Chapter 9
Chapter Nine
Theories of Bonding and Structure
Practice Exercises
9.1
(a)
(b)
Carbon tetrabromide, CBr4, should have a tetrahedral shape (Figure 9.4) because it has
four bonding electron pairs around the central atom.
Arsenic pentaflouride, AsF5, should have a trigonal bipyramidal shape (Figure 9.4)
because it has five bonding electron pairs around the central atom.
9.2
SeF6 should have an octahedral shape (Figure 9.4) because it has six bonding electron pairs
around the central atom.
9.3
SbCl5 should have a trigonal bipyramidal shape (Figure 9.4) because, like PCl5, it has five
bonding electron pairs around the central atom.
9.4
HArF should have a linear shape (Figure 9.7) because although it has five electron pairs around
the central Ar atom, only two are being used for bonding.
9.5
IBr2− should have a linear shape (Figure 9.7) because although it has five electron pairs around
the central I atom, only two are being used for bonding.
9.6
SO3 should have trigonal planar shape because it has three electron domains around the central S
atom. Two of the domains are double bonds.
9.7
SF4 is distorted tetrahedral and has one lone pair of electrons on the sulfur, therefore it is polar.
9.8
(a) TeF6 is octahedral, and it is not polar.
(b) SeO2 is bent, and it is polar.
(c) BrCl is polar because there is a difference in electronegativity between Br and Cl.
(d) AsH3, like NH3, is pyramidal, and it is polar.
(e) CF2Cl2 is polar, because there is a difference in electronegativity between F and Cl.
9.9
The H⎯Cl bond is formed by the overlap of the half–filled 1s atomic orbital of a H atom with the
half–filled 3p valence orbital of a Cl atom:
Cl atom in HCl (x = H electron):
x
3s
3p
The overlap that gives rise to the H⎯Cl bond is that of a 1s orbital of H with a 3p orbital of Cl:
9.10
The half–filled 1s atomic orbital of each H atom overlaps with a half–filled 3p atomic orbital of
the P atom, to give three P⎯H bonds. This should give a bond angle of 90°.
P atom in PH3 (x = H electron):
x
3s
x
x
3p
The orbital overlap that forms the P⎯H bond combines a 1s orbital of hydrogen with a 3p orbital
of phosphorus (note: only half of each p orbital is shown):
9-1
Chapter 9
z
y
x
9.11
BF3 uses sp2 hybridized orbitals since it has only three bonding electron pairs and no lone pairs of
electrons.
The sp2 hybrid orbitals on the B, x = F electron
x
x
x
2p
sp 2
9.12
BeF2 uses sp hybridized orbitals since it has only two bonding electron pairs and no lone pairs of
electrons.
The sp hybrid orbitals on the Be; x = F electron
x
x
2p
sp
9.13
SiH4 uses sp3 hybridized orbitals since it has four bonding electron pairs and no lone pairs of
electrons.
x
x
x
x
sp 3
9.14
Since there are five bonding pairs of electrons on the central phosphorous atom, we choose sp3d
hybridization for the P atom. Each of phosphorus's five sp3d hybrid orbitals overlaps with a 3p
atomic orbital of a chlorine atom to form a total of five PCl single bonds. Four of the 3d atomic
orbitals of P remain unhybridized.
9.15
VSEPR theory predicts that AsCl5 will be trigonal bipyramidal. Since there are five bonding pairs
of electrons on the central arsenic atom, we choose sp3d hybridization for the As atom as a
trigonal bipyramid. Each of arsenic's five sp3d hybrid orbitals overlaps with a 3p atomic orbital
of a chlorine atom to form a total of five As⎯Cl single bonds. Four of the 4d atomic orbitals of
As remain unhybridized.
9.16
Using VSEPR theory we find that XeF4 has eight valence electrons for the Xe and seven for each
F, so there are six electron domains around the Xe. To accommodate the six electron domains, the
Xe needs six hybrid orbitals, so the hybridization is sp3d2
F
F
Xe
F
F
PCl3 has four pair of electrons around the central atom, P, so the hybridization is sp3
BrCl3 has five pair of electrons around the central atom, Br, so the hybridization is sp3d
9.17
(a)
(b)
9.18
NH3 is sp3 hybridized. Three of the electron pairs are use for bonding with the three hydrogens.
The fourth pair of electrons is a lone pair of electrons. This pair of electrons is used for the
formation of the bond between the nitrogen of NH3 and the hydrogen ion, H+.
9.19
Since there are six bonding pairs of electrons on the central phosphorous atom, we choose sp3d2
hybridization for the P atom. Each of phosphorus’s six sp3d2 hybrid orbitals overlaps with a 3p
9-2
Chapter 9
atomic orbital of a chlorine atom to form a total of six P–Cl single bonds. Three of the 3d atomic
orbitals of P remain unhybridized.
P atom in PCl6– (x = Cl electron):
x
x
x
x
x
x
3d
sp3d 2
The ion is octahedral because six atoms and no lone pairs surround the central atom.
–
Cl
Cl
Cl
P
Cl
Cl
Cl
9.20
Atom 1has three electron domains: sp2
Atom 2 has four electron domains: sp3
Atom 3 has three electron domains: sp2
There are 10 σ bonds and 2 π bonds in the molecule.
9.21
Atom 1has two electron domains: sp
Atom 2 has three electron domains: sp2
Atom 3 has four electron domains: sp3
There are 9 σ bonds and 3 π bonds in the molecule.
9.22
CN– has 10 valence electrons and the MO diagram is similar to that of C and N. The bond order
of the ion is 3 and this does agree with the Lewis structure.
σ∗2pz
π∗ 2px
π∗ 2spy
σ2pz
π2px
π2py
σ∗2s
σ2s
9-3
Chapter 9
9.23
NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 9.1 for O2,
except that one fewer electron is employed at the highest energy level
σ∗2pz
π∗ 2px
π∗ 2spy
π2px
π2py
σ2pz
σ∗2s
σ2s
(8 bonding e ) − (3 antibonding e ) = 5
Bond order =
−
−
2
2
The bond order is calculated to be 5/2.
9.24
The hybridization of the N in NO3– is sp2. This leaves one extra p orbital on the nitrogen which is
available for π bonding. The three oxygen atoms are equivalent, and only one can form a π bond
at a time with the N, so there are three resonance structures over the three N⎯O bonds. The
atomic orbitals used in the delocalized molecular orbital are the p orbitals on the N and O that are
not involved in forming the σ bond framework.
9.25
For calcium, the valence band is formed by the 4s orbitals and the conduction band is formed by
the 3d orbitals.
9.26
Non-metals have the largest band gap, metals have the smallest, or non-existent band gap, and
semi-conductors have a band gap between the two. Magnesium is smaller than germanium which
is smaller than sulfur.
9.27
(a)
(b)
(c)
The carbon atoms in diamonds have a tetrahedral structure, so the hybridization is sp3.
The carbon atoms in graphite have a trigonal planar structure, so the hybridization is sp2.
The carbon atoms in buckyballs have a trigonal planar structure, so the hybridization is
sp2.
Review Questions
9.1
(a)
(b)
(c)
See Figure 9.4. The angles are 120°.
See Figures 9.4 and 9.5 The bond angles are 109.5°.
See Figures 9.4 and 9.10. The bond angles are 90°.
9.2
(a)
(b)
See Figure 9.4. The bond angle is 180°.
See Figures 9.4 and 9.7. The bond angles are 120° between the equatorial bonds and
180° between the two axial bonds.
9.3
VSEPR Theory is based on the principle that adjacent electron pairs repel one another, and that
these destabilizing repulsions are reduced to a minimum when electron pairs stay as far apart as
possible.
9.4
An electron domain is a region in space where electrons can be found. Nonbonding electrons and
double bonds are considered electron domains and influence the structure of the molecule.
9-4
Chapter 9
9.5
In HCHO, there are three bonding domains around the C, one bonding domain around each H,
and one bonding domain around O and two nonbonding domains around O.
9.6
(a)
90° bond angles
(b)
120° bond angle between equatorial bonds and 180° bond angle between axial bonds
(c)
90° bond angles between equatorial bonds and 90° bond angles between equatorial bonds
and axial bond
9.7
(a)
(b)
(c)
(d)
Planar triangular, otherwise known as trigonal planar
Octahedral
Tetrahedral
Trigonal bipyramidal
9.8
Polar molecules attract one another, and that influences the physical and chemical properties of
substances.
9.9
A bond's dipole moment is depicted with an arrow having a + sign on one end, where the "barb"
of the arrow is taken to represent the location of the opposing negative charge of the dipole:
9.10
A molecule will be polar if the bonds are polar and these polar bonds are arranged in such a
geometry that the dipoles of the bonds do not cancel one another.
9.11
A molecule having polar bonds will be nonpolar only if the bond dipoles are arranged so as to
cancel one another's effect.
9.12
O
O
S
The individual bond dipoles do not cancel one another.
9.13
Orbital overlap occurs when orbitals from different atoms share the same space. This overlap
provides a more stable region for the electrons, which find themselves under the stabilizing
influence of the positive charge of two nuclei.
9.14
The greater the orbital overlap, the greater the bond energy.
9.15
This is the same as the HF molecule, shown in Figure 9.16. A half–filled valence p orbital of Br
overlaps with the half–filled 1s orbital of the hydrogen atom.
9.16
Hybrid orbitals provide better overlap than do atomic orbitals, and this results in stronger bonds.
9.17
These are shown in Figures 9.21.
9-5
Chapter 9
sp3d
(b)
sp3d2
9.18
(a)
9.19
Elements in period 2 do not have a d subshell in the valence level.
9.20
The Lewis structure shows the connectivity of the atoms and the lone pairs of the electrons on the
molecule. This information can be used by VSPEPR theory to determine the structure of the
molecule. From the VSEPR structure, the hybridization of the central atom can be predicted.
9.21
The VSEPR model gives the shape of the electron domains around the central atom; this
information is used to predict the hybridization of the atom.
9.22
90°
9.23
sp3 C atom
sp3 N atom
sp3 O atom
There are zero, one and two lone pairs, respectively, on these atoms when sp3 hybridization is
utilized to form bonds.
9.24
(a)
H
c
H
H
H
(b)
N
H
H
H
(c)
O
H
H
9.25
This angle would have had to be 90°, the angle between one atomic p orbital and another.
9-6
Chapter 9
9.26
(a)
See Figure 9.25.
Valence shell for boron
2s
(b)
2p
See Figure 9.32
Valence shell for carbon
2s
2
sp hybridized boron atom
sp2
2p
2
sp hybridized carbon atom
sp2
2p
3
2p
3 2
9.27
sp d – trigonal bipyramid
9.28
The ammonium ion has a tetrahedral geometry, with bond angles of 109.5°.
+
H
H
N
sp d – octahedral
H
H
9.29
The geometry of the boron BCl3 is trigonal planar; after the addition of the water molecule, it
becomes tetrahedral. The hybridization changes from sp2 to sp3. The geometry of the oxygen in
the water molecule changes from bent to trigonal pyramidal after coordination to the boron atom.
Its hybridization does not change, but remains at sp3.
9.30
σ bond – The electron density is concentrated along an imaginary straight line joining the nuclei
of the bonded atoms.
π bond – The electron density lies above and below an imaginary straight line joining the bonded
nuclei.
9.31
The characteristic side–to–side overlap of p atomic orbitals that characterizes π bonds is
destroyed upon rotation about the bond axis. This is not the case for a σ bond, because regardless
of rotation, a σ bond is still effective at overlap.
9.32
See Figure 9.32
9.33
See Figure 9.35.
9.34
Two resonance hybrid structures need to be drawn.
sp2 hybrid orbitals are used for each carbon atom. Consequently, each bond angle is 120°. See
Figure 9.45.
9.35
If an electron is forced to occupy the higher energy MO, the molecule loses stability and the bond
is made weaker than if an electron, or a pair of electrons, occupies the lower energy (bonding)
MO.
Bonding MO
Antibonding MO
They are called antibonding orbitals because there is a nodal plane between the nuclei.
9-7
Chapter 9
9.36
The overlap will create a bonding MO.
9.37
There are two regions of overlap between the two 3d orbitals, above and below the internuclear
axis, so this forms a π orbital.
9.38
In the hypothetical molecule He2, both the bonding and the antibonding MO are doubly occupied,
and the net bond order is zero, as shown in Figure 9.39. H2 has a bond order of 1
9.39
As shown in Table 9.1, the two highest energy electrons in dioxygen occupy the doubly
degenerate π–antibonding level:
↑
↑
π*2p x
π*2p y
Since their spins are unpaired, the molecule is paramagnetic.
9.40
As shown in Table 9.1, the bond order of Li2 is 1.0. The bond order of Be2 would be zero. Yes,
Be2+ could exist since the bond order would be 1/2.
9.41
As bond order increases, bond energy (and strength) increases.
9.42
See Figure 9.40.
9.43
Both VB and MO theory have wave mechanics as their theoretical basis. In each theory, bonds
are considered to arise from the overlap of orbitals. Valence bond theory uses the overlap of
atomic orbitals while molecular orbital theory uses the molecular orbitals to describe the bonding.
9.44
Lewis structures do not explain how atoms share electrons, nor do they explain why molecules
adopt particular shapes. Also, electrons are known to be delocalized, a fact which only MO
theory addresses from the beginning. Also, odd-electron systems can be more effectively
discussed with MO theory.
9.45
VB theory views atoms coming together with their orbitals already containing specific electrons.
The bonds that are formed according to VB theory do so by the overlap of orbitals on neighboring
atoms, and this is accompanied by the pairing (sharing) of the electrons that are contained in the
orbitals.
MO theory, on the other hand, considers a molecule to be a collection of positive nuclei
surrounded by a set of molecular orbitals, which, by definition, belong to the molecule as a
whole, rather than to any specific atom. The electrons of the molecule are distributed among the
molecular orbitals according to the same rules that govern the filling of atomic orbitals.
9.46
A delocalized MO is one that extends over more than two nuclei. Benzene is planar and each
carbon atom has a half-filled p orbital perpendicular to the plane of the molecule. These orbitals
overlap to form a π bonding system throughout the molecule.
9.47
Delocalization increases stability.
9.48
Delocalization energy is a term used in MO theory to mean essentially the same thing as the term
resonance energy, which derives from VB theory. They both represent the additional stability
associated with a spreading out of electron density.
9-8
Chapter 9
9.49
A conduction band consists of any band of atomic energy levels that is continuous throughout the
solid and empty or partially filled with electrons.
A valence band are very closely spaced energy levels of valence electrons, particularly in metals.
9.50
Conduction bands require that the orbitals be half filled or empty so that electrons may move
through them.
9.51
For conductors, metals, the valence band and the conduction band are the same so electrons can
easily move throughout the metal.
For insulators, non-metals, the valence band is filled and the gap between the valence band and
conduction band is very large so electrons cannot move into the conduction band.
For semiconductors, the valence band is filled but the separation between the valence band and
the conduction band is small so electrons can easily be promoted into the conduction.
9.52
Electrical conductivity increases in semiconductors with increasing temperature because the
increasing temperature causes the electrons to move more rapidly, and have enough energy to
move through the conduction band.
9.53
The 2s band is completely filled and much lower in energy than the conduction band. It is not
possible to excite electrons from the 2s band into the conduction band. Calcium’s valence band is
formed from the 4s electrons and the conduction band is formed from the 3d orbitals which
overlap with the valence band, so the electrons can move through the conduction band.
9.54
An allotrope is a different structure or physical form of an element. An isotope is a form of an
atom that differs from other atoms of the same element in the number of neutrons in the nucleus.
An example is H and D, two of the isotopes of hydrogen.
9.55
Period 2 elements form strong π bonds because the atom size is small which enables atoms to
approach each other very closely. Period 3 elements (and all other elements) are too large to
accommodate the close proximity required to form strong π bonds. Because the period 3 and
higher elements are large and form only weak π bonds, it is much more effective to form only σ
bonds than to form a σ bond and a π bond.
9.56
Because the halogens need only a single electron to complete their valence shell, these elements
do not require the formation of π bonds. The formation of a single σ bond creates an extremely
stable diatomic molecule.
9.57
Of the nonmetals, only the noble gases exist in nature as isolated atoms.
9.58
Diamond is a network covalent solid in which each carbon atom is the corner of a tetrahedron.
Each carbon atom is bonded to four other carbon atoms using sp3 hybrid orbitals.
9.59
Graphene is a monolayer of graphite. The carbon atoms are sp2 hybridized and the π-network is
formed by the overlap of neighboring p orbitals to form a π-type bonding throughout the layer.
Carbon forms sheets of sp2 – sp2 sigma (σ) bonds.
9.60
The weak π bonds that exist between the sheets of sp2 bonded carbon are easily broken. The
lubricating properties of graphite are a result of the weak attraction between parallel sheets.
9.61
C60, or buckminsterfullerene, resembles a soccer ball. It consists of hexagonal and pentagonal
arrangements of carbon atoms. This arrangement naturally assumes a spherical geometry. The
molecule is named in honor of R. Buckminster Fuller, an architect. Other roughly spherical
arrangements of carbon atoms are also possible. While not perfectly spherical, they resemble
buckminsterfullerene and are given the general name of fullerenes.
9-9
Chapter 9
9.62
Carbon nanotubes have the same molecular structure as graphite, but instead of forming sheets,
the layer of carbon curves and forms a tube.
9.63
Silicon has a molecular structure identical to diamond, i.e., a network covalent solid arranged in a
tetrahedral manner. Silicon, unlike carbon, does not form multiple bonds. Consequently, there is
no graphite-like allotrope of silicon.
9.64
White phosphorous, P4, is a tetrahedron. See Figure 9.52.
9.65
The different allotropes of phosphorous are white phosphorous, red phosphorous, and black
phosphorous. White phosphorous is P4 tetrahedra, see Figure 9.52. Red phosphorous are chains
of P4 tetrahedra, see Figure 9.53. Black phosphorous is a layered structure in which each
phosphorous atom is covalently bonded to three others in the same layer.
9.66
The bond angle in P4 is 60°. Using only p orbitals, a bond angle of 90° is expected. Because the
observed bond angle is much less, the molecule is extremely reactive.
9.67
Red phosphorus consists of tetrahedra joined as in Figure 9.53. While it too is used in explosives,
it is much less reactive than white phosphorus.
9.68
Black phosphorus, like graphite, exists as parallel sheets of atoms.
9.69
The allotropes of oxygen are dioxygen, O2, and ozone, O3.
9.70
O
O
O
The molecule is nonlinear and uses sp2 hybrid orbitals. There is a lone pair of electrons on the
central oxygen, consequently, ozone is indeed a polar molecule. Additionally, the formal charges
suggest that O3 is polar.
9.71
9.72
The presence of ozone in the upper atmosphere shields the earth from harmful ultraviolet
radiation.
S8
Review Problems
9.73
9.74
(a)
(b)
(c)
(d)
(e)
Bent (central N atom has two single bonds and two lone pairs)
Planar triangular (central C atom has three bonding domains—a double bond and two
single bonds)
T-shaped (central I atom has three single bonds and two lone pairs)
Linear (central Br atom has two single bonds and three lone pairs)
Planar triangular (central Ga atom has three single bonds and no lone pairs)
(a)
(b)
(c)
(d)
(e)
Trigonal pyramidal
Tetrahedral
Tetrahedral
Nonlinear (bent)
Linear
9-10
Chapter 9
9.75
(a)
(b)
(c)
(d)
(e)
Bent (central F atom has four electron domains; two are lone pair)
Trigonal bipyramidal (central As atom has five electron domains)
Trigonal pyramidal (central As atom has four electron domains; one is a lone pair)
Trigonal pyramidal (central Sb atom has four electron domains; one is a lone lair)
Bent (central Se atom has four electron domains; two are lone pair)
9.76
(a)
(b)
(c)
(d)
(e)
Distorted tetrahedral
Octahedral
Nonlinear, bent
Tetrahedral
Tetrahedral
9.77
(a)
(b)
(c)
(d)
(e)
Tetrahedral (central I atom has four electron domains)
Square planar (central I atom has six electron domains; two are lone pair)
Octahedral (central Te atom has six electron domains)
Tetrahedral (central Si atom has four electron domains)
Linear (central I atom has five electron domains; three are lone pair)
9.78
(a)
(b)
(c)
(d)
(e)
Linear
Square planar
T–shaped
Trigonal pyramidal
Planar triangular
9.79
BrF4+
The central Br atom has five electron domains, one is a lone pair
9.80
PF3
9.81
All angles are 120°, since there are three electron domains around each carbon atom.
9.82
180°
There are two electron domains around each carbon atom so the molecule is linear
9.83
(a)
(c)
(e)
109.5°
120°
120°
(b)
(d)
109.5°
109.5°
9.84
(a)
(c)
(e)
180°
180°
109.5°
(b)
(d)
109.5°
109.5°
9.85
The ones that are polar are (a), (b), and (c). The last two have symmetrical structures, and
although individual bonds in these substances are polar bonds, the geometry of the bonds serves
to cause the individual dipole moments of the various bonds to cancel one another.
9.86
Two of these substances have planar triangular structures that are not polar because the individual
bond dipole moments cancel one another: SO3 and BCl3. Two of these molecules have pyramidal
structures, and are, therefore, polar: PBr3 and AsCl3. ClF3 is T–shaped and, therefore, polar.
9.87
All are polar. (a), (b), (c) and (e) have asymmetrical structures, and (d) only has one bond, which
is polar.
9.88
H2S is bent and, therefore, polar. BeH2 is nonpolar since it is linear. SCN– and CN– are linear, but
have dipoles that do not cancel out. BrCl3 is T-shaped and, therefore, polar.
9.89
In SF6, although the individual bonds in this substance are polar bonds, the geometry of the bonds
is symmetrical which serves to cause the individual dipole moments of the various bonds to
cancel one another. In SF5Br, one of the six bonds has a different polarity so the individual
dipole moments of the various bonds do not cancel one another.
9-11
Chapter 9
9.90
In CCl4, although the individual bonds in this substance are polar bonds, the geometry of the
bonds is symmetrical which serves to cause the individual dipole moments of the various bonds
to cancel one another. In CH3Cl, one of the four bonds has a different polarity so the individual
dipole moments of the various bonds do not cancel one another.
9.91
This is shown in Figure 9.18 for F2 . Since Cl2 is in the same family the bonding will be similar,
using the n = 3 shell rather than the n = 2 shell. We can diagram it showing the orbitals of one of
the chlorine atoms:
Each Cl atom (x = an electron from the other Cl atom):
9.92
The 1s atomic orbitals of the hydrogen atoms overlap with the mutually perpendicular p atomic
orbitals of the selenium atom.
Se atom in H2Se (x = H electrons):
x
4s
9.93
x
4p
Atomic Be:
2s
2p
Hybridized Be: (x = a Cl electron)
x
x
2p
sp
Beryllium uses sp hybrid orbitals.
9.94
Sn atom in SnCl4 (x = a Cl electron)
x
x
x
x
sp 3
Tin uses sp3 hybrid orbitals.
9.95
Sb atom in SbCl5 (x = a Cl electron)
Antimony uses sp3d hybrid orbitals.
9.96
Te atom in TeF6 (x = an F electron)
x
x
x
x
x
x
sp3d 2
Tellurium uses sp3d2 hybrid orbitals.
9-12
4d
Chapter 9
9.97
(a)
There are three bonds to the central Cl atom, plus one lone pair of electrons. The
geometry of the electron pairs is tetrahedral so the Cl atom is to be sp3 hybridized:
_
O
O
Cl
O
(b)
There are three atoms bonded to the central sulfur atom, and no lone pairs on the central
sulfur. The geometry of the electron pairs is that of a planar triangle, and the
hybridization of the S atom is sp2:
O
S
O
(c)
O
Two other resonance structures should also be drawn for SO3.
There are two bonds to the central O atom, as well as two lone pairs. The O atom is to be
sp3 hybridized, and the geometry of the electron pairs is tetrahedral.
O
F
9.98
(a)
F
Six Cl atoms surround the central Sb atom in an octahedral geometry, and the
hybridization of Sb is sp3d2.
_
Cl
Cl
Cl
Sb
Cl
Cl
Cl
(b)
Three F atoms are bonded to the P atom, plus the P atom has a lone pairs of electrons.
This requires the P atom to be sp3 hybridized, and the geometry of the electron pairs is
tetrahedral.
(c)
The central Xe atom is bonded to four F atoms, plus it has two lone pairs of electrons.
The geometry of the electron pairs is trigonal bipyramidal. This requires sp3d2
hybridization of Xe.
9-13
Chapter 9
F
F
Xe
F
F
9.99
(a)
There are three bonds to As and one lone pair at As, requiring As to be sp3 hybridized.
The Lewis diagram
Cl
As
Cl
Cl
The hybrid orbital diagram for As: (x = a Cl electron)
x
sp
(b)
x
x
3
There are three atoms bonded to the central Cl atom, and it also has two lone pairs of
electrons. The hybridization of Cl is thus sp3d.
The Lewis diagram
F
Cl
F
F
The hybrid orbital diagram for Cl: (x = a F electron)
x
x
x
3d
sp3d
9.100
(a)
Sb has three bonds to Cl atoms and one lone pair and is therefore sp3 hybridized.
The Lewis diagram
Cl
Sb
Cl
Cl
The hybrid orbital diagram for Sb: (x = a Cl electron)
x
x
x
sp3
(b)
Se has two bonds to Cl atoms and two lone pairs of electrons, requiring it to be sp3
hybridized.
The Lewis diagram
Se
Cl
Cl
9-14
Chapter 9
The hybrid orbital diagram for Se: (x = a Cl electron)
x
x
sp3
9.101
We can consider that this ion is formed by reaction of SbF5 with F–. The antimony atom accepts
a pair of electrons from fluoride:
Sb in SbF6–: (xx = an electron pair from the donor F–)
xx
sp 3d 2
9.102
This is an octahedral ion with sp3d2 hybridized tin:
Sn atom in SnCl62– (xx = an electron pair from the donor Cl–)
x
x
x
x xx xx
3d
sp 3d 2
9.103
(a)
N in the C=N system:
2p
sp2
2p
sp2
sp2
sp2
(b)
sigma bond
pi bond
C
N
C
(c)
H
120 o
H
9.104
(a)
120 o
C
N
120o
120 o H
sp hybridized N atom:
sp
2p
9-15
N
Chapter 9
The σ bond:
(b)
C
N
The π bonds:
C
N
For HCN, the only difference from (b) is the formation of another σ bond when an H 1s
orbital overlaps the C sp hybrid orbital.
The HCN bond angle should be 180°.
(c)
(d)
9.105
Each carbon atom is sp2 hybridized, and each C–Cl bond is formed by the overlap of an sp2
hybrid of carbon with a p atomic orbital of a chlorine atom. The C=C double bond consists first
of a C–C σ bond formed by "head on" overlap of sp2 hybrids from each C atom. Secondly, the
C=C double bond consists of a side–to–side overlap of unhybridized p orbitals of each C atom, to
give one π bond. The molecule is planar, and the expected bond angles are all 120°.
9.106
The bonding in phosgene is the same as that diagrammed for H2CO in Figure 9.34, except that H
atom 1s orbitals are replaced by Cl atom to form sp2⎯ p σ bonding.
9.107
1. sp3
2. sp
3. sp2
4. sp2
9.108
1. sp3
2. sp2
3. sp3
4. sp3
9.109
1.
2.
3.
4.
One σ bond
One σ bond and two π bonds
One σ bond
One σ bond and one π bond
9.110
1.
2.
3.
4.
One σ bond
One σ bond and one π bonds
One σ bond and one π bond
One σ bond
9-16
Chapter 9
9.111
The two unpaired electrons are located in the π antibonding molecular orbitals.
σ2pz*
π2px*, π2py*
π2px, π2py
σ2pz
σ2s*
σ2s
The net bond order is (8 Bonding e– – 4 Antibonding e–)/2 = 2.
9.112
σ2pz*
π2px*, π2py*
σ2pz
π2px, π2py
σ2s*
σ2s
The net bond order is (8 Bonding e– –2 Antibonding e–)/2 = 3.
9.113
(a)
(b)
(c)
(i)
O2: BO = 2
(i) O2
(i) O2+
O2+: BO = 2.5 (ii)
(ii) O2–
(ii) O2
O2: BO = 2
O2–: BO = 1.5
9.114
(a)
(b)
(c)
(i)
C2: BO = 2
(i) C2+
(i) C2
C2+: BO = 1.5 (ii)
(ii) N2+
(ii) N2
N2: BO = 3
N2+: BO = 2.5
9.115
NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 9.1 for O2,
except that one fewer electron is employed at the highest energy level
The bond order for NO is calculated to be 5/2:
9-17
Chapter 9
(8 bonding e ) − (3 antibonding e ) = 5
Bond Order =
−
−
2
2
The bond order for NO+ is calculated to be 3, since one electron is removed from an antibonding
orbital:
(8 bonding e ) – (2 antibonding e ) = 3
Bond order =
−
−
2
If we remove one electron to form NO+, the bond order becomes 3 (there are only two
antibonding electrons). The larger bond order indicates a shorter bond length. The bond energy
of NO is lower than the bond energy of NO+.
9.116
NO has 11 valence electrons, and the MO diagram is similar to that shown in Table 9.1 for O2,
except that one fewer electron is employed at the highest energy level
The bond order for NO is calculated to be 5/2, see Review Problem 9.115.
The bond order for NO– is calculated to be 2.0, since one electron is added to an antibonding
orbital:
(8 bonding e ) – (4 antibonding e ) = 2
Bond order =
−
−
2
The smaller bond order indicates a longer bond length. The bond energy of NO is higher than the
bond energy of NO–.
9.117
All of the molecules or ions are paramagnetic except N2. O2+, O2, O2–, and NO are paramagnetic
9.118
B2 and C2– are paramagnetic.
Additional Exercises
9.119
σ∗s
1s
2p
2p
σs
2s
Hydrogen
2s
Oxygen
There are two electrons in bonding MOs and three electrons in nonbonding MOs.
The net bond order is 1.
9-18
Chapter 9
9.120
σ∗p
B
N
π∗ p
2p
2p
σp
πp
σ∗s
2s
2s
σs
The molecule would be diamagnetic and the net bond order in the molecule would be 2.0.
9.121
Planar triangular
9.122
Pyramidal structures have one lone pair electrons, and therefore five valence electrons: Group 5A
X
Cl
Cl
Cl
Trigonal planar structures have no lone pairs of electrons, and therefore three valence electrons:
Group 3A
Cl
X
Cl
Cl
T-shaped molecules have two lone pairs of electrons and three bonded pairs, and therefore seven
valence electrons: Group 7A
Cl
X
Cl
Cl
It is unlikely that X would be in Group 6A because it would have an unpaired electron since X
would have six valence electrons and three would be used for bonding with the Cl, and that would
leave three remaining electrons.
9.123
Since the angles are nearer to 90° than to 109°, we must use atomic rather than hybrid orbitals on
antimony.
9.124
(a)
(b)
(c)
(d)
(e)
Changes in molecular geometry
planar triangular to tetrahedral
trigonal bipyramidal to octahedral
T-shaped to square planar
trigonal pyramidal to trigonal bipyramidal
linear to planar
9-19
Changes in hybridization
sp2 to sp3
sp3d to sp3d2
sp3d to sp3d2
sp3 to sp3d
sp to sp2
Chapter 9
9.125
Structure (d)
9.126
The normal C–C–C angle for an sp3 hybridized carbon atom is 109.5°. The 60° bond angle in
cyclopropane is much less than this optimum bond angle. This means that the bonding within the
ring cannot be accomplished through the desirable "head on" overlap of hybrid orbitals from each
C atom. As a result, the overlap of the hybrid orbitals in cyclopropane is less effective than that
in the more normal, noncyclic propane molecule, and this makes the C–C bonds in cyclopropane
comparatively weaker than those in the noncyclic molecule. We can also say that there is a
severe "ring strain" in the molecule.
9.127
(a)
(b)
(c)
PF3 is a pyramidal molecule and uses sp3 hybrid orbitals. The expected bond angle is
109.5°.
Using the unhybridized p orbitals, we would anticipate a bond angle of 90°.
The observed bond angle is almost exactly the average of the bond angles listed in parts
(a) and (b) above. So, neither hybrid orbitals nor unhybridized atomic orbitals explain the
observed bond angle.
9.128
The arrangement around an sp2 hybridized atom may have 120° bond angles (which well
accommodates a six–membered ring, the geometrical arrangement around an atom that is sp
hybridized must be linear, as in C – C ≡ C – C systems. The orbitals forming the second π bond
in benzyne are not parallel as in acetylene, but they lie at 120° to the bond axes.
9.129
The arrangement of the atoms is trigonal bipyramidal.
Recall that the bond angle between equatorial atoms is 120°. The bond angle from the equatorial
position to the axial position is 90°. Due to the smaller bond angles, the atoms in the axial
positions create more repulsion. The structure with the least amount of total repulsion is
preferred; the statement implies that the more electronegative atoms create less repulsion;
therefore the more electronegative atom should be placed in the axial position. Since fluorine is
more electronegative than chlorine, the F atoms will be in the axial positions and the Cl atoms
will be in the equatorial positions. The molecule is non–polar.
F
Cl
P
Cl
Cl
F
9.130
The lone pair of electrons repels the Te–F bonding pairs, causing the F–Te–F angles to be smaller
than in the ideal trigonal bipyramid.
F
This angle less than 180 o
F
Te
This angle less than 120o
F
F
In BrF5, the lone pair is located perpendicular to the plane made up of four of the F atoms, giving
a square pyramid geometry to the molecule. The angle between the four F in the plane and the F
above the plane will be less than 90° due to lone pair repulsion. The F-Br-F angles in the plan
will remain 90°.
9-20
Chapter 9
9.131
The double bonds are predicted to be between S and O atoms. Hence, the Cl⎯S⎯Cl angle
diminishes under the influence of the S=O double bonds.
O
S
Cl
strong repulsion
O
Cl
weak repulsion
9.132
This is a π bond, since overlap is side to side rather than the end to end. Also, consider that no
bond rotation is possible here without breaking the bond since overlap occurs both above and
below the bond axis.
This diagram shows bonding combination
–
+ +
+
– –
and this one shows the antibonding combination
9.133
–
+ –
+
– +
Only the px orbital can form a π bond with dxz, if the internuclear axis is the z axis.
Multi-Concept Problems
9.134
(a)
(b)
(c)
The O–O–N bond angle is about 109.5° and the O–N–O bond angle is around 120°.
sp2
The nitrate ion benefits from three stable resonance structures (or electron delocalization)
which stabilize it. The peroxynitrite ion does not.
9.135
The polarity in the N–H bond, as measured by the difference in electronegativities between N
(3.1) and H (2.1) adds to the polarity created by the sp3 hybridized lone pair on the N atom.
However, the higher electronegativity of fluorine (4.1) causes the N–F bond polarity to oppose
the polarity associated with the hybridized lone pair on the N atom. Thus in the first case, a net
dipole exists in the molecule, whereas in the second case, the polarity of the N–F bonds cancel
the polarity of the hybrid lone pair.
9.136
(a)
The C–C single bonds are formed from head–to–head overlap of C atom sp2 hybrids.
This leaves one unhybridized atomic p orbital on each carbon atom, and each such atomic
orbital is oriented perpendicular to the plane of the molecule.
9-21
Chapter 9
(b)
(d)
9.137
Sideways or π type overlap is expected between the first and the second carbon atoms, as
well as between the third and the fourth carbon atoms. However, since all of these
atomic p orbitals are properly aligned, there can be continuous π type overlap between all
four carbon atoms.
The situation described in part (b) is delocalized. We expect completely delocalized π
type bonding among the carbon atoms. Therefore, the center C-C bond has some π
character and is thus shorter than a normal C-C single bond.
The compound is of the form PnBrm. From the mass of AgBr formed we can determine the mass
of Br in the original compound. The mass of Br will also allow the determination of the number
of moles of Br in the phosphorus compound. Then, using the original mass of the phosphorus
compound, we can determine the mass and moles of P. Using this information, we can determine
the empirical formula of the compound.
1mol AgBr
1 mol Br
×
= 2.705 × 10−3 mol Br
187.77 g
mol AgBr
Mass of Br = 2.705 × 10−3 mol Br × 79.904 g = 0.216 g Br
mol
0.508 g AgBr ×
Mass of P = 0.244 g PmBrn – 0.216 g Br = 0.028 g P
Moles of P = 0.028 g P × 1 mol P = 9.04× 10–4
30.97 g
Empirical formula:
P0.000904 Br0.002705 or PBr3
Since P has five valence electrons, forming PBr3 results in three P⎯Br bonds and a lone-pair of
electrons on P. The geometry of the compound is trigonal pyramid and the hybridization is sp3.
Since the P⎯Br bonds are polar the geometry of PBr3 results in a polar molecule.
9-22