Chemical Bonding II:
Molecular Shapes,
Valence Bond Theory,
and Molecular
Orbital Theory
Review Questions
10.1 J
The properties of molecules are directly related to their shape. The sensation of taste, immune
response, the sense of smell, and many types of drug action all depend on shape-specific
interactions between molecules and proteins.
According to VSEPR theory, the repulsion between electron groups on interior atoms of a
molecule determines the geometry of the molecule.
The five basic electron geometries are
(1)
Linear, which has two electron groups.
(2)
Trigonal planar, which has three electron groups.
(3)
Tetrahedral, which has four electron groups.
(4)
Trigonal bipyramid, which has five electron groups.
(5)
Octahedral, which has six electron groups.
An electron group is defined as a lone pair of electrons, a single bond, a multiple bond, or
even a single electron.
H—C—H
ijj^^jl
(a) Linear geometry \
\ (b) Trigonal planar geometry I
109.5=
Tetrahedral geometry I
Equatorial chlorine
Axial chlorine
"P—Cl:
\
Trigonal bipyramidal geometry 1
369
I Octahedral geometry I
370
Chapter 10 Chemical Bonding II
The electron geometry is the geometrical arrangement of the electron groups around the central atom.
The molecular geometry is the geometrical arrangement of the atoms around the central atom.
The electron geometry and the molecular geometry are the same when every electron group bonds two
atoms together. The presence of unbonded lone-pair electrons gives a different molecular geometry and
electron geometry.
(a)
Four electron groups give tetrahedral electron geometry, while three bonding groups and one lone
pair give a trigonal pyramidal molecular geometry.
(b)
Four electron groups give a tetrahedral electron geometry, while two bonding groups and two lone
pairs give a bent molecular geometry.
(c)
Five electron groups give a trigonal bipyramidal electron geometry, while four bonding groups and
one lone pair give a seesaw molecular geometry.
(d)
Five electron groups give a trigonal bipyramidal electron geometry, while three bonding groups and
two lone pairs give a T-shaped molecular geometry.
(e)
Five electron groups gives a trigonal bipyramidal electron geometry, while two bonding groups and
three lone pair give a linear geometry.
(f)
Six electron groups give an octahedral electron geometry, while five bonding groups and one lone
pair give a square pyramidal molecular geometry.
(g)
Six electron groups give an octahedral electron geometry, while four bonding groups and two lone
pairs gives a square planar molecular geometry.
Larger molecules may have two or more interior atoms. When predicting the shapes of these molecules,
determine the geometry about each interior atom and use these geometries to determine the entire threedimensional shape of the molecules.
To determine if a molecule is polar, do the following:
1.
Draw the Lewis structure for the molecule and determine the molecular geometry.
2.
Determine whether the molecule contains polar bonds.
3.
Determine whether the polar bonds add together to form a net dipole moment.
Polarity is important because polar and nonpolar molecules have different properties. Polar molecules interact strongly with other polar molecules, but do not interact with nonpolar molecules, and vice versa.
10.8
According to valence bond theory a chemical bond results from the overlap of two half-filled orbitals with
spin-pairing of the two valence electrons.
10.9
According to valence bond theory, the shape of the molecule is determined by the geometry of the overlapping orbitals.
10.10
In valence bond theory, the interaction energy is usually negative (or stabilizing) when the interacting
atomic orbitals contain a total of two electrons that can spin-pair.
10.11
Hybridization is a mathematical procedure in which the standard atomic orbitals are combined to form
new atomic orbitals called hybrid orbitals. Hybrid orbitals are still localized on individual atoms, but they
have different shapes and energies from those of standard atomic orbitals. They are necessary in valence
bond theory because they correspond more closely to the actual distribution of electrons in chemicallybonded atoms.
10.12
Hybrid orbitals minimize the energy of the molecule by maximizing the orbital overlap in a bond.
Chapter 10 Chemical Bonding II
373
10.29
Nonbonding orbitals are atomic orbitals not involved in a bond and will remain localized on the atom.
10.30
In Lewis theory, a chemical bond is the transfer or sharing of electrons represented as dots. Lewis theory allows
us to predict the combination of atoms that form stable molecules, and the general shape of a molecule.
Lewis theory is a quick way to predict the stability and shapes of molecules based on the number of
valence electrons. However, it does not deal at all with how the bonds that we make are formed. Valence
bond theory is a more advanced bonding theory that treats electrons in a quantum-mechanical manner. A
quantitative approach is extremely complicated but a qualitative approach allows an understanding of
how the bonds are formed. In valence bond theory, electrons reside in quantum-mechanical orbitals localized on individual atoms. When two atoms approach each other, the electrons and nucleus of one atom
interact with the electron and nucleus of the other atom. If the energy of the system is lowered, a chemical bond forms. So, valence bond theory portrays a chemical bond as the overlap of two half-filled atomic
orbitals. The shape of the molecule can be predicted from the geometry of the overlapping orbitals. Also,
valence bond theory explains the rigidity of the double bond. However, valence bond theory falls short
in explaining certain phenomenon such as magnetism and certain bond properties. Valence bond theory
treats the electrons as if they reside in the quantum-mechanical orbitals that we calculate for an atom. This
is an oversimplification that is partially compensated for by introducing the concept of hybridization. An
even more complex quantum-mechanical model is molecular orbital theory. In molecular orbital theory, a
chemical bond occurs when the electrons in the atoms can lower their energy by occupying the molecular orbitals of the resultant molecule. The chemical bonds in MO theory are not localized between atoms,
but spread throughout the entire molecule. Molecular orbital theory uses trial functions to solve the
Schrodinger equation for the molecules. In order to determine how well the trial function works, you calculate the energy, trying to minimize the energy. However, no matter how "good" your guess, you can
never do better than nature at minimizing energy. These minimum-energy calculations for orbitals must
be done by computer.
All three of these models have strengths and weaknesses, none is "correct." What information you need,
depends on which approach you use.
Problems by Topic
VSEPR Theory and Molecular Geometry
/
•
10.31
Four electron groups: A trigonal pyramidal molecular geometry has three bonding groups and one lone pair
of electrons, so there are four electron pairs on atom A.
10.32
~N.
Three electron groups: A trigonal planar molecular geometry has three bonding groups and no lone pairs of
electrons so there are three electron pairs on atom A.
1
10.33
I (a)
4 total electron groups, 4 bonding groups, 0 lone pairs
A tetrahedral molecular geometry has four bonding groups and no lone pairs. So, there are four total
electron groups, four bonding groups, and pairs.
(b)
5 total electron groups, 3 bonding groups, 2 lone pairs
A T-shaped molecular geometry has three bonding groups and two lone pairs. So, there are five total
electron groups, three bonding groups, and two lone pairs.
(c)
6 total electron groups, 5 bonding groups, 1 lone pairs
A square pyramidal molecular geometry has five bonding groups and one lone pair. So, there are six
total electron groups, five bonding groups, and one lone pairs.
(a)
6 total electron groups, 6 bonding groups, 0 lone pairs
An octahedral molecular geometry has six bonding groups and no lone pairs. So, there are six total
electron groups, six bonding groups, and no lone pairs.
(b)
6 electron groups, 4 bonding groups, 2 lone pairs
A square planar molecular geometry has four bonding groups and two lone pairs. So, there are six
total electron groups, four bonding groups, and two lone pairs.
/
V
/
10.34
374
Chapter 10 Chemical Bonding II
(c)
5 electron groups, 4 bonding groups, 1 lone pair
A seesaw molecular geometry has four bonding groups and one lone pair. So, there are five total electron groups, four bonding groups, and one lone pair.
PF3:
Electron geometry-tetrahedral; molecular geometry-trigonal pyramidal; bond angle = 109.5°
Because of the lone pair, the bond angle will be less than 109.5°.
Draw a Lewis structure for the molecule:
PF3 has 26 valence electrons.
• c •
•
r •
-p
••
-F
••
Determine the total number of electron groups around the central atom:
There are four electron groups on P.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are three bonding groups and one lone pair.
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Four electron groups give a tetrahedral electron geometry; three bonding groups and one lone
pair give a trigonal pyramidal molecular geometry; the idealized bond angles for tetrahedral
geometry are 109.5°. The lone pair will make the bond angle less than idealized.
(b)
SBr2:
Electron geometry-tetrahedral; molecular geometry-bent; bond angle = 109.5°
Because of the lone pairs, the bond angle will be less than 109.5°.
Draw a Lewis structure for the molecule:
has 20 valence electrons.
Determine the total number of electron groups around the central atom:
There are four electron groups on S.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are two bonding groups and two lone pairs.
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Four electron groups give a tetrahedral electron geometry; two bonding groups and two lone
pair give a bent molecular geometry; the idealized bond angles for tetrahedral geometry are
109.5°. The lone pairs will make the bond angle less than idealized.
(0
CHC13: Electron geometry-tetrahedral; molecular geometry-tetrahedral; bond angle = 109.5°
Because there are no lone pairs, the bond angle will be 109.5°.
Draw a Lewis structure for the molecule:
CHC13 has 26 valence electrons.
••
Cl-
/~i
Cl ••
: ci:
Determine the total number of electron groups around the central atom:
There are four electron groups on C.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are four bonding groups and no lone pairs.
Chapter 10 Chemical Bonding II
375
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Four electron groups give a tetrahedral electron geometry; four bonding groups and no lone
pairs give a tetrahedral molecular geometry; the idealized bond angles for tetrahedral geometry are 109.5°; however, because the attached atoms have different electronegativities the
bond angles are less than idealized.
(d)
CS2:
Electron geometry-linear; molecular geometry-linear; bond angle = 180°
Because there are no lone pairs, the bond angle will be 180°.
Draw a Lewis structure for the molecule:
C$2 has 16 valence electrons.
Determine the total number of electron groups around the central atom:
There are two electron groups on C.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are two bonding groups and no lone pairs.
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Two electron groups give a linear geometry; two bonding groups and no lone pairs give a linear
molecular geometry; the idealized bond angle is 180°. The molecule will not deviate from this.
10.36
(a)
CF4:
Electron geometry-tetrahedral; molecular geometry-tetrahedral; bond angle = 109.5°
Draw a Lewis structure for the molecule:
has 32 valence electrons.
Determine the total number of electron groups around the central atom:
There are four electron groups on C.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are four bonding groups and no lone pairs.
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Four electron groups give a tetrahedral electron geometry; four bonding groups and no lone
pairs give a tetrahedral molecular geometry; idealized tetrahedral bond angles for tetrahedral
geometry are 109.5°.
(b)
NFj.
Electron geometry-tetrahedral; molecular geometry-trigonal pyramidal; bond angle = 109.5°
Because of the lone pair, the bond angle will be less than 109.5°.
Draw a Lewis structure for the molecule:
NF3 has 26 valence electrons.
N
Determine the total number of electron groups around the central atom:
There are four electron groups on N.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are three bonding groups and one lone pair.
Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:
Chapter 10 Chemical Bonding II
10.38
377
ClOa will have the smaller bond angle because lone pair-bonding pair repulsions are greater than bonding pair-bonding pair repulsions.
Draw the Lewis structures for both structures:
has 26 valence electrons.
C1O4 ~ has 32 valence electrons.
o
••
••
•• oii ••
There are three bonding groups and
There are four bonding groups and
one lone pair.
no lone pairs.
Both have four electron groups, but the lone pair in QO^ ~ will cause the bond angle to be smaller because
of the lone pair-bonding pair repulsions.
S?4
Draw a Lewis structure for the molecule:
SF4 has 34 valence electrons.
••
:F :
.:F—•• s — F".•
•
•
F
•
r .
Determine the total number of electron groups around the central atom:
There are five electron groups on S.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are four bonding groups and one lone pair.
Use Table 10.1 to determine the electron geometry and molecular geometry:
The electron geometry is trigonal bipyramidal so the molecular geometry is seesaw.
Sketch the molecule:
l\
(b)
C1F3
Draw a Lewis structure for the molecule:
C1F3 has 28 valence electrons.
••
:F :
: F
ci
F •
Determine the total number of electron groups around the central atom:
There are five electron groups on CI.
378
Chapter 10 Chemical Bonding II
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are three bonding groups and two lone pairs.
Use Table 10.1 to determine the electron geometry and molecular geometry:
The electron geometry is trigonal bipyramidal so the molecular geometry is T-shaped.
Sketch the molecule:
-Cl
(c)
IF2
Draw a Lewis structure for the ion:
IF-> ~ has 22 valence electrons.
:F
••
i
-F
••
Determine the total number of electron groups around the central atom:
There are five electron groups on I.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are two bonding groups and three lone pairs.
Use Table 10.1 to determine the electron geometry and molecular geometry:
The electron geometry is trigonal bipyramidal so the molecular geometry is linear.
Sketch the ion:
[F
(d)
IBr4
I
Draw a Lewis structure for the ion:
IBr4 " has 36 valence electrons.
Br«
••
Br-
•
Br :
Determine the total number of electron groups around the central atom:
There are six electron groups on I.
Determine the number of bonding groups and the number of lone pairs around the central atom:
There are four bonding groups and two lone pairs.
Use Table 10.1 to determine the electron geometry and molecular geometry:
The electron geometry is octahedral so the molecular geometry is square planar.
Sketch the ion:
Br
Br
Br
Br
Chapter 10 Chemical Bonding II
(a)
N2
Draw the Lewis structure:
Atom
Number of
Electron Groups
LeftN
Right N
2
2
Number of
Lone Pairs
1
1
Molecular
Geometry
Linear
Linear
Sketch the molecule:
N==N
(b)
Draw the Lewis structure:
N
N
•H
Atom
Number of
Electron Groups
LeftN
Right N
3
3
Number of
Lone Pairs
1
1
Molecular
Geometry
Bent
Bent
Sketch the molecule:
/ \
N
H
(c)
N2H4 Draw the Lewis structure:
••
••
Atom
Number of
Electron Groups
Number of
Lone Pairs
Molecular
Geometry
LeftN
4
Right C
4
1
1
Trigonal pyramidal
Trigonal pyramidal
Sketch the molecule:
H
10.43
(a)
Four pairs of electrons give a tetrahedral electron geometry. The lone pair would cause lone
pair-bonded pair repulsions and would have a trigonal pyramidal molecular geometry.
Chapter 10 Chemical Bonding II
385
Atom
Number of
Electron Groups
Number of
Lone Pairs
Molecular
Geometry
LeftC
Center C
0
4
3
4
0
0
2
Right C
4
0
Tetrahedral
Trigonal Planar
Bent
Tetrahedral
Sketch the molecule:
H
(c)
NH2CO2H Draw the Lewis structure and determine the geometry about each interior atom:
H
••
-o
N.
••
Number of
Electron Groups
Atom
N
4
3
4
C
O
Number of
Lone Pairs
1
Trigonal Pyramidal
0
2
Trigonal Planar
Bent
Molecular
Geometry
Sketch the molecule:
O
H
\
cular Shape and Polarity
Draw the Lewis structure for CC>2 and CCLj determine the molecular geometry and then the polarity.
•PJ •
0
Number of electron groups on C
Number of lone pairs
Molecular geometry
2
0
linear
•• uPI••
4
0
tetrahedral
386
Chapter 10 Chemical Bonding II
Even though each molecule contains polar bonds, the sum of the bond dipoles gives a net dipole of zero for
each molecule.
The linear molecular geometry of CC>2 will have bond vectors that are equal and opposite. *
^
The tetrahedral molecular geometry of CCLj will have bond vectors that are equal and have a net dipole of zero.
10.48
Draw the Lewis structure of CHsF determine the molecular geometry and then the polarity.
••
•
•
' F •
H
I
C
H
H
Number of electron groups on C
4
Number of lone pairs
0
Molecular geometry
tetrahedral
The molecule is tetrahedral but is polar because the C - H bond dipoles are different from the C - F bond dipoles.
Because the bond dipoles are different, the sum of the bond dipoles is NOT zero. Therefore, the molecule is polar.
The tetrahedral molecular geometry of Cti^F will have unequal bond vectors so the molecule will have a
net dipole.
10.49
(a)
PF3 - polar
Draw the Lewis structure and determine the molecular geometry:
The molecular geometry from Exercise 35 is trigonal pyramidal.
Determine if the molecule contains polar bonds:
The electronegativities of P = 2.1 and F = 4. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecule is trigonal pyramidal, the three dipole moments sum to a nonzero net
dipole moment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole moments
add to determine polarity.
(b)
SBr2 - nonpolar
Draw the Lewis structure and determine the molecular geometry:
The molecular geometry from Exercise 35 is bent.
Determine if the molecule contains polar bonds:
The electronegativities of S = 2.5 and Br = 2.8. Therefore the bonds are nonpolar.
Even though the molecule is bent, since the bonds are nonpolar, the molecule is nonpolar.
388
Chapter 10 Chemical Bonding II
Determine if the molecule contains polar bonds:
The electronegativities of H = 2.1 and S = 2.5. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is bent, the two dipole moments sum to a nonzero net dipole
moment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole moments add
to determine polarity.
(a)
C1O3 " - polar
Draw the Lewis structure and determine the molecular geometry:
Four electron pairs, with one lone pair give a trigonal pyramidal molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of Cl = 3.0 and O = 3.5. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is trigonal pyramidal, the three dipole moments sum to a
nonzero net dipole moment. The molecule is polar. See Table 10.2 p. 415 in text to see how
dipole moments add to determine polarity.
(b)
SC12- polar
Draw the Lewis structure and determine the molecular geometry:
Four electron pairs with two lone pairs give a bent molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of S = 2.5 and Cl = 3.0. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is bent, the two dipole moments sum to a nonzero net dipole
moment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole moments add
to determine polarity.
(c)
SC14 - polar
Draw the Lewis structure and determine the molecular geometry:
••
: ci :
: ci — s —
ici :
Five electron pairs with one lone pair give a seesaw molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of S = 2.5 and Cl = 3.0. Therefore the bonds are polar.
Chapter 10 Chemical Bonding II
389
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is seesaw, the four equal dipole moments sum to a nonzero
net dipole moment. The molecule is polar.
The seesaw molecular geometry will not have offsetting bond vectors.
A
(d)
BrCls - nonpolar
Draw the Lewis structure and determine the molecular geometry.
.. ! ci
CL
ci:
: aSix electron pairs with one lone pair gives square pyramidal molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativity of Br = 2.8 and Cl = 3.0. The difference is only 0.2, therefore the bonds
are nonpolar. Even though the molecular geometry is square pyramidal, the five bonds are
nonpolar so there is no net dipole. The molecule is nonpolar.
10.52
(a)
SiCl4 - nonpolar
Draw the Lewis structure and determine the molecular geometry:
: ci:
: ••
c"—si — ••ci:
i
101
Four electron pairs with no lone pairs give a tetrahedral molecular geometry.
.•
Determine if the molecule contains polar bonds:
The electronegativities of Cl = 3.0 and Si = 1.8. Therefore the bonds are polar.
i
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is tetrahedral, the four equal dipole moments sum to a zero
net dipole moment. The molecule is nonpolar. See Table 10.2 p. 415 in text to see how dipole
moments add to determine polarity.
(b)
CF2C12 - polar
Draw the Lewis structure and determine the molecular geometry:
: ••ci
I
••
:•• :
Four electron pairs with no lone pairs give a tetrahedral molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of C = 2.5, F = 4.0, and Cl = 3.0. Therefore the bonds are polar.
390
Chapter 10 Chemical Bonding II
Determine whether the polar bonds add together to form a net dipole:
Even though the molecular geometry is tetrahedral, which normally yields a nonpolar molecule, the four dipole moments sum to a nonzero net dipole moment because of the different
electronegativities of Cl and F. The molecule is polar. See Table 10.2 p. 415 in text to see how
dipole moments add to determine polarity.
(c)
SeF6 - nonpolar
Draw the Lewis structure and determine the molecular geometry:
:F
••
-/.'
••
:F
Six electron pairs with no lone pairs give an octahedral molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of Se = 3.0 and F = 4.0. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is octahedral, the six equal dipole moments sum to a zero net
dipole moment. The molecule is nonpolar. See Table 10.2 p. 415 in text to see how dipole
moments add to determine polarity.
(d)
- polar
Draw the Lewis structure and determine the molecular geometry:
••
• c-
•
: ••F , ••F :
..\ I S»
••
s '
••
Six electron pairs with one lone pair give square pyramidal molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of I = 2.0 and F = 4.0. Therefore the bonds are polar.
Determine whether the polar bonds add together to form a net dipole:
Because the molecular geometry is square pyramidal, the five dipole moments sum to a
nonzero net dipole moment. The molecule is polar.
The square pyramid structure has offsetting bond vectors in the equatorial plane, but not in
the axial positions.
/N
y*
.<•" '*',.
ce Bond Theory
(a)
Be 2s2
No bonds can form. Beryllium contains no unpaired electrons, so no bonds can form without hybridization.
Chapter 10 Chemical Bonding II
10.54
10.55
391
(b)
P 3s23p3 Three bonds can form. Phosphorus contains three unpaired electrons, so three bonds can
form without hybridization.
(c)
F 2s22p5 One bond can form. Fluorine contains one unpaired electron, so one bond can form without hybridization.
(a)
B 2s22p1 One bond can form. Boron contains one unpaired electron, so one bond can form without
hybridization.
(b)
N 2s22p3 Three bonds can form. Nitrogen contains three unpaired electrons, so three bonds can form
without hybridization.
(c)
O 2s22p4 Two bonds can form. Oxygen contains two unpaired electrons, so two bonds can form without hybridization.
PH3
P
[ED
3s
l i t ) I (T)| (T)|
3p
The unhybridized bond angles should be 90°. So, without hybridization, there is good agreement between
valence bond theory and the actual bond angle of 93.3°.
10.56
SF2
s
TTJ
lltimimi
3s
FI
3p
TT]
2s
F2
Umtimi
2p
TT]
UTUTimi
2s
2p
The unhybridized bond angles should be 90°. So, without hybridization, there is not very good agreement
between valence bond theory and the actual bond angle of 98.2°.
392
Chapter 10 Chemical Bonding II
10.57
2s22p2
C
TTTT
TTTITIT
2s22p2
C
10.58
Till
I
2s
10.60
sp2
Only sp2 hybridization of this set of orbitals has a remaining p orbital to form a -IT bond.
sp3 hybridization utilizes all 3 p orbitals.
sp3d2 hybridization utilizes all 3 p orbitals and 2 d orbitals.
sp3d
sp3d hybridization utilizes an s orbital, 3 p orbitals, and d orbital. Since 5 orbitals are used, 5 hybrid
orbitals form and 5 bonds can form.
Hybridization utilizes an s orbital and 3 p orbitals. Four orbitals are used, so 4 hybrid orbitals form
and 4 bonds can form.
Hybridization utilizes an s orbital and 2 p orbitals. Three orbitals are used, so 3 hybrid orbitals form.
This allows 3 cr and 1 TT bond to form for a total of 4 bonds formed.
sp3
sp2
(a)
CC14
Write the Lewis structure for the molecule:
-8 •
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
Sketch the molecule and label the bonds:
o-C(sp3)-Cl(/>)
Chapter 10 Chemical Bonding II
(b)
NH3
393
Write the Lewis structure for the molecule:
H
H
N
••
H
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
Sketch the molecule and label the bonds:
I Lone pair in N spi
(c)
OF2
Write the Lewis structure for the molecule:
••
! F
••
••
O
••
••
F•
••
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
Sketch the molecule and label the bonds:
I Lone pair in O s/>3
(d)
CC>2
Write the Lewis structure for the molecule:
Use VSEPR to predict the electron geometry:
Two electron groups around the central atom give a linear electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Linear electron geometry has sp hybridization.
394
Chapter 10 Chemical Bonding II
Sketch the molecule and label the bonds:
<rC<sp)-0(p)
10.62
(a)
CH2Br2 Write the Lewis structure for the molecule:
H
••
: Br••
H
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
Sketch the molecule and label the bonds:
(b)
SO2
Write the Lewis structure for the molecule:
••
••
o= s — o :
Use VSEPR to predict the electron geometry:
Three electron groups around the central atom give a trigonal planar electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Trigonal planar electron geometry has sp2 hybridization.
Sketch the molecule and label the bonds:
irS(p)-O(p)
Lone pair
in S(sp2) A
- O(p)
Chapter 10 Chemical Bonding II
(c)
JSTF-3
395
Write the Lewis structure for the molecule:
••
F-
"
••
N-
i
• i
•
r ••
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
Sketch the molecule and label the bonds:
o-N(sp 3 )-F(p)
(d)
Bp3
[B Lone pair in N sp3
Write the Lewis structure for the molecule:
••
:F :
••
: F-
F :
Use VSEPR to predict the electron geometry:
Three electron groups around the central atom give a trigonal planar electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Trigonal planar electron geometry has sp2 hybridization.
Sketch the molecule and label the bonds:
Empty p orbital
o-B( 5 p 2 )-F(p)
COC12 Write the Lewis structure for the molecule:
:0:
•••_ I—"•
Use VSEPR to predict the electron geometry:
Three electron groups around the central atom give a trigonal planar electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Trigonal planar electron geometry has sp2 hybridization.
396
Chapter 10 Chemical Bonding II
Sketch the molecule and label the bonds:
o-C(s/)-0(p)
(b)
BrF5
Write the Lewis structure for the molecule:
F:
••
IBr
: F"
F
••
Use VSEPR to predict the electron geometry:
Six electron pairs around the central atoms gives an octahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Octahedral electron geometry has sp3d2 hybridization.
Sketch the molecule and label the bonds:
Lone pair in lAr(
(c)
Xep2
Write the Lewis structure for the molecule:
••
: F••
••
Xe
• F
••
Use VSEPR to predict the electron geometry:
Five electron groups around the central atom give a trigonal bipyramidal geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Trigonal bipyramidal geometry has sp3d hybridization.
Chapter 10 Chemical Bonding II
397
Sketch the molecule and label the bonds:
3 lone pairs in Xe
5p3t/ equatorial
(d)
13
Write the Lewis structure for the molecule:
••
: ••i -
-1 :
Use VSEPR to predict the electron geometry:
Five electron groups around the central atom give a trigonal bipyramidal geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Trigonal bipyramidal geometry has sp*d hybridization.
Sketch the molecule and label the bonds:
3 lone pairs in 1
sp3d equatorial
10.64
(a)
SO32~ Write the Lewis structure for the ion:
_. 2-
o-
••
••
-o:
Use VSEPR to predict the electron geometry:
Four electron groups around the central atom give a tetrahedral electron geometry.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization.
402
Chapter 10 Chemical Bonding II
N has four electron pairs around the atom, which is tetrahedral electron pair geometry. Tetrahedral electron
pair geometry is sp3 hybridization.
10.68
C - 1 and C - 4 each have three electron groups around the atom, which is trigonal planar electron pair
geometry. Trigonal planar electron pair geometry is sp2 hybridization.
C - 2 and C - 3 each have four electron pairs around the atom, which is tetrahedral electron pair geometry.
Tetrahedral electron pair geometry is sp3 hybridization.
O - 1 and O - 2 each have four electron pairs around the atom, which is tetrahedral electron pair geometry.
Tetrahedral electron pair geometry is sp3 hybridization.
N has four electron pairs around the atom, which is tetrahedral electron pair geometry. Tetrahedral electron
pair geometry is sp3 hybridization.
Orbital Theory
Is + Is constructive interference results in a bonding orbital:
Bonding
molecular
orbital
Isl Is
10.70
Is - Is destructive interference results in an antibonding orbital:
Antibonding
molecular
orbital
Destructive
interference
1s -is
has seven electrons.
t
has nine electrons.
t
<r
tl
T »>
t
n_^
I.
AO
MO
AO
isjfl.
Jti_l-
41
IT
AO
MO
&\s
AO
AO = Atomic Orbital; MO = Molecular Orbital
5 — 4 1
4 — 3 1
Bond order = - = - stable
Bond order = —-— = - stable
Chapter 10 Chemical Bonding II
411
Sketch the molecule and label the bonds:
irC(p)-C(p)
o-C(sp z )-H(s)
C(sp')-C(sp3)
<rC( 5 p 2 )-H(j)
<rC(sp 3 )-H(s)
(c)
CH3SH Write the Lewis structure for the molecule:
H
-S••
H-
H
Use VSEPR to predict the electron geometry:
Four electron groups around the C atom and the S atom give a tetrahedral electron geometry.
Four bonding pairs of electrons around the C give a tetrahedral molecular geometry and two
bonding groups and two lone pairs around the S give a bent molecular geometry.
Determine if the molecule contains polar bonds:
The electronegativities of C = 2.5, S = 2.5, and H = 2.1. The C - H bonds and the S - H bond
will be slightly polar, and the C - S bond will be nonpolar.
Determine whether the polar bonds add together to form a net dipole:
In both molecular geometries the sum of the dipole moments is not zero. The molecule is polar.
Select the correct hybridization for the central atom based on the electron geometry:
Tetrahedral electron geometry has sp3 hybridization on both C and S.
Sketch the molecule and label the bonds:
o-S(sp 3 )-H(s)
2 lone paii in sp
on S
S(sp3)-C(jp3)
crC(sp})-H«
<7C(*p 3 )-H(s)
senne
O.
H
O:
N
H
. •o•'
C - 1 and C - 3 each have four electron groups around the atom. Four electron pairs give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding pairs and
zero lone pairs give a tetrahedral molecular geometry.
412
Chapter 10 Chemical Bonding II
C - 2 has three electron groups around the atom. Three electron pairs give a trigonal planar geometry; trigonal planar geometry has sp2 hybridization. Three bonding pairs and zero lone pairs give a
trigonal planar molecular geometry.
N has four electron groups around the atom. Four electron pairs give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Three bonding pairs and one lone pair give
a trigonal pyramidal molecular geometry.
O -1 and O - 2 each have four electron groups around the atom. Four electron pairs give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Two bonding pairs and
two lone pairs give a bent molecular geometry.
H
H
(b)
asparagine
H
H
N1
1
H
H
IQ:
d
C.
C3
O
••
H
H
c= o
4
••
C - 1 and C - 3 each have four electron groups around the atom. Four electron groups give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding groups
and zero lone pairs give a tetrahedral molecular geometry.
C - 2 and C - 4 each have three electron groups around the atom. Three electron groups give a trigonal planar geometry; trigonal planar geometry has sp2 hybridization. Three bonding pairs and zero
lone groups give a trigonal planar molecular geometry.
N - 1 and N - 2 each have four electron groups around the atom. Four electron groups give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Three bonding groups
and one lone pair give a trigonal pyramidal molecular geometry.
O has four electron groups around the atom. Four electron groups give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Two bonding groups and two lone pairs
give a bent molecular geometry.
H
Chapter 10 Chemical Bonding II
(c)
413
cysteine
H
H
H
N
d
**
I
H
<L
CL
O
H
**
H
C -I and C - 3 each have four electron groups around the atom. Four electron pairs give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding pairs and
zero lone pairs give a tetrahedral molecular geometry.
C - 2 has three electron groups around the atom. Three electron groups give a trigonal planar geometry; trigonal planar geometry has sp2 hybridization. Three bonding groups and zero lone pairs give
a trigonal planar molecular geometry.
N has four electron groups around the atom. Four electron groups give a tetrahedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Three bonding groups and one lone pair
give a trigonal pyramidal molecular geometry.
O and S have four electron groups around the atom. Four electron groups give a tetrahedral electron
geometry; tetrahedral electron geometry has sp3 hybridization. Two bonding groups and two lone
pairs gives bent molecular geometry.
H
(a)
H
cytosine
H
N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral
electron geometry and sp3 hybridization. Three bonding pairs of electrons and one lone pair give a
trigonal pyramidal molecular geometry.
C - 2 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar molecular geometry.
N - 3 has two bonding groups of electrons and one lone pair; three electron groups give a trigonal
planar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bent
molecular geometry.
C - 4 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonal
planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
C - 5 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonal
planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
414
Chapter 10 Chemical Bonding II
C - 6 has three bonding pairs of electrons and zero lone pairs; three electron groups give a trigonal
planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
N outside the ring has three bonding groups of electrons and one lone pair; four electron groups give
a tetrahedral electron geometry and sp3 hybridization. Three bonding groups of electrons give a trigonal pyramidal molecular geometry.
(b)
adenine
NH2
(b)
N - 1 has two bonding groups of electrons and one lone pair; three electron groups give a trigonal
planar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bent
molecular geometry.
C - 2 has three bonding groups of electrons and zero lone groups; three electron groups give a trigonal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
N - 3 has two bonding groups of electrons and one lone pair; three electron groups give a trigonal
planar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bent
molecular geometry.
C - 4 has three bonding groups of electrons and zero lone groups; three electron groups give a trigonal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
C - 5 has three bonding groups of electrons and zero lone groups; three electron groups give a trigonal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
C - 6 has three bonding groups of electrons and zero lone groups; three electron groups give a trigonal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
N - 7 has two bonding groups of electrons and one lone pair; three electron groups give a trigonal
planar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bent
molecular geometry.
C - 8 has three bonding groups of electrons and zero lone groups; three electron groups give a trigonal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planar
molecular geometry.
N - 9 has three bonding groups of electrons and one lone pair; four electron groups give a tetrahedral
electron geometry and sp3 hybridization. Three bonding groups of electrons give a trigonal pyramidal molecular geometry.
N outside the ring has three bonding groups of electrons and one lone pair; four electron groups give
a tetrahedral electron geometry and sp3 hybridization. Three bonding groups of electrons give a trigonal pyramidal molecular geometry.
(c)
thymine
Chapter 10 Chemical Bonding II
415
N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral
electron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidal
molecular geometry.
C - 2 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
N - 3 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral electron geometry and sp3 hybridization. Three bonding pairs and one lone pair give a trigonal pyramidal
molecular geometry.
C - 4 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
C - 5 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
C - 6 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
C outside the ring has four bonding pairs of electrons and zero lone pairs; four electron pairs give
a tetrahedral electron geometry and sp3 hybridization. Four bonding pairs of electrons give a
tetrahedral molecular geometry.
(d)
guanine
543
9/
^lM
N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral
electron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidal
molecular geometry.
C - 2 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
N - 3 has two bonding pairs of electrons and one lone pair; three electron pairs give a trigonal planar
electron geometry and sp2 hybridization. Two bonding pairs and one lone pair give a bent molecular
geometry.
C - 4 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
C - 5 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
C - 6 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
N - 7 has two bonding pairs of electrons and one lone pair; three electron pairs give a trigonal planar
electron geometry and sp2 hybridization. Two bonding pairs and one lone pair give a bent molecular
geometry.
C - 8 has three bonding pair of electrons and zero lone pair; three electron pairs give a trigonal
planar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar
molecular geometry.
416
Chapter 10 Chemical Bonding H
N - 9 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral
electron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidal
molecular geometry.
N outside the ring has three bonding pairs of electrons and one lone pair; four electron pairs give a
tetrahedral electron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal
pyramidal molecular geometry.
10.87
4 TT bonds; 25 a bonds; the lone pair on the Os and N - 2 occupy sp2 orbitals; the lone pairs on N - 1, N - 3,
and N - 4 occupy sp3 orbitals.
caffeine
10.88
5 TT bonds; 21 a bonds
aspirin
'V
There is rotation around the bond from C -1 to the ring and from C -1 to OH bond. There is rotation around
the O - 2 to the ring bond and around the O - 2 to C - 2 bond. There is rotation around the C - 2 to C - 3 bond.
The C -1 to O - 1 bond is rigid, the ring structure is rigid, and the C - 2 to O - 3 bond is rigid.
10.89
(a)
Water soluble: The 4 C - OH bonds, the C = O bond, and the C - O bonds in the ring, make the molecule polar. Because of the large electronegativity difference between the C and O, each of the bonds
will have a dipole moment. The sum of the dipole moments does NOT give a net zero dipole moment,
so the molecule is polar. Since it is polar, it will be water soluble.
(b)
Fat soluble: There is only one C - O bond in the molecule. The dipole moment from this bond is not
enough to make the molecule polar because of all of the nonpolar components of the molecule. The
C - H bonds in the structure lead to a net dipole of zero for most of the sites in the molecule. Since
the molecule is nonpolar, it is fat soluble.
(c)
Water soluble: The carboxylic acid function (COOH group) along with the N atom in the ring make
the molecule polar. Because of the electronegativity difference between the C and O and the C and
N atoms, the bonds will have a dipole moment and the net dipole moment of the molecule is NOT
zero, so the molecule is polar. Since the molecule is polar, it is water soluble.
(d)
Fat soluble: The two O atoms in the structure contribute a very small amount to the net dipole
moment of this molecule. The majority of the molecule is nonpolar because there is no net dipole
moment at the interior C atoms. Because the molecule is nonpolar it is fat soluble.