e2f10 - Version PREVIEW Exam 1 JOHNSON(53140 This print-out...

Version PREVIEW – Exam 1 – JOHNSON – (53140) 1
This print-out should have 30 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
LDE Identifying Bonds 005
001 10.0 points
Rank the labeled bonds in the molecule
below from least to most polar.
C
c
b
N
H
e
O H
a
S
d
S
H
1. c < e < b < a < c
2. c < d < e < a < b
3. d < c < e < b < a correct
4. c < d < e < b < a
5. d < c < b < e < a
Explanation:
Bonds a, b, c, d, and e have a ΔEN of 1.4,
1.0, 0.5, 0.0, and 0.9, respectively.
Brodbelt 013 322
002 10.0 points
Which of the following statements about po-
larity is false?
1. Linear molecules can be polar.
2. Polar molecules must have a net dipole
moment.
3. Lone (unshared) pairs of electrons on the
central atom play an important role in influ-
encing polarity.
4. CCl
4
is a polar molecule. correct
5. Dipole moments can “cancel”, giving a
net non-polar molecule.
Explanation:
The Lewis Dot structure for CCl
4
is
C
·
·
· ·
··
Cl
· ·
·
·
··
Cl
·
·
· ·
·
·
Cl
·
· ·
·
··
Cl
The molecule has tetrahedral electronic and
molecular geometry. The C - Cl bond is polar,
but becuase of the symmetry of the molecule,
the individual dipole moments cancel. The
molecule is therefore nonpolar.
ChemPrin3e T03 39
003 10.0 points
Which of the following is NOT polar?
1. COCl
2
2. ClF
3
3. BrO
−
3
4. CH
+
3
correct
5. O
3
Explanation:
COCl
2
(the central atom is C) is polar be-
cause there are two different bonds: C=O
and C-Cl which have different sized dipoles
so their effects do not cancel. The other in-
correct choices have either 3, 4 or 5 RHED:
one or more of these are lone pairs on the
central atom which causes the polar bonds to
be placed in positions where the dipole of at
least one bond is not opposing another, caus-
ing the species to be polar. In CH
+
3
there are
no lone pairs; the C-H bonds are at 120
◦
so
their effects cancel and the ion is nonpolar.
ChemPrin3e T03 22
004 10.0 points
Which of the following has bond angles
slightly less than 109
◦
?
1. NH
+
4
2. BrO
−
3
correct

Version PREVIEW – Exam 1 – JOHNSON – (53140) 2
3. ClO
−
4
4. BH
−
4
5. PO
3 −
4
Explanation:
While all structures have four regions of
electron density around the central atom, only
BrO
−
3
also has one of these regions as a lone
pair. Since the lone pair requires more room,
the other bonds are repelled away from it
slightly, folding them up rather like an um-
brella and resulting in angles slightly less than
109.5
◦
.
Brodbelt 013 320
005 10.0 points
The sp
3
hybridization has what percent s
character and what percent p character?
1. 50%; 50%
2. 33%; 67%
3. 67%; 33%
4. 75%; 25%
5. 25%; 75% correct
Explanation:
s =
1
4
= 25%
p =
3
4
= 75%
Brodbelt 013 308
006 10.0 points
NF
−
2
has (2, 3, 4) regions of high electron
density and (2, 4, 6) bonded electrons.
1. 2; 2
2. 4; 4 correct
3. 2; 4
4. 4; 6
5. 2; 6
6. 4; 2
7. 3; 4
8. 3; 6
9. 3; 2
Explanation:
N = 8 × 3 = 25
A = 5 + (7 × 2) + 1 = 20
S = 24 - 20 = 4
First, draw the molecule. The number of
regions of HED equals the number of bonds
and the number of lone pairs. There are 4
shared, or bonded electrons, giving 2 regions
of HED. The 2 F atoms have an octet. This
means that 16 of the 20 available electrons are