NAME:____________________________
Fall 2008
INSTRUCTIONS:
Student Number:______________________
Chemistry 1000 Midterm #1B
____/ 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
/6
/2
/4
/3
/7
/4
/4
/3
/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)
The diagram below represents a mixture of sulfur atoms and oxygen molecules in a
closed container.
Draw the contents of the container after the mixture reacts as completely as possible
according to the equation:
2 S + 3 O2  2 SO3
[1 mark]
(b)
Assume that a beaker of pure water has been boiling for 30 minutes. What is in the
bubbles in the boiling water?
[1 mark]
Water vapour
2.
[5 marks]
(a)
Rubidium has two naturally occurring isotopes, 85 Rb and 87 Rb. Which of these
two isotopes is more abundant? Briefly explain your answer.
[2 marks]
85
Rb is more abundant
The average atomic mass of rubidium is 85.4678 u. 85 Rb has an atomic mass
of ~85 u; 87 Rb has an atomic mass of ~87 u. Since 85.4678 u is closer to the
atomic mass of 85 Rb, there is more 85 Rb than 87 Rb.
(b)
How many protons, neutrons and electrons are there in a
85
Rb atom?
[1.5 marks]
___37___ protons, ___48___ neutrons, ___37___ electrons
(c)
How many protons, neutrons and electrons are there in a
87
Rb + cation?
[1.5 marks]
___37___ protons, ___50___ neutrons, ___36___ electrons
NAME:____________________________
3.
Student Number:______________________
Give the name and symbol for each of the elements below:
[6 marks]
name
symbol
i.
Z = 12
magnesium
Mg
ii.
Z = 14
silicon
Si
iii.
Z = 16
sulfur
S
iv.
Z = 18
argon
Ar
v.
Z = 24
chromium
Cr
vi.
Z = 28
nickel
Ni
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
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
88
39
La-Lu
Ac-Lr
24
26
28
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
30
112.411
Hg
80
Tl
81
Pb
82
Bi
83
Po
84
At
85
83.80
Rn
86
NAME:____________________________
4.
Student Number:______________________
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.
5.
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.
YES
/
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.
YES
/
NO
The values for l are
different, so the two
electrons described are in
different orbitals.
NO
Both electrons have all
four quantum numbers the
same. They cannot
therefore belong to the
same atom.
NO
The values for m l are
different, so the two
electrons described are in
different 3d orbitals.
n = 3, l = 1, m l = 0, m s = +½
and
n = 3, l = 1, m l = 0, m s = -½
n = 3, l = 2, m l = 1, m s = +½
and
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 = 2, m l = 2, m s = +½
and
n = 3, l = 2, m l = 1, m s = -½
Briefly, justify your
answer.
YES
/
NAME:____________________________
6.
Student Number:______________________
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.
n = 3, l = 1, m l = +1
(a)
(b)
draw a p orbital
n = 3, l = 1, m l = -1
draw a p orbital along a
different axis than your answer
to part (a)
n = 3, l = 2, m l = +1
(c)
draw a d orbital
7.
Consider an atom of titanium (Ti) in the ground state.
(a)
Draw an orbital occupancy diagram (aka “orbital box diagram”) for an atom
of titanium in the ground state.
[3 marks]
1s
2s
2p
3s
3p
core electrons
[7 marks]
4s
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 ti tanium ion. Justify your answer.
[2 marks]
Ti +4
Removing four electrons from Ti gives Ti 4+ with a noble gas electron
configuration.
Note that the 4s electrons are lost before the 3d electron s, so the
electron configuration of Ti 2+ would be [Ar]3d 2 not [Ar]4s 2 .
NAME:____________________________
8.
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 2
YES
/
NO
carbon
1s 1 2s 2
YES
/
NO
This atom is an excited state. Ground
state would be 1s 2 2s 1 .
1s 2 2s 2 2p 5
YES
/
NO
fluorine
1s 2 2s 2 3s 2
YES
/
NO
This atom is an excited state. Ground
state would be 1s 2 2s 2 2p 2 .
9.
Consider an atom of tin (Sn) in the ground state.
[4 marks]
(a)
Write the electron configuration for a ground state atom of tin using line
notation with the noble gas abbreviation.
[2 marks]
[Kr] 5s 2 4d 10 5p 2
(b)
Give a valid set of quantum numbers for the highest energy electron in a
ground state atom of tin.
[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:______________________
10.
Consider a calcium atom (Ca) and an calcium cation (Ca2+).
(a)
Which of these two species would you expect to have a larger radius?
Ca
[1 mark]
(b)
Justify your answer to part (a).
[3 marks]
[2 marks]
Ca and Ca 2+ have the same number of protons; however, Ca 2+ has two fewer
electrons.
The loss of these two electrons decreases the shielding of all electrons in Ca 2+
(relative to Ca). As such, the valence electrons in Ca 2+ are more strongly
attracted to the nucleus, so the ion is smaller than the neutral atom .
11.
[4 marks]
(a)
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, fluorine, oxygen, potassium
(b)
Explain your answer to part (a), making sure th at your explanation 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 energ y required to 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 potassium ([Ar] 4s 1 ), the first electron removed comes from the 4s orbital
while the second comes from a 3p orbital. The second ionization energy of
potassium 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 fluorine (1s 2 2s 2 2p 5 ) and
oxygen (1s 2 2s 2 2p 4 ), both the first and second electrons removed come from
2p orbitals.
NAME:____________________________
Student Number:______________________
12.
A far-infrared laser delivers light with a power of 0.075 Watts (0.075 J/s).
The wavelength of the light delivered is 1000 m (assume 4 sig. fig.).
[7 marks]
(a)
Calculate the frequency of the light. Choose a unit that will give you a
numerical answer between 0.1 and 1000.
[3 marks]
c  
m
6
s  10 m  2.998  1011 s 1  2.998  1011 Hz
 

1000 m
1m
1GHz
  2.998  1011 Hz  9
 299 .8GHz
10 Hz
c
(b)
2.9979  10 8
Calculate the energy of one photon of light. Report your answer in Joules.
[1 mark]
E  h

J 

11
E   6.626  10 34
 2.998  10 Hz
Hz


 22
E  1.986  10 J
(c)

How many photons would the laser deliver in 30 seconds of operation?
Report your answer in moles.
[3 marks]
J
 30 s  2.25 J
(2sig.fig. )
s
1 photon
# photons  2.25 J 
 1.1  10 22 photons
1.986  10 22 J
1mol
n photons  1.1  10 22 photons 
 0.019 mol
6.02214  10 23 photons
Etotal  0.075
NAME:____________________________
Student Number:______________________
13.
Consider a Li 2+ cation in the ground state.
(a)
Calculate the energy of the lowest energy photon that can be absorbed by a
Li 2+ cation in the ground state.
[7 marks]
[9 marks]
Li 2+ starts in the ground state, so n initial = 1.
The lowest energy photon that Li 2+ in the n = 1 state can absorb will excite
the ion into the n = 2 state, so n final = 2.
Li 2+ has three protons, so Z = 3.

Z 2  
Z2 


E photon  E final  Einitial    RH 


R

H
2 
2 






n
n
final
initial





32  
32 
E photon    RH  2     RH  2 
2  
1 

 9

E photon    RH    9 RH 
 4

3
E photon  6  RH
4
3

E photon  6  2.179  10 18 J
4

E photon  1.471  10 17 J
(b)
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 energies 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  h
p  mv
n2
rn  a0
Z
Z2
E n   RH 2
n
Ek 
1

h
p
x  p 
h
4
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