Answer the following questions. You have 20 minutes, and you may not use a calculator. You may use the periodic
table at the back of the book.
1.
2.
VSEPR predicts an SbF5 molecule will be which of the following shapes?
a.
tetrahedral
b.
trigonal bipyramidal
c.
square pyramid
d.
trigonal planar
e.
square planar
The shortest bond would be present in which of he following substances?
a.
b.
I2
CO
3.
c.
CCl4
d. O22–
e.
SCl2
Which of the following does not have one or more π bonds?
4.
a.
H2O
b. HNO3
c.
O2
d. N2
e.
NO2–
Which of the following is polar?
5.
a.
SF4
b. XeF4
c.
CF4
d. SbF5
e.
BF3
Resonance structures are needed to describe the bonding in which of the following?
a.
H2O
b. ClF3
c.
HNO3
d. CH4
e.
NH3
For questions 6 and 7, pick the best choice from the following:
f.
ionic bonds
g.
hybrid orbitals
h.
resonance structures
i.
hydrogen bonding
j.
van der Waals attractions
6.
An explanation of the equivalent bond lengths of the nitrite ion is:
7.
Most organic substances have low melting points. This may be because, in most cases, the intermolecular forces are:
8.
Which of the following has more than one unshared pair of valence electrons on the central atom?
a.
b.
c.
d.
e.
BrF5
NF3
IF7
ClF3
CF4
9.
What is the expected hybridization of the central atom in a molecule of TiCl4? This molecule is tetrahedral.
a.
sp3d2
b. sp3d
c.
sp
d. sp2
e.
sp3
10. The species in the following set do not include which of the following geometries?
SiCl4, BrF4–, C2H2, TeF6, NO3–
a.
square planar
b.
tetrahedral
c.
octahedral
d.
trigonal pyramidal
e.
linear
11. The only substance listed below that contains ionic, σ, and π bonds is:
a.
b.
c.
d.
e.
Na2CO3
HClO2
H2O
CO2
NaCl
For problems, 12–14 choose a molecule from the following list:
f.
C2
g. F2
h. B2
i.
O2
j.
Ne2
12. The paramagnetic molecule with a bond order of two.
13. The diamagnetic molecule with no antibonding electrons.
14. The paramagnetic molecule with antibonding electrons.
15. The electron pairs point toward the corners of which geometrical shape for a molecule with sp2 hybrid orbitals?
a.
trigonal planar
b.
octahedron
c.
trigonal bipyramid
d.
trigonal pyramid
e.
tetrahedron
16. Regular tetrahedral molecules or ions include which of the following?
a.
b.
c.
d.
CH4
SF4
NH4+
I, II, and III
e.
I and III only
f.
I only
g.
I and II only
h.
II only
17. Which molecule or ion in the following list has the greatest number of unshared electrons around the central atom?
a.
CF4
b. ClF3
c.
BF3
d. NH4+
e.
IF5
18. Which of the following molecules is the least polar?
a.
b.
c.
d.
e.
PH3
CH4
H2O
NO2
HCl
19. What types of hybridization of carbon are in the compound 1,4-butadiene, CH2CHCHCH2?
a.
sp3
b. sp2
c.
sp
d.
I and II
e.
I, II, and III
f.
I and III
g.
I only
h.
II only
20. Which of the following molecules is the most polar?
a.
C2H2
b. N2
c.
CH3I
d. BF3
e.
NH3
21. Which of the following processes involves breaking an ionic bond?
a.
b.
c.
H2(g) + Cl2(g) → 2 HCl(g)
I2(g) → 2 I(g)
Na(s) → Na(g)
d.
e.
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
2 KBr(s) → 2 K(g) + Br2(g)
Answers and Explanations
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
B— The Lewis (electron-dot) structure has five bonding pairs around the central Sb and no lone pairs. VSEPR predicts this number of
pairs to give a trigonal bipyramidal structure.
B—All the bonds except in CO are single bonds. The CO bond is a triple bond. Triple bonds are shorter than double bonds, which are
shorter than single bonds. Drawing Lewis structures might help you answer this question.
A—Answers B–E contain molecules or ions with double or triple bonds. Double and triple bonds contain π bonds. Water has only single
(σ) bonds. If any are not obvious, draw a Lewis structure.
A—The VSEPR model predicts all the other molecules to be nonpolar.
C—All the other answers involve species containing only single bonds. Substances without double or triple bonds seldom need
resonance structures.
C—Resonance causes bonds to have the same average length.
E—Many organic molecules are nonpolar. Nonpolar substances are held together by weak London dispersion forces.
D—Lewis structures are required. You do not need to draw all of them. A and B have one unshared pair, while C and E have no
unshared pairs. D has two unshared pairs of electrons.
E—Tetrahedral molecules are normally sp3 hybridized.
D—SiCl4, is tetrahedral. BrF4– is square planar. C2H2 is linear. TeF6 is octahedral. NO3–is trigonal planar. If you are uncertain about any
of these, Lewis structures and VSEPR are needed.
A—Only A and E are ionic. The chloride ion has no internal bonds, so σ and π bonds are not possible. 12–14––Sketch a molecular
orbital energy-level diagram. Use the same diagram to save time, unless it becomes too messy.
D
A
D
A—This hybridization requires a geometrical shape with three corners.
B—One or more Lewis structures may help you. I and III are tetrahedral, and II is an irregular tetrahedron (see-saw).
B—A has 0. B has 2. C and D have 0. E has 1. You may need to draw one or more Lewis structures.
B—All the molecules are polar except B.
E—The structure is:
All the carbon atoms have one double and two single bonds. This combination is sp 2.
20. E—Drawing one or more Lewis structures may help you. Only C and E are polar. Only the ammonia has hydrogen bonding, which is
very, very polar.
21. E—C is metallic bonding. All the others involve covalently bonded molecules.