NAME:___________________________
Fall 2008
INSTRUCTIONS:
1.
Section:_____
Student Number:__________________
Chemistry 2000 Midterm #1C
____/ 48 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.
Consider the bonding in nitrosyl fluoride, NOF (nitrogen is the central atom) according to
valence bond theory.
(a)
:F:
..
(b)
sp
[7 marks]
Draw a Lewis structure for NOF.
.. ..
N ..O
What is the hybridization of the nitrogen atom in NOF?
2
(c)
Clearly indicate which atomic orbitals combine to make each σ bond in NOF.
σ:N(sp2)+F(2p)
:F:
..
(d)
:F:
..
.. ..
N ..O
Clearly indicate which
NOF.
.. ..
N ..O
π:N(2pz)+O(2pz)
σ:N(sp2)+O(2p)
atomic orbitals combine to make each π bond in
NAME:___________________________
Section:_____
Student Number:__________________
2.
(a)
[17 marks]
4-
Draw a qualitative valence molecular orbital diagram for the BC ion. Label all orbitals on
your diagram and include electrons.
3σ*
1π*
2p
2p
Energy
3σ
1π
2σ*
2s
2s
2σ
B2(b)
BC4-
C2-
Make an educated guess about the energy sequence of the MOs. Explain the sequence you are
choosing and why there is potential ambiguity?
Between the 2σ and 3σ, mixing can occur (the 2σ will be lowered while the 3σ MO will be raised).
Since for B2 and C2 mixing is significant enough to raise the 3σ MO above the 1π MO, BC4- should
have the same MO sequence.
(c)
According to your MO diagram, what is the bond order for this ion?
2.5
(c)
Draw the MOs that correspond to bonding electron pair(s) beside your diagram. Indicate which
orbital you are drawing.
(d)
What would you expect for the bond order for a neutral BC molecule?
There will be one less electron in the MO diagram, hence the bond order will decrease to 1.5.
NAME:___________________________
3.
Section:_____
Student Number:__________________
(a) Draw the valence molecular orbital diagram for LiH. (Li: 2s atomic orbital energy: -5.39 eV;
H: 1s atomic orbital energy: -13.61 eV). Name the orbitals and fill them with the correct
number of electrons.
[7 marks]
1σ*
2s
1s
Energy
Li
LiH
2σ
H
(b) What can you deduce about the nature of the Li-H bond from the molecular orbital diagram?
Since the hydrogen 1s orbital is so much lower than the 2s orbital from Li, the 2σ MO is closer to
the 1s of H and is much closer in nature to H than to Li. As a consequence the Li-H bond
should be very polar with the bonding electron pair being mainly on hydrogen.
(actually LiH is considered and ionic compound, with Li+ and H- ions)
NAME:___________________________
Section:_____
Student Number:__________________
4.
The valence π molecular orbital diagram shown below can be used to describe the π bonding in
nitryl fluoride, NO2F (nitrogen is the central atom).
[11 marks]
1π∗
E
2π
1π
(a)
i.
Draw the Lewis structure of NO2F, including all valid resonance structures.
..
:F
..
+
N
.. :O :
O
.. :
(b)
+
N
:O
.. :
-
ii.
Label the MOs in the above diagram.
i.
Fill the π-MOs in the diagram above with the appropriate number of π electrons for the
NO2F.
What is the average π bond order for each N-O bond in NO2F according to the MO
diagram?
0.5
ii.
iii.
(c)
..
:F
..
..
O:
What is the average bond order for each N-O bond in the NO2F according to MO
theory?
1.5
On the π-MO diagram above, draw pictures of all three π-MO.
NAME:___________________________
Section:_____
Student Number:__________________
5.
In band theory, compounds with a band gap which is smaller than 0.1 eV (which is very small)
can essentially be considered metallic conductors.
[6 marks]
(a)
Explain the band structure of metals and why metals are electric conductors.
(b)
•
Atomic orbitals from metal atoms combine to form MOs.
•
These MOs form bands with very small energy spacings between the individual MOs.
•
Conduction band (empty) and valence band (full) overlap.
•
As a result, electrons can be easily excited from the valence to the conduction band.
Why are compounds with a band gap of less than 0.1 eV not considered semiconductors?
A band gap smaller than 0.1 eV allows for facile excitation of electrons from the valence to the
conduction band. The thermal energy that is needed for excitation is minimal.
NAME:___________________________
1
Section:_____
Student Number:__________________
CHEM 1000 Standard Periodic Table
18
4.0026
1.0079
H
He
1
2
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
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