BOOK PROBLEMS C 7
14-16. H2+ has 1 electron (In H2, each H contributes 1 e-. That’s 2e-. The +
indicates 1 e- removed. 2-1 = 1e(See the molecular orbital diagram on p 224 in Munowitz)
H2+
s
2s
2s
s
s
1s
1s

s
Put the one electron into the lowest energy level. That’s a bonding orbital. H2+
has ½ of a sigma () bond.(so the bond order is ½). That electron is unpaired, so
H2+ is in fact paramagnetic.
If you consider H2, with 2 electrons (1 each from the H’s), again, add electrons
starting in the lowest energy level. Each energy level can accommodate 2
electrons, paired spin up-spin down (Remember the Pauli exclusion principle…2
electrons in the same orbital will have the same n, l, and ml quantum numbers,
so their spin quantum numbers (ms) will have to be +1/2 and –1/2.
H2
s
2s
2s
s
s
1s
1s

s
H2 has 2 electrons in a  bonding orbital. 2 shared electrons is a single covalent
bond. This will be shorter than the ½ bond in H2+.
17. If you consider B2, with 10 electrons (5 each from the Borons), again, add
electrons starting in the lowest energy level. Each energy level can accommodate
2 electrons, paired spin up-spin down (Remember the Pauli exclusion
principle…2 electrons in the same orbital will have the same n, l, and ml
quantum numbers, so their spin quantum numbers (ms) will have to be +1/2 and
–1/2. (see molecular orbital diagram 7-11b on p 232)
B2
2px
2pz
2py
2px 2py 2pz
2px 2py 2pz
 2px
2py

2pz

s
2s
2s

s

s
1s
1s

s
B2 starts with 2 electrons in a  1s bonding orbital. The next 2 electrons go into
a * 1s anti-bonding orbital. The fully occupied bonding & anti-bonding orbitals
cancel each other out. NO bond yet. The same thing happens with the next 4
electrons, which go into the  2s bonding orbital & the  *2s bonding orbital,
respectively. Still no bond.
The last 2 e- go into the 2px and 2py orbitals. They are equal energy, so each gets
half filled. The  orbitals are bonding, and there are 2 electrons, so that’s a single
bond (bond order = 1). There are 2 unpaired e-, so the system is paramagnetic.
19. (I’m not showing the s and s orbitals, in the interests of space, but they are filled)
B2+ 9 e-
B2 10 e-
2px
2py

2px
2pz
2px
2py
B2- 11 e-
2py

2pz
2px
2py
2px
2pz
2py


2pz
2py
2pz
2px
2pz



 s
 s
s





s
s
s
B2+ will have: ½ bond
B2 will have: 1 bond
B2- will have: 1½ bond
longest bond length
intermediate bond length
shortest bond length

lowest bond dissociation energy
intermediate bond dissociation energy
highest bond dissociation energy
20.
C2+ 11 e-
C2 12 e-
2px
2py
C2- 13 e-
2px
2pz
2py
2px
2pz
2py
2pz


2px
2py


2pz
2py
2px 
2pz

2px 
2py
2pz



s
s
s



s
s
s
C2+ will have: 1½ bonds
C2 will have: 2 bonds
C2- will have: 2½ bonds
longest bond length
intermediate bond length
shortest bond length
lowest bond dissociation energy
intermediate bond dissociation energy
highest bond dissociation energy
GOOD QUESTION TO WORK OUT: How many bonds will the molecular
orbital theory predict that N2 will have. Show on an MO diagram.
21. For diatomic molecules with atoms larger than N, the 2px has lower energy than
2py and 2pz .
O2+
15 e
2px
2py

2pz 

2py
O217 e-
O2
16 e

 2pz 
2px
2py
2pz 


2py

 2px


 2pz 

2py
2pz


2py
 2pz
2px

2px

2px

s
s
s



s
s
s
O2+ will have: 2½ bonds
O2 will have: 2 bonds
O2- will have: 1½ bonds
shortest bond length
intermediate bond length
longest bond length

highest bond dissociation energy
intermediate bond dissociation energy
lowest bond dissociation energy
Note that the MO model does a better job of representing the “reality” of O2 than
the Lewis dot structure. The Lewis structure shows O2 with a double bond and
diamagnetic (no unpaired electrons). The DATA, which is what the model has
to predict, shows that O2 does have double bond character, but is paramagnetic
(has unpaired electrons). The MO model predicts that O2 is paramagnetic. A
better model of reality.
23. Electronegativity is the measure of the tendency of a combined atom (an
atom that’s part of a molecule) to attract a shared pair of electrons (that’s
electrons in a bond, holding two atoms together) to itself.
Electronegativity, the measure of the tendency of a combined atom to attract
a shared pair of electrons to itself, increases from lower left to upper right on
the periodic table. Fluorine is the most electronegative atom. Linus Pauling
assigned relative electronegativities to the atoms, as shown in Figure 7.18, p
244 in the text.
H
2.1
Li Be
1.0 1.5
Na
0.9
B C N O
2.0 2.5 3.0 3.5
P
Al
2.1 S
F
4.0
Cl
3.0
K
0.8 Ca
Br
2.8
Rb
0.8 Sr Y Zr
I
2.5
Cs
0.7 Ba
Tl
4f1
At
2.2
4f14
A molecule is polar if there is a substantial DIFFERENCE in Electronegativity
between the atoms. For example, CsF is VERY polar (ΔElectronegativity = 4.0 – 0.7 =
3.3). The electrons in the Cs-F bond spend much of their time closer to F than Cs.
The F has more negative charge character. On the other hand, F2 is NONPOLAR
(ΔElectronegativity = 4.0 – 4.0 = 0) The electrons in the F-F bond spend equal time
with each fluorine.
What’s “substantial difference in Electronegativity”? well, Cesium fluoride
((ΔElectronegativity = 3.3) is certainly way polar. C-H bonds (ΔElectronegativity = 2.5 – 2.1
= 0.4) are considered relatively nonpolar. A C-N bond (ΔElectronegativity = 2.5 – 3.0
= 0.5) is considered polar. A C-O bond (ΔElectronegativity = 2.5 – 3.5 = 1.0) is
considered polar, as is an O-H bond ((ΔElectronegativity = 3.5 – 2.1 = 1.4)
There is no hard and fast line between polar and nonpolar. There is a gradient of
polarities.
For NO, ΔElectronegativity = 3.0– 3.5 = 0.5 The O has partial negative (δ-) character.
And the N a partial positive (δ+)
For BeN, ΔElectronegativity = 1.5– 3.0 = 1.5 Be δ+N δFor PN, ΔElectronegativity = 2.1 - 3.0 = 0.9 P δ+N δNitrogen monoxide < Phosphorus nitride < Beryllium nitride
32-37. Look back at chapter 2 to recall how to draw Lewis Structures. You can’t
do molecular geometry or polarity without it.
If you can count, you can predict molecular shapes, bond angles, and
hybridizations.
Number of
things around
central atom
Shape
Bond angle
hybridization
180
180
120
sp
sp2
109
90 & 120
sp3
sp3d
90
sp3d2
(to give everything
around the central
atom the maximum
amount of room)
(a lone pair : of
electrons or a =
double bond count as
1 thing)
1
2
3
Linear
Linear
Trigonal
planar
Tetrahedral
Trigonal
bipyramidal
octahedral
4
5
6
32. Ammonium ion:
+
H
N
H
H
H
4 things around a central atom, tetrahedral, 109o
bond angles, N is sp3 hybridized.
33-34. Sulfur trioxide:
O
S
O
O
+ 2 other resonance structures. What are they? What is resonance?
What’s a better way to represent the reality of sulfur trioxide?
3 things around a central atom, trigonal planar, 120o bond angles, S is sp2 hybridized,
Each Sδ+-Oδ- bond is polar, but the whole molecule is nonpolar, since the polarities cancel
out.
35. Sulfite ion:
2..
S
O
O
O
There are 4 things around the central sulfur atom, 3 O’s and a lone pair of e-. Therefore,
the ion is tetrahedral, (or pyramidal if you pretend you can’t see the lone pair); 109o bond
angles, S is sp3 hybridized. Each Oxygen, of course, has an octet of electrons, 3 lone
pairs and the bonding pair, sp O is sp3 hybridized and tetrahedral as well. Each Sδ+-Oδbond is polar, and the whole ion is polar, since it is an ion.
36. Boron trifluoride has an sp2 hybridized Boron (and sp3 hybridized fluorines), and is
trigonal planar. Notice that B doesn’t have an octet. It has space to accept a pair of
electrons (so it is a Lewis acid). Ammonia has an sp3 hybridized N, tetrahedral, and has a
pair of electrons to donate (It is a Lewis base).
F
F
F
B
F
F
B
F
**
**
N
H
N
H
H
H
H
H
The complex is tetrahedral about both N &
B. , with 109o bond angles
B is now sp3 hybridized. N remains sp3
37. The geometry of ethane is tetrahedral about both C’s. The molecule is considered
nonpolar, tho’ there is some difference in electronegativity between Cδ- & Hδ+
H
H
C
H
H
C
H
H
The C’s are sp3 hybridized.
F
F
Sb
F
F
43a.
Trigonal bipyramidal. Sb is sp3d hybridized. F’s are sp3 hybridized.
The bonds are polar, and the whole molecule is non-polar.
F