NAME:___________________________
Fall 2012
Section:_____
Student Number:__________________
Chemistry 2000 Midterm #1A
INSTRUCTIONS:
____/ 55 marks
1) Please read over the test carefully before beginning. You should have 5
pages of questions and a periodic table.
2) If you need extra space, use the bottom of the periodic table page, indicate
that you are doing so next to the question and clearly number your work.
3) If your work is not legible, it will be given a mark of zero.
4) Marks will be deducted for incorrect information added to an otherwise
correct answer.
5) You have 90 minutes to complete this test.
1.
Consider the bonding in hydrogen cyanide, HCN (carbon is the central atom, bonded to H and
N) according to valence bond theory.
[11 marks]
(a)
Draw a Lewis structure for HCN.
H‒C≡N:
(b)
What is the hybridization of the carbon atom in HCN?
sp
(c)
Clearly indicate which atomic orbitals combine to make each  bond in HCN.
:C(sp)+H(1s)
H‒C≡N:
(d)
:C(sp)+N(2pz)
Clearly indicate which atomic orbitals combine to make each  bond in HCN.
:C(2py)+N(2py)
H‒C≡N:
:C(2px)+N(2px)
NAME:___________________________
Section:_____
Student Number:__________________
2.
[16 marks]
(a)
Draw a qualitative valence molecular orbital diagram for diatomic F2. Label all orbitals on
your diagram and include electronsin AOs and MOs.
(b)
According to your MO diagram, what is the bond order for this ion?
1
(c)
Draw the filled MOs beside your diagram. It must be clear which orbital corresponds to each
drawing.
(d)
What would you expect for the bond order for a cationic F2+ molecule?
There will be one electron less in the anti-bonding MOs, hence the bond order will increase to
1.5.
NAME:___________________________
3.
Section:_____
Student Number:__________________
The alkali metals form ozonides, which contain the paramagnetic O3 anion. [10 marks]
(a) What conclusion can you draw from the fact that these salts are paramagnetic?
The fact that ozonides are paramagnetic indicates that these salts (the anion) contain at
least one unpaired electron.
(b) The energy level diagram for the -molecular orbitals of neutral O3 is given below. Draw
the Lewis diagram (with all resonance structures) of O3 AND fill the -MO diagram with the
correct number of -electrons.
(c) Fill the -MO diagram for the O3 anion with the correct number of -electrons (the extra
electron in O3 is a -electron). How does the O-O bond order changes with going from neutral
O3 to the O3 anion?
The ozonide anion has an additional electron compared to neutral ozone. This electron goes into
the antibonding pi orbital, therefore reducing the O-O bond order.
Ozone: bond order (sigma and pi) per O-O bond: 1.5
Ozone: bond order (sigma and pi) per O-O bond: 1.25
NAME:___________________________
4.
Section:_____
Student Number:__________________
The valence  molecular orbital diagram shown below can be used to describe the  bonding in
the planar cyclic C4H4.
[11 marks]
3π*
2π (non-bonding)
1π
(a)
(b)
i.
Draw the Lewis structure of C4H4, including all valid resonance structures.
ii.
Label the MOs in the above diagram.
i.
Fill the π-MOs in the diagram above with the appropriate number of  electrons for
C4H4.
What is the average π bond order for each C‒C bond in C4H4 according to the MO
diagram?
0.25
ii.
iii.
What is the average total bond order for each C‒C bond in the C4H4 according to MO
theory?
1.25
(c)
On the π-MO diagram above, draw the picture of the lowest lying π-MO.
(d)
Draw the picture of ONE of the degenerate (E = 0) MOs?
NAME:___________________________
Section:_____
Student Number:__________________
5.
Molecular orbital theory for extended solids is called band theory.
[7 marks]
(a)
Explain the difference between the conduction band and the valence band. What happens when
these bands overlap?

The bands are formed when atomic orbitals from metal atoms combine to form MOs.

Conduction band (empty) and valence band (full) overlap.

If they overlap, electrons can be easily excited from the valence to the conduction band.

In that case the solid is a metal with metallic conduction.
(b)
Explain the difference between intrinsic and extrinsic semiconductors.

For semiconductors, the valence and conduction bands do not overlap, but there is a band
gap between them.

For intrinsic semiconductors electrons can be excited from the valence band to the
conduction band (thermally or photochemically) to result in conduction.

In extrinsic semiconductors, the material is doped by an electron donor or electron
acceptor, which reduces the energy necessary for electron excitation.
NAME:___________________________
1
Section:_____
Student Number:__________________
CHEM 2000 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
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
2
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
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)
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
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
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
107
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
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
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré