CEM 141 Mock Exam 3 Questions – SS25 - KEY

This mock exam is intended to give you an idea of some of the different types of questions that
could be asked on an exam. It does not cover all of the material that may appear on your exam.

Exam 3 will cover all of the following learning objectives: 2.26 – 2.31, 3.5–3.13, 4.1–4.11
and 4.16 – 4.17.

Multiple Choice Questions

1. At room temperature, substance A is a solid and substance B is a gas. A and B react
according to the following equation: A + 3B  C
At room temperature, what state of matter do you predict for substance C?

a. Solid, because C is heavier than A or B.

b. Gas, because C is three parts B (which is a gas) and one part A (which is a solid).
c. Liquid, because C will be intermediate between a solid and a gas.
d. We cannot predict the state of matter because we don’t know about the
bonding/interactions present within C.

2. To the right is an incomplete molecular orbital diagram for a

diatomic molecule. The electrons are shown in the atomic
orbitals, but not within the molecular orbitals. Add the
electrons to the molecular orbitals. How many bonds exist
between the atoms in this diatomic molecule?

a. 0 bonds
b. 1 bond
c. 2 bonds
d. 3 bonds
e. 4 bonds
f. 5 bonds

3. What is the best explanation for why metals are ductile?

a. Valence electrons carry charge anywhere within molecular orbitals that span the metal.
b. Photons are absorbed and re-emitted.
c. Metallic bonds involve delocalized electrons whereas covalent bonds involve localized
electrons.
d. When the nuclei move relative to one another, the valence electrons act as a glue holding
them together.

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4. Carbon can exist in many forms known as allotropes. Two such allotropes are diamond and
graphite. Unlike diamond, graphite conducts electricity. This is because
a. In graphite, there are hybrid orbitals used for bonding that contain electrons.
b. In diamond, all of the electrons are in sigma bonds which allow conductivity of
electricity.
c. In graphite, there are unhybridized p-orbitals that extend over the entire sheet forming a
delocalized pi (π) bonding network where electrons are free to move.
d. In diamond, the bonding electrons have enough energy to move and can therefore
conduct electricity.

5. Boron nitride is the second hardest substance known (next to diamond). It has a very high
melting point but does not conduct electricity. What types of bonding do you think is present
in boron nitride?

a. There is covalent bonding within two-dimensional layers of alternating boron and

nitrogen molecules, with London dispersion forces between the layers.
b. There is covalent bonding within boron nitride molecules (BN), and London dispersion
forces between the molecules.
c. There is a three dimensional network of metallic bonds between boron and nitrogen
atoms.
d. There are covalent bonds between boron and nitrogen atoms which exist in a 3D network.
e. It is not possible to predict the type of bonding from a consideration of bulk properties.

6. Compare the two molecular representations shown

to the right.
These structures represent
I. the same molecule
II. isomers
III. different molecules that are not isomers
Because
IV. free rotation around the bond between the two carbon atoms makes the structures the
same.
V. rotation is not possible around a double bond without breaking it.
VI. these structures have different numbers of carbon and hydrogen atoms.

a. I and IV c. II and IV e. III and VI

b. I and V d. II and V

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7. Which do you predict has the higher boiling point and why?
I. Cl2
II. Ar
Because
III. When this substance boils, covalent bonds are overcome.
IV. When this substance boils, LDFs are overcome.
V. Covalent bonds are stronger than LDFs.
VI. The strength of the LDFs depends on the size of the electron cloud.

a. I, III, V c. I, IV, VI e. II, IV, VI

b. I, III, IV, V d. II, III, V f. II, III, IV, V

8. Draw the Lewis structure for sulfur dichloride, SCl2. How many lone pair of electrons are on
the sulfur atom?

a. 0 lone pair c. 2 lone pair e. 4 lone pair

b. 1 lone pair d. 3 lone pair

9. Draw the Lewis structure for a nitrogen molecule, N2. When nitrogen melts, what
interaction(s) are overcome?
a. London dispersion forces
b. London dispersion forces and covalent bonds
c. London dispersion forces and hydrogen bonds
d. London dispersion forces, hydrogen bonds, and covalent bonds
e. Covalent bonds

10. A Lewis structure for nitrous oxide (laughing gas) is shown to the
right. What is the formal charge on the nitrogen atom in the middle
of the molecule?

a. 0 c. +2 e. –2
b. +1 d. –1

11. For the structure to the right, what is the electron center
geometry for the oxygen atom that the arrow is pointing to?

a. linear d. trigonal planar

b. bent e. trigonal pyramid
c. tetrahedral

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12. Use the Lewis structure in question 11 to answer this question. What is the shape around the
oxygen atom that the arrow is pointing to?

a. linear d. trigonal planar

b. bent e. trigonal pyramid
c. tetrahedral

13. Two Lewis structures are shown below. Are these molecules the same or are they isomers?

a. Isomers, because the molecules have the same formula but different atom connectivity.
b. Isomers, because they have different numbers of each kind of atom.
c. The same, because they have the same formula and the same atom connectivity.
d. The same, because they both contain only carbon and hydrogen atoms.

14. Which bonds allow free rotation around the bonded atoms and why?
I. Sigma bonds
II. Pi bonds
Because
III. The orbitals overlap side-to-side.
IV. The orbitals overlap end-to-end.
V. The overlap doesn’t change upon rotation.
VI. The overlap breaks upon rotation.

a. I, III, V c. I, IV, V e. II, III, V g. II, IV, V

b. I, III, VI d. I, IV, VI f. II, III, VI h. II, IV, VI

15. The Lewis structure for ethene (C2H4) is shown below. What is the hybridization and the
shape around the carbon labelled with the arrow?
a. sp3 and tetrahedral
b. sp2 and tetrahedral
c. sp3 and linear
d. sp2 and trigonal planar
e. sp2 and trigonal pyramid
f. sp3 and trigonal pyramid

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16. Dipole-dipole interactions and London dispersion forces are two types of intermolecular
forces. For molecules of about the same size
a. London dispersion forces are the stronger force due to the temporary, fluctuating dipoles.
b. Dipole-dipole interactions are the stronger force due to the presence of permanent
dipoles.
c. For molecules of the same size, London dispersion forces and dipole-dipole interactions
are about the same strength.
d. It is impossible to tell which interaction is stronger from the information given.

17. Draw the Lewis structure for difluoroborane (HBF2). Is this a polar molecule?
I. Yes
II. No
III. Sometimes
Because
IV. The bond dipoles add together to create a molecular dipole.
V. The bond dipoles cancel each other out when they are added together.
VI. The bond dipoles may cancel each other out or create a molecular dipole
depending on how the Lewis structure is drawn.

a. I and IV c. II and IV e. III and VI

b. I and V d. II and V

18. Which element is more electronegative?

I. N
II. P
Because
III. the bonding electrons are closer to the nucleus.
IV. the bonding electrons experience a higher effective nuclear charge.
V. phosphorus has more electrons than nitrogen.

a. I and III c. II and IV e. II and V

b. I and IV d. II and III

19. What are the strongest intermolecular forces present in dimethyl ether (CH3OCH3) in the
liquid state?

a. London dispersion forces c. Hydrogen bonding interactions

b. Dipole-dipole interactions d.
Covalent bonds

20. Three molecules of methanol (CH3OH) are shown

below. Which of the labeled interactions are
hydrogen bonding interactions?
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 a. I
b. II
c. II and III
d. IV
e. I, II, and III
f. III

21. Use the data table for the first eight ionization energies of a particular element shown below
to predict and explain which element you would expect to show this trend in successive
ionization energies.
I. Fluorine
Ionization Ionization Energy
II. Oxygen
Number (kJ/mol)
Because 1st 1,314
III. core electrons are more strongly attracted to the 2nd 3,388
nucleus than valence electrons. 3rd 5,300
IV. valence electrons release more energy when 4th 7,469
they are removed from the atom. 5th 10,990
6th 13,327
a. I and III c. II and III 7th 71,330
b. I and IV d. II and IV 8th 84,078

Short Answer Questions

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1. a. Draw a molecular level picture of the bonding in diamond. Show at least 8 atoms. What
do the components of your drawing represent?

The black circles represent carbon atoms and the lines

represent covalent bonds. Each atom uses sp3 hybrid
orbitals to form covalent bonds to four other carbon
atoms. The bond angles are about 109 degrees.

b. How does the model of bonding in diamond you have drawn explain each of these
properties of diamond?

i) does not conduct electricity

The model of bonding in diamond uses the idea that electrons are localized between the
two nuclei that participate in the bond. The electrons cannot move out of this area and
therefore are not mobile (as they are in metals). To conduct electricity, charged particles
(electrons) must be able to move freely in molecular orbitals that extend throughout the
whole structure.

ii) has a very high melting point

To melt diamond and allow the particles to move relative to one another, the strong C-C
covalent bonds must be broken, since the structure is a 3D network of these bonds.
Substances that exist as molecules tend to have lower melting points because it is the
intermolecular forces between the molecules that are overcome rather than the actual
covalent bonds.
c. How is the model of bonding you described for diamond different from that in metals
(draw molecular level picture of the bonding in metals to help explain differences)?

d.
e.
Positively charged core of neutral atom
(nucleus and core electrons)
f.
Electron “sea” (mobile electrons in
delocalized molecular orbitals)

2. Use this table of data to support your argument about the nature of bonds and interactions
that are disrupted as each substance indicated below melts.

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 Elemental H2 He Li(s) Be(s) B(s) C(s) N2 O2 F2 Ne
form

Melting point 13.81 K 0.95 K 453.7 K 1560 K 2348 K 3823 K 63.15 K 54.36 K 53.53 K 24.56
K

Boiling point 20.28 K 4.22 K 1615 K 2744 K 4273 K 4098 K 77.36 K 90.20 K 85.03 K 27.07
K

Name hydrogen helium lithium beryllium boron carbon nitrogen oxygen fluorine neon

Your answer should include a claim about the relevant bonding and/or interactions that are
disrupted during melting for the following elements supported by evidence and the reasoning
that connects the claim and evidence.

a. Nitrogen (N2):
The boiling point and melting point of N2 are low. This means that when nitrogen
changes state the interactions between the particles that are overcome must be quite
weak. Nitrogen exists as diatomic molecules held together by covalent bonds. Clearly the
bonds are not breaking when nitrogen melts or boils (otherwise the temperature would be
much higher – ie more energy would be needed). The interactions that are being
overcome are London dispersion forces that exist between separate molecules making the
molecules stick together.

b. Carbon (diamond):
Diamond has a high melting point meaning that very strong interactions must be broken
when diamond melts. Diamond is made up of carbon atoms each covalently bonded to 4
other carbon atoms. To melt diamond, these covalent bonds must be broken which
requires a lot of energy (a high temperature).

c. Why is the melting point of diamond so much higher than nitrogen?

Melting diamond requires breaking strong covalent bonds and melting nitrogen requires
overcoming much weaker London dispersion forces between the nitrogen (N2) molecules.
The stronger the interaction, the more energy (and higher temperature) is needed to
overcome it.
3. For the molecules HCN and CH3NH2

a. Draw the Lewis structure of each.

b. Give the hybridization for each atom bonded to more than one other atom.

C (in HCN) is sp
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 C (in CH3NH2) is sp3
N (in CH3NH2) is sp3

c. What is the electron pair geometry for each atom bonded to more than one other atom?

C (in HCN) is linear

C (in CH3NH2) is tetrahedral

N (in CH3NH2) is tetrahedral

d. What is the molecular shape and bond angle for each atom bonded to more than one other
atom?

C (in HCN) is linear, 180 degrees

C (in CH3NH2) is tetrahedral, 109 degrees

N (in CH3NH2) is trigonal pyramidal, about 109 degrees

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4. a. Draw a Lewis structure for C3H8 and another for CH3OCH3.

b. What intermolecular forces exist between molecules of each compound?

C3H8: LDFs

CH3OCH3: LDFs and dipole-dipole interactions

c. Which one will have the higher boiling point? Why?

CH3OCH3 has the higher boiling point because stronger intermolecular forces (dipole-
dipole interactions) will take more energy to overcome. Boiling requires energy to
separate one molecule from another molecule (overcome intermolecular forces).

d. Draw three molecules of CH3OCH3 showing how the strongest type of intermolecular
forces act between the molecules.

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5a. The first ionization energy of an element in the third row (elements Na to Ar) of the periodic
table is 578 kJ/mol, its 2nd IE is 1820 kJ/mol, the 3rd IE is 2750 kJ/mol, the 4th IE is 11600
kJ/mol, and the 5th IE is 14842 kJ/mol. Which element do you think this is?

Aluminum

b. Explain your reasoning. Be sure to state the scientific principles you are using in your
explanation and why the data supports your argument.

The “big jump” in ionization energies lies between the 3rd and 4th ionization energies. This
means that three electrons are relatively easy to remove and the 4th must come out of the
core. Core electrons are more difficult to remove because they are closer to the nucleus.
Therefore, the attractive electrostatic forces holding these electrons in the atom are stronger
and they take more energy to remove.

c. How does this data support the idea that electron energies are quantized?

If electrons could exist with any energy level (that is if their energies were not quantized), any
amount of energy might remove an electron. The fact that ionization energies are constant
and replicable for any given electron means that this amount of energy must be added to the
atom to eject the electron. (This is a similar argument to the idea that emission and
absorption spectra are evidence for quantized energy levels – except here the electrons are
ejected from the atom rather than moving between energy levels.)

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Possibly useful information:

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