NAME:____________________________
Fall 2008
INSTRUCTIONS:
Student Number:______________________
Chemistry 1000 Midterm #1A
____/ 60 marks
1) Please read over the test carefully before beginning. You should have
8 pages of questions and a formula/periodic table sheet.
2) If your work is not legible, it will be given a mark of zero.
3) Marks will be deducted for incorrect information added to an
otherwise correct answer.
4) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
5) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
6) You may use a calculator.
7) You have 90 minutes to complete this test.
Confidentiality Agreement:
I agree not to discuss (or in any other way divulge) the contents of this exam until after 8pm
Mountain Time on Wednesday, October 8th, 2008. I understand that, if I were to break this
agreement, I would be choosing to commit academic misconduct and that is a serious offense
which will be punished. The minimum punishment would be a mark of 0/60 on this exam and
removal of the “overwrite midterm mark with final exam mark” option for my grade in this
course; the maximum punishment would include expulsion from this university.
Signature: ___________________________
Course: CHEM 1000 (General Chemistry I)
Semester: Fall 2008
The University of Lethbridge
Date: _____________________________
Question Breakdown
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
/2
/5
/3
/4
/4
/3
/4
/2
/6
/7
/4
/7
/9
Total
/ 60
NAME:____________________________
Student Number:______________________
1.
As promised, two questions from the Chemical Concepts Inventory on the first
MasteringChemistry assignment:
[2 marks]
(a)
Salt is added to water and the mixture is stirred until no more salt dissolves. The salt that
does not dissolve is allowed to settle out. What happens to the concentration of salt in
solution if water evaporates until the volume of the solution is half the original volume?
Assume that the temperature is constant.
[1 mark]
The concentration stays the same.
(solution
is
saturated;
concentration cannot increase;
some salt precipitates)
(b)
A 1.0 gram sample of solid iodine is placed in a tube, and the tube is sealed after all of
the air is removed. The tube and the solid iodine together weigh 27.0 grams.
The tube is then heated until all of the iodine sublimes and the tube is filled with iodine
gas. What is the total mass after heating?
[1 mark]
27.0 g
2.
(a)
[5 marks]
63
65
Copper has two naturally occurring isotopes, Cu and Cu. Which of these
two isotopes is more abundant? Briefly explain your answer.
[2 marks]
63
Cu is more abundant
The average atomic mass of copper is 63.546 u. 63 Cu has an atomic mass of
~63 u; 65 Cu has an atomic mass of ~65 u. Since 63.546 u is closer to the
atomic mass of 63 Cu, there is more 63 Cu than 65 Cu.
(b)
How many protons, neutrons and electrons are there in a
63
Cu atom?
[1.5 marks]
___29___ protons, ___34___ neutrons, ___29___ electrons
(c)
How many protons, neutrons and electrons are there in a
65
Cu 2+ cation?
[1.5 marks]
___29___ protons, ___36___ neutrons, ___27___ electrons
NAME:____________________________
Student Number:______________________
3.
Consider an oxygen atom (O) and an oxide anion (O2-).
(a)
Which of these two species would you expect to have a larger radius?
O 2[1 mark]
(b)
Justify your answer to part (a).
[3 marks]
[2 marks]
O and O 2- have the same number of protons; however, O 2- has an additional 2
electrons. These electrons increase the shielding of all electrons in O 2(relative to O). As such, the valence electrons in O 2- are less strongly
attracted to the nucleus, so the ion is larger than the neutral atom.
4.
(a)
[4 marks]
Which of the elements listed below will have the greatest difference between
its first ionization energy and its second ionization energy? Circle your
choice.
[1 mark]
beryllium, carbon, lithium, nitrogen
(b)
Explain your answer to part (a), making sure that your expla nation clearly
indicates your understanding of the terms first and second ionization energy.
[3 marks]
First ionization energy is the energy required to remove an electron from a
neutral atom. Second ionization energy is the energy required t o remove an
electron from a +1 cation.
The difference between the first and second ionization energies will depend
on how much more difficult it is to remove the second electron than the first
electron. The second ionization energy will be particularly higher if the
second electron comes from a different subshell than the first.
In lithium (1s 2 2s 1 ), the first electron removed comes from the 2s orbital
while the second comes from the 1s orbital. The second ionization energy of
lithium is therefore *much* higher than the first ionization energy.
In beryllium (1s 2 2s 2 ), on the other hand, both the first and second electrons
removed come from the same orbital – the 2s. In carbon (1s 2 2s 2 2p 2 ) and
nitrogen (1s 2 2s 2 2p 3 ), both the first and second electrons removed come from
2p orbitals.
NAME:____________________________
5.
Student Number:______________________
For each of the following pairs of electrons, indicate whether or not they
could both be in the same orbital according to the Pauli exclusion principle.
Explain each of your decisions in a few words.
[4 marks]
Quantum Numbers of
Two Electrons
Could this be a pair of
electrons in the same orbital?
Circle yes or no.
NO
Both electrons have all
four quantum numbers the
same. They cannot
therefore belong to the
same atom.
NO
n, l and m l are the same,
so the electrons are in the
same orbital. The m s
values are opposite,
indicating opposite spins.
NO
The values for m l are
different, so the two
electrons described are in
different 3p orbitals.
NO
n, l and m l are the same,
so the electrons are in the
same orbital. The m s
values are opposite,
indicating opposite spins.
n = 3, l = 1, m l = -1, m s = +½
and
YES
/
n = 3, l = 1, m l = -1, m s = +½
n = 3, l = 1, m l = -1, m s = +½
and
YES
/
n = 3, l = 1, m l = -1, m s = -½
n = 3, l = 1, m l = 1, m s = +½
and
YES
/
n = 3, l = 1, m l = -1, m s = -½
n = 3, l = 0, m l = 0, m s = +½
and
YES
/
n = 3, l = 0, m l = 0, m s = -½
6.
Briefly, justify your
answer.
Each of the following sets of quantum numbers describes an orbital in the
same atom. Draw one orbital that could be described by each set of quantum
numbers. Be sure to include labeled axes on your picture.
[3 marks]
All pictures must have correct phases and labeled axes.
(a)
n = 4, l = 2, m l = +1
draw a d orbital
(c)
(b)
n = 4, l = 1, m l = -1
draw a p orbital
n = 4, l = 1, m l = +1
draw a p orbital along a different axis than your answer to part (b)
NAME:____________________________
7.
Student Number:______________________
For each of the following electron configurations, indicate whether or not it
correctly describes a ground state atom.
For each incorrect electron
configuration, explain what is wrong with it. For each correct electron
configuration, give the name (not symbol) of the element.
[4 marks]
Electron
Configuration
Could describe a
ground state atom?
Circle yes or no.
If no, why not?
If yes, name the element.
1s 2 2s 2 2p 6 3s 3
YES
/
NO
There can only be two electrons in an s
subshell.
1s 2 2s 2 2p 4
YES
/
NO
oxygen
1s 2 3p 1
YES
/
NO
This atom is an excited state. Ground
state would be 1s 2 2s 1 .
1s 2 2s 2 2p 6 3s 2 3p 10
YES
/
NO
There can only be six electrons in a p
subshell.
8.
How did Heisenberg’s uncertainty principle influence our understanding of
the structure of an atom?
[2 marks]
Heisenberg’s uncertainty principle states that the precision of our knowledge
about a small particle’s position and its momentum are inversely related. If
we have more information about its position, we must have less information
about its momentum (and vice versa).
As a result, the model of the atom in which an electron orbits a nucleus in
circular fashion cannot be correct as we could know the electron’s precise
momentum and position at the same time.
NAME:____________________________
9.
Student Number:______________________
Give the name and symbol for each of the elements below:
[6 marks]
name
symbol
i.
Z = 11
sodium
Na
ii.
Z = 13
aluminium
Al
iii.
Z = 15
phosphorus
P
iv.
Z = 17
chlorine
Cl
v.
Z = 26
iron
Fe
vi.
Z = 30
zinc
Zn
Partial Periodic Table (copied from data sheet)
1
18
1.0079
4.0026
H
He
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
1
11
12
3
4
5
39.0983
40.078
44.9559
47.88
50.9415
54.9380
58.9332
63.546
69.723
72.61
74.9216
78.96
79.904
K
Ca
Sc
Ti
V
Mn
Co
Cu
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
95.94
25
(98)
101.07
27
102.906
106.42
29
107.868
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
37
132.905
38
137.327
40
178.49
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
88
39
La-Lu
Ac-Lr
24
8
26
9
10
28
11
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
104
Db
105
Sg
106
Bh
107
Hs
108
Mt
109
Dt
110
Rg
111
12
30
112.411
13
14
Hg
80
Tl
81
7
8
15
16
Ne
4
7
6
20.1797
3
6
5
2
Pb
82
Bi
83
Po
84
9
10
17
18
At
85
83.80
Rn
86
NAME:____________________________
Student Number:______________________
10.
Consider an atom of scandium (Sc) in the ground state.
(a)
Draw an orbital occupancy diagram (aka “orbital box diagram”) for an atom
of scandium in the ground state.
[3 marks]
1s
2s
2p
3s
3p
[7 marks]
4s
core electrons
3d
valence electrons
(b)
On your diagram, clearly label the core and valence electrons.
[1 mark]
(c)
Is this atom paramagnetic or diamagnetic?
[1 mark]
paramagnetic
(d)
Predict the charge of the most common scandium ion. Justify your answer.
[2 marks]
Sc +3
Losing three electrons gives Sc 3+ a noble gas electron configuration.
Note that the 4s electrons are lost before the 3d electron, so the
electron configuration of Sc + would be [Ar]3d 2 not [Ar]4s 2 .
11.
Consider an atom of tellurium (Te) in the ground state.
[4 marks]
(a)
Write the electron configuration for a ground state atom of tellurium using
line notation with the noble gas abbreviation.
[2 marks]
[Kr] 5s 2 4d 10 5p 4
(b)
Give a valid set of quantum numbers for the highest energy electron in a
ground state atom of tellurium.
[2 marks]
Any one of the following six answers is acceptable:
n = 5, l = 1, m l = +1, m s = +½
n = 5, l = 1, m l = +1, m s = -½
n = 5, l = 1, m l = 0, m s = +½
n = 5, l = 1, m l = 0, m s = -½
n = 5, l = 1, m l = -1, m s = +½
n = 5, l = 1, m l = -1, m s = -½
NAME:____________________________
Student Number:______________________
12.
A microwave oven delivers microwaves with a power of 700 Watts (700 J/s ;
assume 3 sig. fig.). The wavelength of the microwaves delivered is 12.24 cm.
[7 marks]
(a)
Calculate the frequency of the microwaves. Choose a unit that will give you
a numerical answer between 0.1 and 1000.
[3 marks]
c
m
c
s 100 cm 2.449 10 9 s
12 .24 cm
1m
1GHz
2.449 10 9 Hz
2.449 GHz
10 9 Hz
2.9979 10 8
(b)
2.449 10 9 Hz
Calculate the energy of one photon of microwave.
Joules.
E
E
Report your answer in
[1 mark]
h
E
(c)
1
6.626 10
1.623 10
34
24
J
2.449 10 9 Hz
Hz
J
How many photons would the microwave deliver in 1 0 seconds of operation?
Report your answer in moles.
[3 marks]
Etotal
700
J
10 s
s
7.0 10 3 J
1 photon
4.3 10 27 photons
1.623 10 24 J
1mol
4.3 10 27 photons
7.2 10 3 mol
6.02214 10 23 photons
# photons 7.0 10 3 J
n photons
NAME:____________________________
Student Number:______________________
13.
Consider a He + cation in the ground state.
(a)
Calculate the energy of the lowest energy photon that can be absorbed by a
He + cation in the ground state.
[7 marks]
[9 marks]
He + starts in the ground state, so n initial = 1.
The lowest energy photon that He + in the n = 1 state can absorb will excite
the ion into the n = 2 state, so n final = 2.
He + has two protons, so Z = 2.
E photon
E final Einitial
E photon
RH
E photon
RH
Z2
RH
22
22
RH
n final
2
RH
Z2
ninitial
2
22
12
4 RH
E photon 3RH
E photon 3
2.179 10
E photon 6.537 10
(b)
18
18
J
J
How would your answer to part (a) change if the phrase “in the ground state”
was changed to “in the n=3 state”?
You do not need to include a calculation or provide a numerical value in your
answer to part (b) of this question; however, you should briefly explain your
answer verbally (and/or pictorially).
[2 marks]
The energy of the photon would be smaller. As n increases, the en ergies of
the states get closer together, so the energy gap between n = 3 and n = 4 is
smaller than the energy gap between n = 1 and n = 2.
NAME:____________________________
Student Number:______________________
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u)
1.6605 10-27 kg
Avogadro's number
6.02214 1023 mol–1
Bohr radius (a0)
5.29177 10-11 m
Electron charge (e)
1.6022 10-19 C
Electron mass
5.4688 10-4 u
6.626 10-34 J·Hz-1
1.0072765 u
1.0086649 u
2.179 x 10-18 J
2.9979 x 108 m·s-1
Planck's constant
Proton mass
Neutron mass
Rydberg Constant (RH)
Speed of light in vacuum
Formulae
c
E
rn
n2
a0
Z
h
Z2
RH 2
n
En
1
p
h
p
mv
Ek
x
h
4
p
1 2
mv
2
CHEM 1000 Standard Periodic Table
18
1.0079
4.0026
H
He
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
1
3
4
11
12
3
4
5
6
7
8
9
10
11
12
2
20.1797
Ne
5
6
7
8
9
10
13
14
15
16
17
18
39.0983
40.078
44.9559
47.88
50.9415
54.9380
58.9332
63.546
69.723
72.61
74.9216
78.96
79.904
K
Ca
Sc
Ti
V
Mn
Co
Cu
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
95.94
25
(98)
101.07
27
102.906
106.42
29
107.868
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
37
132.905
38
137.327
40
178.49
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
39
La-Lu
Ac-Lr
88
24
Hf
Ta
W
Re
Os
Ir
Pt
Au
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
28
72
(261)
104
89
26
Th
90
Pa
91
U
92
Bh
107
Hs
Mt
Dt
30
112.411
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
Np
Pu
94
Am
95
Cm
96
Rn
86
Rg
108
93
83.80
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré