KEY
Name: ________________________________________________
Student #: ________________________________________________
Chemistry 111 Sect. 3
Exam #3
November 30, 2000
Please put your name and student number on all pages in case they become separated. The
last page of the exam has some useful information. The exam lasts 60 minutes. There are a
total of 100 points.
MULTIPLE CHOICE QUESTIONS
Place an X in the box corresponding to the correct answer.
1. (5 pts) Arrange the following neutral elements in order of increasing atomic radius.
Cl, Cs, Si, Sn, Sr
smallest → largest
o
Cs < Sr < Sn < Si < Cl
X
o
Cl < Si < Sn < Sr < Cs
o
Cl < Cs < Si < Sn < Sr
o
Sr < Sn < Si < Cs < Cl
o
Si < Cl < Sr < Sn < Cs
2. (7 pts) According to the octet rule, the carbon-oxygen bond in the molecule COBr2, in which
carbon is bonded to O and Br atoms, is a:
1. # of valence electrons = 4 + 6 + 14 = 24
o
single bond
2. C is central atom (lower electron affinity than O & Br)
:O:
o
double bond
X
3. Form sigma bonds, then place remaining electrons on
outer atoms
o
triple bond
C
4.
Move
one
O
LP
to
form
a
π
bond
with
C,
leaving
all
o
quadruple bond
:Br: :Br:
..
atoms with zero formal charge. This is not the case if we ..
o
resonance bond try to make a double bond with one of the Br atoms, so
there is no resonance in this molecule.
3. (8 pts) In principle, dinitrogen monoxide, N2O or N≡N–O, can decompose to nitrogen and
oxygen gas:
2 N2O (g) → 2 N2 (g) + O2 (g)
Estimate the enthalpy change for this reaction in kJ from the bond energies on p. 6.
o
X
o
o
o
o
-297
-96
There are 2 mol of N≡N bonds in 2 N2O, 2 mol of N-O bonds in 2 N2O, 2
mol of N≡N bonds in 2 N2 and 1 mol of O=O bonds in O2. The net
enthalpy changes is sum of reactant bonds minus sum of product bonds, or
96
716
4680
∆Hrxn = {(2 mol × 945 kJ/mol) + (2 mol × 201 kJ/mol)} {(2 mol × 945 kJ/mol) + (1 mol × 498 kJ/mol)}
= 2292 kJ - 2388 kJ = -96 kJ
Page 1
KEY
Chemistry 111 Sect. 3
Name: ________________________________________________
Exam #3
Student #: ________________________________________________
November 30, 2000
4. (5 pts) The bond angle a shown below for the molecule acetonitrile is, in degrees:
o
o
oX
o
o
45
The bonding around the first carbon is
tetrahedral, so the bond angle is the
tetrahedral angle of 109.5 °.
90
109.5
a
H
H C
120
C
N:
H
180
5. (5 pts) The element potassium (K) has a ________ first ionization energy and a ________
electronegativity than the element fluorine (F).
o
o
o
oX
o
higher, higher
higher, lower
lower, higher
lower, lower
equal, equal
6. (7 pts) The electron pair geometry around the central atom in H2O is:
o
o
oX
o
o
There are 2 lone pairs and two sigma
bonds (i.e 4 “lumps” of electrons),
meaning that the electron geometry is
tetrahedral.
linear
trigonal planar
tetrahedral
trigonal bipyramidal
octahedral
7. (8 pts) The bond order for a carbon – oxygen bond in carbonate ion (Lewis structure
shown below) is:
o
oX
o
o
o
1
3
There are 3 carbon-oxygen links and
four total bonds between C and O, so
the bond order is an average for this
resonance structure:
4
B.O. = (4 bonds)/(3 links) = 1.33
1.33
2
.. C
:O
..
MULTIPLE MULTIPLE CHOICE QUESTIONS
Page 2
2-
:O:
..
O:
..
KEY
Chemistry 111 Sect. 3
Name: ________________________________________________
Exam #3
Student #: ________________________________________________
November 30, 2000
Place an X in the boxes corresponding to the correct answers. Any number of answers may
be correct, including none of them.
8. (10 pts) For which of the labeled atoms in sulfuric acid, H2SO4 shown below, is the formal charge
equal to zero?
oX
H 1 FC = 1 - ½(2) - 0 = 0
oX
O2
FC = 6 - ½(4) - 4 = 0
o
O3
FC = 6 - ½(2) - 6 = -1
o
O4
FC = 6 - ½(2) - 6 = -1
o
S5
FC = 6 - ½(8) - 0 = +2
4
..
:O: 5
..
..
S
H O
O
..
..
:O:
1 2
..
H
3
9. (10 pts) Which of the following are true?
oX
o
Resonance Lewis structures always include a double or triple bond. Switching between single
bonds doesn’t change
The electronic geometry of a molecule is always the same as its molecular
geometry.
anything.
Lone pairs!
A fluoride ion has a smaller radius than a fluorine atom.
oX
An N≡C bond is shorter than an N–C bond.
o
o
More electrons → larger.
Multiple bonds are always shorter than single
bonds
for the Depends
same paironofbond
atoms.
The polarity of a molecule depends only on its
geometry.
polarity, too.
10. (5 pts) Which of the following molecules is NOT polar?
oX
CCl4 Polar bonds, but symmetric molecule.
o
H2O Polar bonds, and bent, asymmetric molecule.
oX
Cl2
o
HCl Only one polar bond, therefore an asymmetric molecule.
o
NH3 Polar bonds, and trigonal pyramidal, asymmetric molecule.
Nonpolar bonds (same atoms).
SHORT ANSWER QUESTIONS
Answer in the spaces indicated.
Page 3
Chemistry 111 Sect. 3
Exam #3
November 30, 2000
KEY
Name: ________________________________________________
Student #: ________________________________________________
11. (7 pts) Draw the Lewis structure for SO3.
1. # of valence electrons = 6 + 18 = 24
2. S is central atom (lower electron affinity than O)
3. Form sigma bonds, then place remaining electrons on
outer atoms: leaves each O with 3 LPs and S with less
than an octet.
4. Move one O LP to form a pi bond with S.
:O:
S
:O:
..
:O:
..
Note that this is one of three possible resonance
structures that can be drawn.
12. (8 pts) Draw the Lewis structure for SF4.
1. # of valence electrons = 6 + 28 = 34
2. S is central atom (lower electron affinity than F)
3. Form sigma bonds, then place remaining electrons on
outer atoms: leaves each F with 3 LPs and S with the
extra lone pair.
Note that this is a case where the central atom has more
than its octet of electrons.
..
:F
..
..
:F:
.. ..
S F:
..
:F:
..
13. (7 pts) The Lewis structure for PCl3 is given below. What is the molecular geometry of
PCl3 ?
..
:Cl
..
There are four electron “lumps” around the P, so the
electron geometry is tetrahedral. The one lone pair
takes up one of these, so the molecular geometry (atom
positions) is trigonal pyramidal.
..
:Cl:
P:
:Cl:
.. The Lewis structure for SF6 is given below. What is the molecular geometry of
14. (8 pts)
SF6?
..
.. :F:
:F
..
S
..
:F
.. :F:
..
There are six electron “lumps” around the S, so the
electron geometry is octahedral. There are no lone
pairs, so the molecular geometry (atom positions) is
octahedral, just like the electron geometry.
..
F:
..
..
F:
..
Page 4
KEY
Name: ________________________________________________
Student #: ________________________________________________
Chemistry 111 Sect. 3
Exam #3
November 30, 2000
Useful information
c = 3.0 × 108 m/sec
h = 6.6 × 10-34 J sec
Rhc = 2.2 × 10-18 J
me = 9.1 × 10-28 g
a0 = 0.0529 nm
NA = 6.02 × 1023 molecules/mol
2-linear, 3-trigonal planar, 4-tetrahedral
5-trigonal bipyramidal, 6-octahedral
∆Hrxn = Σ(bond ener)react - Σ(bond ener)prod
q = C m ∆T
Bond
H-C
H-H
H-N
H-O
C-C
q = (heat of fusion) m
∆H = qp
∆H0net = Σ (∆H0f)prod - Σ (∆H0f)react
∆H0net = Σ (∆H0rxn)
c = λν
E = hν
E = -Rhc/n2
λ = h/(m v)
Z* = Z - ninner electons
d = m/V
Some useful data
Bond Bond
Bond
energy
(kJ/mol)
C-N
305
C=C
C-O
358
C≡C
N-N
163
C=N
N-O
201
C≡N
O-O
146
C=O
Bond
energy
(kJ/mol)
413
436
391
463
346
Bond
energy
(kJ/mol)
602
835
615
887
732
Bond
Bond
energy
(kJ/mol)
1072
418
945
607
498
C≡O
N=N
N≡N
N=O
O=O
PERIODIC TABLE OF THE ELEMENTS
IA
IIA
IIIB
IVB
VB
VIB
VIIB
VIIIB
IB
IIB
IIIA
IVA
VA
VIA
VIIA VIIIA
1
2
H
He
1.008
4.003
3
4
5
6
7
8
9
Li
Be
B
C
N
O
F
10
Ne
6.939
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
19
20
21
22
K
Ca
Sc
Ti
39.10
40.08
44.96
37
38
39
Rb
Sr
85.47
55
26.98
28.09
30.97
32.07
35.45
39.95
24
25
26
27
28
29
30
31
32
33
34
35
36
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
47.90
50.94
52.00
54.94
55.85
58.93
58.71
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
87.62
88.91
91.22
92.91
95.94
(99)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
137.3
138.9
178.5
181.0
183.8
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
87
88
89
Fr
Ra
Ac
(223)
226.0
227.0
23
Page 5
I
Xe
Chemistry 111 Sect. 3
Exam #3
November 30, 2000
KEY
Name: ________________________________________________
Student #: ________________________________________________
Page 6