bond polarity will cancel and a molecule will be
nonpolar if the molecule is perfectly symmetrical. In this case, the molecule is not
symmetrical, and the polarities of the individual N—Cl bonds do not cancel. The
molecule has only two poles, on negative, in space between the three Cl atoms. The
positive end resides near the nitrogen atom. Again, only partial charges are formed, and
this can be shown by
(d) Sigma bonds form when there is a high electron density between two nuclei. The
line between the two nuclei is called the internuclear axis. The greater the density of
electrons along the internuclear axis, the stronger the bond. Because the reguin in space
is taken up with sigma bonding electrons, the pi bond forms with increased electron
density on either side of the internuclear axis. Each pi bond has two electron clouds, one
on either side of the internuclear axis.
The nature of sigma and pi bonds can be explained on the basis of valence bond theory.
Valence bond theory considers bonds forming from the overlap of orbitals. Direct
overlap of two s-orbitals, and s- and a p-orbital, or two p-orbitals along the internuclear
axis results in a sigma bond. Sidewise overlap of two p-orbitals results in a pi bond.