Name
CHEM 115 EXAM #3
Practice Exam Fall 2013
Circle the correct answer. (numbers 1-9, 3 points each)
1.
Choose the correct statement about the compound SO2.
a.
b.
c.
d.
e.
2.
An expanded octet may occur:
a.
b.
c.
d.
e.
3.
the S—O bonds are ionic in character
the molecule cannot exist since the bonding does not obey the octet rule (typo corrected)
the S atom has a lone pair of electrons
the S—O bonds are different lengths since one is a single and the other a double bond
none of the above
in the first and second period only
in groups 1A, 2A, and 3A
in periods 3 and greater
in all groups except 1A
none of the above
Which chloride should exhibit the most covalent (least ionic) type of bond?
a. NaCl
4.
c. F—F
d. H—I
e. N—Br
b. NO3-
c. CH4
d. PCl5
e. none of the above
b. C≡C
c. O=O
d. C≡ O
_
e. C C
b. H2S, HCN, CO2
c. H2O, CO, H2S
d. H2S, CO, CO2
Which of the following is a tetrahedral molecule (or ion)?
a. CO2
9.
b. N—F
Which of the following sets contains the greatest number of bent molecules?
a. CO2, HCN, O2
8.
e. CaCl2
Which of the following bonds should be the shortest and strongest?
a. C=C
7.
d. BeCl2
Which of the following species has a central atom with sp2 hybridization?
a. CO2
6.
c. BaCl2
Which of the following bonds should be the most polar?
a. N—Cl
5.
b. KCl
b. NO3-
c. CH4
d. PCl5
e. none of the above
Which of the following is a molecule (or ion) that contains at least one π bond?
a. CO2
b. NO3-
c. CH4
d. PCl5
e. none of the above
[note: nitrate also has π bonding, but it is delocalized . . . it has less than a full π bond on any one bond axis]
2
Complete the following as directed (point values as indicated).
(16 points)
10. For the following two species complete the tasks specified below in a - d.
ii. C2H2
i. CH3COOH
a. Draw the best Lewis
structures.
b. Specify the hybridization C in CH3
sp3
of both
the C atoms and the O atom C in CO
sp2
specified in (i)
and the C atoms in (ii)?
O between CO and H
C in CH
sp
sp3
c. Show the energy level
diagram for both the C
atoms and the O atom in (i)
and the C atoms in (ii), fill
the diagrams with the
correct number of valence
electrons, and identify the
types of electrons present
(i.e. σ, π, lone pair)
C in CH3
sp3 (↑↓)(↑↓)(↑↓)(↑↓)
4 σ bonds
bold ↑ are electrons from that C
C in CO
p (↑↓)
sp2 (↑↓)(↑↓)(↑↓)
σ σ σ
bold ↑ are electrons from that C
O between CO and H
sp3 (↑↓)(↑↓)(↑↓)(↑↓)
2 lone pairs & 2 σ bonds
bold ↑ or ↓ are O electrons
C in CH
d. What are the bond angles
around both the C atoms
and the O atom in (i) and
each of the C atoms in (ii)?
C in CH3 about 109.5°
C in CH
180°
π π
p (↑↓) (↑↓)
sp (↑↓)(↑↓)
σ σ
bold ↑ are electrons from
that C
C in CO about 120°
O [the COH angle] a bit < 109.5°
(10 points)
11. (a) Draw Lewis structures for CN-, CO, N2, and NO+. Label all atoms with their formal
charges.
ALL OF THESE HAVE TEN (10) VALENCE ELECTRONS
[:C≡N:]-1
0
:C≡O:
-1
+1
:N≡N:
0
0
[:N≡O:]+
0 +1
(b) Select the species from part (a) that you would expect to have the shortest, strongest
bond. In order to receive credit you must EXPLAIN the reasoning for your selection.
The triple bond between the C and O should be the shortest and strongest. In addition to being a
triple bond, the -1 and +1 formal charges indicate that a strong partial ionic character will exist in
this bond (adding to the strength of the bond).
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(15 points)
12. Fill in the MO energy diagrams for the species listed below using valence electrons ONLY. The
answer the questions that appear under the MO energy diagram of each species.
Ο2
Oxygen
12 e-
Peroxide
Ο 2 −2
σ∗2p
14 e-
σ∗2p
↑
π∗2p
↑
π∗2p
↑↓ π∗2p
↑↓ π∗2p
↑↓
π2p
↑↓
π2p
↑↓
↑↓
↑↓
σ2p
π2p
↑↓
π2p
σ2p
↑↓
σ∗2s
↑↓
σ∗2s
↑↓
σ2s
↑↓
σ2s
Show the calculation of the bond order for
oxygen.
BO = ½ (bonding – antibonding)
BO = ½ (8 - 4) = 2
Show the calculation of the bond order for the
peroxide ion.
BO = ½ (bonding – antibonding)
BO = ½ (8 - 6) = 1
Is oxygen diamagnetic or paramagnetic?
Explain your answer.
Is the peroxide ion diamagnetic or paramagnetic?
Explain your answer.
Molecular oxygen, O2, is paramagnetic
because it possesses unpaired electrons.
The peroxide ion, O22-, is diamagnetic because
it possesses all paired electrons.
Draw the Lewis structure of oxygen.
Does the bond order for oxygen from the
Lewis structure match the bond order from
the MO diagram?
Yes it does!
Draw the Lewis structure for the peroxide ion
Does the bond order for the peroxide ion from
the Lewis structure match the bond order from
the MO diagram?
Yes it does!
Which should have the longer, weaker bond; oxygen or the peroxide ion? Explain your choice!
The peroxide ion is expected to have the longer weaker bond. This is due to two factors.
First, the peroxide ion has a bond order of ONE vs. the bond order of TWO found in
molecular oxygen. Also, the peroxide ion has a formal charge of minus one (-1) on each
oxygen which will result in a repulsive force – lengthening the bond.
4
Answer the following. Provide as much information as required to answer the question clearly.
13.
Balance the following reactions AND draw Lewis symbol structures for part (e):
(a)
2 C6H14 (l) +
19 O2 (g)

12 CO2 (g) +
14 H2O (l)
(b)
5 O2 (g) +
4 NH3 (g)

4 NO (g) +
6 H2O (l)
(c)
3 NO2 (g) +
H2O (l)

2 HNO3 (aq) +
NO (g)
(d)
Fe2O3 (s) +
2 Al (s)

2 Fe (l) +
Al2O3 (s)
Don’t forget to draw the Lewis symbols for each of the compounds in this next reaction!
(e)
CaC2 (s) +
2 H2O (l)
→
C2H2 (g) +
Ca(OH)2 (s)
Ca2+
14.
Write balanced chemical equations for each of the following “word descriptions” of a
chemical process.
(a) Solid magnesium metal reacts with carbon dioxide to produce magnesium oxide and
carbon monoxide.
Mg (s)
+
CO2 (g) 
MgO (s)
+
CO (g)
(b) Solid magnesium oxide reacts with water to form aqueous magnesium hydroxide.
MgO(s)
+
H2O (l) 
Mg(OH)2 (aq)
(c) Liquid bromine reacts (quite violently) with aluminum metal to form aluminum
bromide.
2 Al(s)
+
3 Br2 (l) 
2 AlBr3 (s)
(d) Butane gas combusts completely with oxygen gas to form the expected products.
2 C4H10 (g)
+
13 O2 (g)

8 CO2 (g)
+
10 H2O( g)
(e) Aqueous hydrochloric acid reacts with solid zinc to form zinc chloride and hydrogen.
Zn (s)
+
2 HCl (aq)

ZnCl2 (aq)
+
H2 (g)
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BONUS (for up to 10 points)
Place the correct atomic symbols into the correct blocks for elements 1 through 30.
Label the following correctly on the table as well:
(a) noble gases, (b) alkali metals, (c) halogens, (d) alkaline earth metals, (e) transition metals, (f)
lanthanides, (g) actinides, (h) the coinage metals, and (i) the chalcogens. (fixed another typo)
coinage metals are Cu, Ag, Au (group 11)
chalcogens (“ore formers”) the oxygen family (group 16) principally O, S, Se
See the next page for a chart of shape and name info you should also study. In addition to
knowing the names, bond angles, etc. you should also know the hybridization type for each
category. Below you will find a quick listing.
#RHED
2
3
4
5
6
hybridization
sp
sp2
sp3
sp3d
sp3d2
RHED = region of high electron density
Note: other options exist beyond those in this table . . . but we’ll not be testing all the way up to
a 16 coordinate system that employs all of the s, p, d and f orbitals from a single atom!
6