Version 084 – Exam 2 – Lyon – (52380)
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001
(part 1 of 4) 10.0 points
Draw the Lewis structure for the following
hydrocarbon molecule. The carbons are num
bered one to four starting with the far left
carbon as one.
CH
2
CCHCH
3
What is the molecular shape of carbon 4?
1.
trigonal planar
2.
tetrahedral
correct
3.
trigonal pyramidal
4.
angular
5.
bipyramidal
6.
linear
7.
square planar
Explanation:
The molecule has the structure
C
C
C
C
Carbon 4 has bonds to one carbon and three
hydrogens or RHED = 4. This is tetrahedral
molecular geometry.
002
(part 2 of 4) 10.0 points
What is the hybridization of carbon 2?
1.
sp
correct
2.
sp
2
3.
sp
4
4.
sp
2
d
5.
sp
3
6.
sp
3
d
2
7.
sp
3
d
Explanation:
The molecule has the structure
C
C
C
C
Carbon 2 has bonds only to two other carbons
(double bonds).
This is RHED = 2 or
sp
hybridization.
003
(part 3 of 4) 10.0 points
What are the bond angles of carbon 1?
1.
less than 109.5
◦
2.
109.5
◦
3.
120
◦
correct
4.
180
◦
5.
90
◦
, 120
◦
, 180
◦
6.
90
◦
, 120
◦
7.
90
◦
, 180
◦
Explanation:
The molecule has the structure
C
C
C
C
Carbon 1 has bonds to one carbon and two
hydrogens or RHED = 3.
This is trigonal
planar molecular geometry with bond angles
of 120 degrees.
004
(part 4 of 4) 10.0 points
Carbon 3 has
1.
2 sigma bonds and 2 pi bonds.
2.
no sigma bonds and 4 pi bonds.
3.
1 sigma bonds and 3 pi bonds.
4.
4 sigma bonds and no pi bonds.
5.
3 sigma bonds and 1 pi bond.
correct
6.
2 sigma bonds and 1 pi bonds.
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Version 084 – Exam 2 – Lyon – (52380)
2
Explanation:
The molecule has the structure
C
C
C
C
Carbon 3 has bonds to two other carbons
(one double bond) and one hydrogen. Double
bonds are composed of one sigma and one
pi bond, therefore there is one sigma bond
from the hydrogen, one sigma bond from the
carbon, and one sigma with one pi from the
double bond for a total of 3 sigma bonds and
one pi bond.
005
10.0 points
What is the hybridization at the atom indi
cated by the arrow in the following molecule?
H
H
H
H
C
O
·
· ·
·
O
·
·
··
H
O
·
·
··
C
O
·
·
·
·
C
H
H
H
Acetylsalicylic acid (Aspirin)
1.
sp
2.
sp
3
d
3.
sp
3
d
2
4.
sp
3
5.
sp
2
correct
Explanation:
The C has RHED = 3 so three hybrid or
bitals are needed. The hybridization
sp
2
pro
vides 3 orbitals.
006
10.0 points
Write the electrondot notation for the ele
ment P.
1.
P
2.
None of these
3.
P
4.
P
5.
P
6.
P
correct
7.
P
8.
P
9.
P
Explanation:
According
to
its
electron
configuration
([Ne] 3
s
2
3
p
3
), P has 2 + 3 = 5 valence elec
trons.
The electrondot notation for P is
therefore
P
007
10.0 points
The first ionization potential of the elements
B, C, and N (atomic numbers 5, 6, and 7)
steadily increases, but that of O is less than
that of N. The best interpretation of the lower
value for O is that
1.
there is more shielding of the nuclear
charge in O than in B, C, or N.
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 Spring '07
 Fakhreddine/Lyon
 Chemistry, Atom, Electrons, Chemical bond

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