Answers to Problems
VSEPR examples
SeF 2
basis structure:
tetrahedron
reported structure:
bent
SeF 2
6+2(7)=20 vse
S
F
F
A X2 E2
bond angles:
CO 3
e.m.l.t. 109.5 deg.
2-
basis structure:
triangle
reported structure:
bent
CO3
2-
4+3(6)+2=24 vse
bond angles:
A X3 Eo
2-
O
C
O
O
l.t. 120 deg.
XeO 4
basis structure:
tetrahedron
reported structure:
tetrahedron
XeO4
O
8+4(6)=32 vse
A X4 E 0
bond angles:
Xe
O
O
O
109.5 deg
ClF 5
basis structure:
octahedron
F
7+5(7)=42 vse
reported structure:
bond angles:
square-based
pyramid
90 and 180 deg.
F
ClF5
A X5 E1
F
Cl
F
F
Br 3 1 1-
Br
basis structure:
Br 31-
trigonal bipyramid
Br
7+2(7)+1=22 vse
reported structure:
linear
Br
A X2 E3
bond angle:
180 deg.
SO 3 2 basis structure:
tetrahedral
2SO3 2S
reported structure:
bond angles:
trigonal pyramid
6+3(6)+2=26 vse
O
O
O
A X3 E1
l.t. 109.5 deg.
CO2
basis structure:
CO2
linear
4+2(6)=16 vse
reported structure:
O
C
O
linear
A X2 E0
bond angle:
180 deg.
KrF 4
basis structure:
KrF4
octahedral
F
8+4(7)=36 vse
F
reported structure:
bond angles:
square
90 and 180 deg.
A X4 E 2
F
Kr
F
SF 4
basis structure:
trigonal bipyramid
reported structure:
teeter totter
bond angles:
O3
F
SF4
6+4(7)=34 vse
90, 120, and 180 deg.
S
F
F
A X4 E 1
F
(ozone)
basis structure:
triangle
O3
reported structure:
bent
6+2(6)=18 vse
bond angle
O
l.t. 120 deg.
A X2 E1
v.
a.
O
CHCl3
4 + 1 + 3(7) = 26 vse
C has 4 bonded atoms and no unshared electron pairs, so n = 4, m = 0
VSEPR formula:
AX4 E0
basis str. tetrahedron
reported str: TETRAHEDRON
angle = 109.5 deg.
Cl are the most EN atoms present. Determine the location of their
common center. C and H are more electropositive. Determine the
location of their charge center.
H
H
C
Cl
C
Cl
Cl
Cl
Cl
Cl
Do the centers of negative and positive charges coincide? ANS: NO
So CHCl3 would be expected to have a permanent dipole moment.
b.
NH 4 1+
5 + 4(1) - 1 = 8 vse
N has 4 bonded atoms and no unshared electron pairs, so n = 4, m = 0
VSEPR formula:
AX4 E0
basis str. tetrahedron
O
reported str: TETRAHEDRON
angle = 109.5 deg.
N is the most EN atom present. So the center of negative charge is the N
atom. H are the more electropositive atoms present. Determine their
common center.
H
H
1+
1+
1+
N
H
H
H
H
H
H
Do the centers of negative and positive charges coincide? ANS: YES
So NH 4 1+ would NOT be expected to have a permanent dipole moment.
c.
SO 2
3(6) = 18 vse
S has 2 bonded atoms and one unshared electron pairs, so n = 2, m = 1
VSEPR formula:
AX2 E1
basis str. triangle
reported str. BENT
angle = L.T. 120 deg.
S
O
O
O
O
CO 3 24 + 3(6) + 2 = 24 vse
C has 3 bonded atoms and no unshared electron pairs, so n = 3, m = 0
VSEPR formula:
AX3 E0
basis str. triangle
reported str. TRIANGLE
angle = 120 deg.
d.
O
O
2-
2C
O
O
O
O
2-
e.
IF5
7 + 5(7) = 42 vse
I has 5 bonded atoms and one unshared electron pair, so n = 5, m = 1
VSEPR formula:
AX5 E1
basis str. octahedron
reported str: SQUARE-BASED PYRAMID
angles, 90
F
F
F
F
F
F
I
F
vi.
a.
H2
TWO electrons in
BOND ORDER =
H2
H2
σ
F*
* anti-bonding m.o.
1s
σ
1s
bonding m.o.
F
2 - 0
2
= 1
w /b diamagnetic
1-
a THREE electron system
ONE electron in
σ
* anti-bonding m.o.
TWO electrons in
σ
BOND ORDER =
2 - 1
2
H2
F
a TWO electron system
NO electrons in
b.
F
F
1-
bonding m.o.
w /b paramagnetic
= 1/ 2
F*
1s
1s
F
c.
H2
1+
a ONE electron system
NO electrons in
σ
* anti-bonding m.o.
ONE electron in
σ
bonding m.o.
BOND ORDER =
H2
d.
1+
He 2
1 - 0
2
F*
1s
1s
F
= 1/ 2
w /b paramagnetic
a FOUR electron system
TWO electrons in
σ
* anti-bonding m.o.
TWO electrons in
σ
bonding m.o.
2 - 2
BOND ORDER =
2
F*
1s
1s
F
= 0
He 2 has never been observed.
vii.
First-row-of-eight atoms have 2 s and 2 p valence shells, which can accommodate
maxima of TWO and SIX electrons respectively, for a total of EIGHT per atom.
Molecular orbital energy level diagrams for first-row-of-eight diatomic molecules and ions
must therefore accommodate up to SIXTEEN vse. These m.o. diagrams are just a little
more complex than those used above.
In fact two diagrams are used for these cases b/c there is a change in m.o. energy levels
between B 2 and C 2 , and the remaining members N 2 , O 2 , F 2 and Ne 2 .
F*
vii.
a.
B2
2(3) = 6 vse
BOND ORDER = (4 - 2) / 2 = 1
PARAMAGNETIC
( 2 unpaired electrons)
(Use Fig. A)
B*
2p
2p
F
B
F*
2s
2s
F
F*
b.
C2
2(4) = 8 vse
B*
2p
2p
F
BOND ORDER = (6 - 2) / 2 = 2
B
F*
DIAMAGNETIC
(Use Fig. A)
2s
2s
F
F*
c.
N2
2(5) = 10 vse
BOND ORDER = (8 - 2) / 2 = 3
DIAMAGNETIC
(Use Fig. B)
B*
2p
2p
B
F
F*
2s
2s
F
F*
d.
O2
2(6) = 12 vse
BOND ORDER = (8 - 4) / 2 = 2
PARAMAGNETIC
(2 unpaired electrons)
(Use Fig. B)
B*
2p
2p
B
F
F*
2s
2s
F
Refer to TABLE 2.
(on class web site) for a listing of bond orders, B D E , magnetic
properties, and bond lengths for remaining diatomic molecules and
ions in exercise ix.