Name: ____KEY______________
CHEM&141_EXAM III_F2008_CHAPTERS 7-10_105 Points
*Valuable equations and constants are at the bottom of the last page*
1. The total number of f-orbital electrons in americium are: (4 points)
a. 14
b. 20
c. 21
d. None of the above
2. Which atom would be expected to have the lowest electron affinity?
(4 points)
a. F
b. O
c. N
d. C
e. B
3. How many resonance structures does SO3-2 have? (4 points)
a. 2
b. 3
c. 5
d. 6
4. Which of the following quantum numbers is invalid? (4 points)
a. n = 3, l = 2, ml = 1, ms = -1/2
b. n = 2, l = 1, ml = 2, ms = +1/2
c. n = 4, l = 3, ml = 0, ms = 0
5. What is the formal charge of Al in AlBr63-? (4 points)
a. +3
b. -1
c. -3
d. +1
6. Which species is paramagnetic? (4 points)
a. Zn
b. Zn2+
c. Ni
d. Cr6+
____________/24 points total
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7. The 4th principle energy level has _____ electrons when filled to
capacity according to the Periodic Table. (4 points)
a. 8
b. 18
c. 32
d. None of the above
8. In VSEPR theory, the geometries are based on trying to minimize the
repulsions of ______. (4 points)
a. Atoms
b. Electrons
c. electron pairs
d. none of the above
9. Which of the following transitions of an electron will yield a
wavelength of light in the visible red region? (4 points)
a. n = 3 to n = 2
b. n = 6 to n = 2
c. n = 6 to n = 1
10.Which of the following is the least electronegative? (4 points)
a. Si
b. Ge
c. Tl
d. Pb
11.Which of the following has the highest ionization energy? (4 points)
a. He
b. Hec. Li
d. Li-
___________/20 points total
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12.Give the electron configurations for the following: (4 points each –
total 16 points)
a. Cu
[Ar]4s13d10
b. Hf
[Xe]6s24f145d2
c. Ra2+
[Rn]
d. Se2[Kr]
13.Give the: a) Lewis dot structures (including resonance), b)
hybridizations of the central atom, c) electron and molecular
geometries of the central atom, and d) bond angles for the following
species (10 points each – total 30 points):
O3
Lewis structure:
O
O
O
O
O
O
Hybridization:
sp2
Electron geometry:
Trigonal planar
Molecular geometry:
bent
Bond angle:
<120
____________/26 points total
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HOSO2OH
Lewis structure:
O
H
O
S
O
H
O
Hybridization:
sp3
Electron geometry:
tetrahedral
Molecular geometry:
tetrahedral
Bond angle:
109.5
HN(CH3)3+
Lewis structure:
+
H
H
H
C
N
CH3
H
CH3
Hybridization:
sp3
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Electron geometry:
tetrahedral
Molecular geometry:
tetrahedral
Bond angle:
109.5
____________/20 points total
14.Draw the net dipole moment for the following molecule (hint: N is
more electronegative than H): (5 points)
H
N
H
H
15.Radiation in the ultraviolet region of the electromagnetic spectrum is
quite energetic. It is this radiation that causes dyes to fade and your
skin to develop a sunburn. If you are bombarded with 1.00 mol of
photons with a wavelength of 375 nm, what amount of energy, in
kilojoules per mole of photons, are you being subjected to? (10 points)
E photon  h  h
c

 (6.626 10
34
J  s)
2.998 108 m
375nm 
s
1m
1 109 nm
 5.30 10 19 J
photon
5.30 1019 J
1kJ
6.022 10 23 photons


 319 kJ
mol photons
photon
1000 J
1mol photons
_____________/15 points total
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Ephoton = h; h = 6.626 x 10-34 Js
c = ; c = 2.998 x 108 m/s
FC = VE – [LE + ½ BE]
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