CHAPTER 9 Problems: 4, 7, 10, 16, 20, 22, 28, 30, 36, 40, 52, 56, 66, 81, 84, 92, 104,
107
4) Discuss the basic features of the VSEPR model. Explain why the magnitude of the repulsion decreases in the following order: lone pair - lone pair > lone pair - bonding > bonding - bonding.
The VSEPR (Valence Shell-Electron Pair Repulsion) model is based on the idea that electron containing regions in the valence shell of an atom will repel one another.
Therefore, they will arrange themselves so that the regions are as far apart from one another as possible.
Because bonding electrons are localized between the two atoms bonded together, while lone pair electrons are in a more diffuse cloud of charge, lone pair electrons tend to repel other electron containing regions more strongly than bonding electrons. This leads to deviations from the “pure” geometries predicted using VSEPR. The effect of these small differences in repulsion is to decrease the bond angles in the molecular structures of species. The effect, however, is usually small.
7) Predict the geometries of the following species using the VSEPR model: a) PCl
3 b) CHCl
3 c) SiH
4 d) TeCl
4
The Lewis structures for each molecule are given below, which will be of use in finding the geometries. molecule
PCl
3
CHCl
3
SiH
4
TeCl
4 electron cloud
geometry tetrahedral tetrahedral tetrahedral trigonal bipyramid molecular
geometry trigonal planar tetrahedral tetrahedral see-saw
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10) Predict the geometry of the following molecules or ions using the VSEPR model. a) CH
3
I c) H
2
S e) SO
4
2b) ClF
3 d) SO
3
The Lewis structures for each molecule or ion are given below, which will be of use in finding the geometries. molecule
CH
ClF
3
3
I electron cloud
geometry tetrahedral trigonal bipyramid molecular
geometry tetrahedral
T-shaped
H
2
S
SO
3 tetrahedral trigonal planar nonlinear trigonal planar
SO
4
2tetrahedral
16) The bonds in beryllium hydride (BeH
2
) are polar, but the permanent dipole moment of the molecule is zero. Explain.
The molecular geometry for BeH
2 tetrahedral
is linear. Each polar bond is shown below.
While each Be - H bond is polar, the contributions to the overall polarity of the molecule cancel (as the vectors are the same magnitude by in opposite directions) and so the molecule is overall nonpolar.
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20) Which of the molecules below is polar? You may assume bonds between different colors of atoms are polar bonds.
All of the molecules are polar except SF
4
Br
2
, where the dipole moments for the individual S - F and S - Br bonds all cancel.
In CH
2
F
2
the contributions to the overall polarity from the two C - F bonds do not cancel. The same is true of the two Xe - N bonds (likely triple bonds) in XeN
2
. In OCS the central carbon is bonded to two different atoms, and so exact cancellation of the polar bonds will not occur even though the bonds point in opposite directions.
22) Use valence bond theory to explain the bonding in Cl
2
and HCl. Show how the atomic orbitals overlap when a bond is formed.
In Cl
2
the bonding is from the overlap of two 3p orbitals, one from each of the chlorine atoms, each of which contains one electron. In HCl the bonding is from the overlap of the 1s orbital of hydrogen with a 3p orbital from chlorine, with one electron initially in each orbital.
28) Describe the bonding scheme of the AsH
3
molecule in terms of hybridization.
The Lewis structure for AsH
3
is given below.
There are four electron containing regions around the central atom, and so we would expect sp
3
hybridization for the central As atom. Note that because of the lone pair electrons, we would expect the H - As - H bond angle to be smaller than 109.5

, the value for pure tetrahedral geometry. In fact, the experimental value for the H - As - H bond angle in AsH
3
is 91.8

, indicating quite a lot of distortion from pure tetrahedral geometry.
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30) Describe the changes in hybridization (if any) of the Al atom in the following reaction
AlCl
3
+ Cl 
AlCl
4
-
The appropriate Lewis structures are given below.
The hybridization for Al goes from sp
2
(in AlCl
3
) to sp an additional electron containing region in AlCl
4
-
.
3
(in AlCl
4
-
) due to the presence of
36) What are the hybrid orbitals on the carbon atoms in the following molecules a) H
3
C-CH
3 c) CH
3
-C=C-CH
2
OH e) CH
3
COOH b) H
3
C-CH=CH
2 d) CH
3
CH=O
The Lewis structures for each molecule are given below. Note that the number of electron containing regions determines the hybridization (two regions = sp; three regions
= sp 2 ; and four regions = sp 3 ).
For a, both carbons are sp
3
hybridization. For b, the left carbon is sp
3
hybridization, and the two other carbons are sp
2
hybridization. In c carbons 1 and 4 are sp
3
hybridization, and carbons 2 and 3 are sp hybridization. In d the left carbon is sp 3 hybridization and the right carbon is sp
2
hybridization. Finally, in e the left carbon is sp
3
hybridization and the right carbon is sp
2
hybridization.
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40) How many sigma and pi bonds are there in each of the following molecules?
The Lewis structures are given below. Note that in counting sigma and pi bonds, all single bonds are sigma bonds. For double bonds there is one sigma bond and one pi bond, and for triple bonds there is one sigma bond and two pi bonds.
So a b c
4 sigma bonds and 0 pi bonds
5 sigma bonds and 1 pi bond
7 sigma bonds and 3 pi bonds
52) Use molecular orbital theory to explain why the Be
2
molecule does not exist.
The MO diagram for Be
2
is given below.
This diagram corresponds to the electron configuration (

2s
)
2
(

*
2s
)
2
, and so a bond order of 0. Since the bond order is zero, there is no decrease in energy in forming Be
2
out of two beryllium atoms, and so it does not exist. Note that the Be
2+
cation (bond order =
1
/
2
) could form from a Be atom and Be + ion.
56) Explain why the bond order in N
2
is greater than in N
2
+
, but the bond order in O
2
is less than in O
2
+
.
If we use the MO chart in the book (Fig. 9.18, page 398) we can write electron configurations for each species above
N
2
(

2s
)
2
N
2
+
(

2s
)
2
(

*
2s
)
2
(

*
2s
)
2
(

2p
)
4
(

2p
(

2p
)
2
)
4
(

2p
)
1
Bond order = (8-2)/2 = 3
Bond order = (7-2)/2 = 2
1
/
2
So N
2
has a higher bond order (and so a stronger bond) than N
2
+
.
O
2
(

2s
)
2
(

*
2s
)
2
O
2
+
(

2s
)
2
(

*
2s
)
2
(

2p
)
2
(

2p
)
4
(

*2p)
2
(

2p
)
1
(

2p
)
4
(

*2p)
1
Bond order = (8-2)/2 = 2
Bond order = (8-3)/2 = 2
1
So O
2
has a lower bond order (and so a weaker bond) than O
2
+
.
/
2
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66) Describe the bonding in the nitrate ion (NO
3
-
) in terms of delocalized molecular orbitals.
The nitrate ion can be represented in terms of three equivalent resonance structures, as indicated below
The presence of these resonance structures suggests that the bonding in the NO
3
-
ion is best represented as a central nitrogen atom making single bonds with three oxygen atoms, using sp 2 hybridization on all of the atoms. An additional multicentered pi bond forms making use of the p z
orbitals of the nitrogen and three oxygen atoms perpendicular to the plane of the ion.
81) Which of the following molecules or ions is linear? a) ICl
2
b) IF
2
+ c) OF d) SnI
2
2 e) CdBr
2
The Lewis structures for these molecules or ions are given below.
Based on these Lewis structures we can construct the following table. molecule electron cloud
geometry molecular geometry
ICl
2
-
IF
2
+ trigonal bipyramid tetrahedral linear nonlinear
OF
2
SnI
2 tetrahedral trigonal planar
CdBr
2 linear
So only ICl
2
-
and CdBr
2
have linear molecular geometries. nonlinear nonlinear linear
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84) Cyclopropane (C
3
H
6
) has the shape of an equilateral triangle in which each carbon atom is bonded to two hydrogen atoms and two other carbon atoms. Cubane (C
8
H
8
) has the shape of a cube in which each carbon atom is bonded to one hydrogen atom and three other carbon atoms at each corner of the cube. a) Draw Lewis structures for these molecules. b) Compare the C-C-C angles in these molecules to the values predicted from sp
3 hybridization of a carbon atom.
The bond angle for pure sp 3 hybridization is 109.5

. The C - C - C bond angle in cyclopropane is 60.

, and the C - C - C bond angle in cubane is 90.

. c) Based on your answer in b, would you expect the above molecules to be easy to make? Why or why not?
I would expect them to be difficult to make (and highly reactive) because they require the formation of bonding between carbon atoms with C - C - C bond angles much different than those preferred based on the hybridization, placing a lot of strain on the molecule.
92) Write the ground state electron configuration for B
2
. Is the molecule diamagnetic or paramagnetic? Why?
The MO diagram for B
2
is given below.
There are two unpaired electron spins in the

2p
molecular orbitals, and so B
2
is paramagnetic.
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104) Which of the following molecules or ions possesses a permanent dipole moment? a) ClF
2
+ c) IF
4
+ b) ClF
2
d) IF
4
-
Drawings for each of these ions, indicating their geometry, are given below.
Based on these diagrams ions a and c have a permanent dipole moment, and ionsb and d do not.
107) The compound TCDD (2,3,7,8-tetrachlorodibenzo-pdioxin) is highly toxic.
The molecule gained considerable notoriety in 2004 when it was implicated in the attempted murder of a Ukranian politician. a) Describe the geometry of the molecule. Does it have a permanent dipole moment? Explain.
The molecule is planar due to the presence of the two benzene rings.
There are four polar C - Cl bonds, and four polar C - O bonds, but in both cases the contributions from the set of four bonds to the overall polarity of the molecule cancel.
Therefore the molecule is nonpolar b) How many pi bonds and how many sigma bonds are there in TCDD?
There are 20 sigma bonds and 6 pi bonds (though a better description of the bonding would be in terms of multicentered pi bonds.
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