Chapter 9
Practice Exercises
9.1
SeF6 should have an octahedral shape (Figure 9.4) because is has six electron pairs around the central atom.
9.2
SbCl5 should have a trigonal bipyramidal shape (Figure 9.4) because, like PCl5, it has five electron pairs
around the central atom.
9.3
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.
9.4
I3− 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.
9.5
XeF4 should have a square planar shape (Figure 9.10) because although it has six electron pairs around the
central Xe atom, only four are being used for bonding.
9.6
In SO32–, there are three bond pairs and one lone pair of electrons at the sulfur atom, and as shown in Figure
9.5, this ion has a trigonal pyramidal shape.
In CO32–, there are three bonding domains (2 single bonds and 1 double bond) at the carbon atom, and as
shown in Figure 9.4, this ion has a planar triangular shape.
In XeO4, there are four bond pairs of electrons around the Xe atom, and as shown in Figure 9.5, this
molecule is tetrahedral.
In OF2, there are two bond pairs and two lone pairs of electrons around the oxygen atom, and as shown in
Figure 9.5, this molecule is bent.
9.7
SF4 is distorted tetrahedral and has one lone pair of electrons on the sulfur, therefore it is polar.
9.8
a) SF6 is octahedral, and it is not polar.
b) SO2 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
183
Chapter 9
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):
z
y
x
9.11
BCl3 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 = Cl 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
Since there are five bonding pairs of electrons on the central phosphorous atom, we choose sp3d
hybridization for the P atom. Each of phosphorous’s five sp3d hybrid orbitals overlaps with a 3p atomic
orbital of a chlorine atom to form a total of five P–Cl single bonds. Four of the 3d atomic orbitals of P
remain unhybridized.
9.14
sp3
x
x
x
x
sp 3
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
sp3d2
9.17
(a)
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 phosphorous’s six sp3d2 hybrid orbitals overlaps with a 3p 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):
sp3
x
sp3d
(b)
x
x
x
x
x
3d
sp3 d 2
The ion is octahedral because six atoms and no lone pairs surround the central atom.
184
Chapter 9
9.20
atom 1: sp2
atom 2: sp3
atom 3: sp2
There are 10 σ bonds and 2 π bonds in the molecule.
9.21
atom 1: sp
atom 2: sp2
atom 3: sp3
There are 9 σ bonds and 3 π bonds in the molecule.
9.22
CN– has 10 valence electons 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
9.23
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
The bond order is calculated to be 5/2:
2
Review Questions
9.1
(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.
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)
(c)
9.3
An electron domain is a region in space where electrons can be found.
185
Chapter 9
9.4
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.
9.5
(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
9.6
(a)
(b)
(c)
(d)
Planar triangular, otherwise known as trigonal planar
Octahedral
Tetrahedral
Trigonal bipyramidal
9.7
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.
9.8
Polar molecules attract one another, and that influences the physical and chemical properties of substances.
9.9
Dissymmetric molecules are polar only if there is a difference is the electronegativity of the atoms.
9.10
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:
9.11
O
O
S
The individual bond dipoles do not cancel one another.
9.12
A molecule having polar bonds will be nonpolar only if the bond dipoles are arranged so as to cancel one
another's effect.
9.13
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.
9.14
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.
186
Chapter 9
9.15
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.
9.16
According to VB theory, spin–paired electrons are shared between atoms via the overlap of orbitals from
the different atoms.
9.17
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.
9.18
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.
9.19
Hybrid orbitals provide better overlap than do atomic orbitals, and this results in stronger bonds.
9.20
These are shown in Figures 9.19.
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
(a)
H
c
H
H
H
(b)
N
H
H
H
(c)
O
H
H
9.23
Elements in period 2 do not have a d subshell in the valence level.
9.24
sp3d – trigonal bipyramid
9.25
This angle would have had to be 90°, the angle between one atomic p orbital and another.
sp3d2 – octahedral
187
Chapter 9
9.26
See Figure 9.15.
Valence shell for boron
2s
Valence shell for carbon
2p
2s
sp2 hybridized boron atom
sp2
9.27
90°
9.28
sp3 C atom
2p
sp2 hybridized carbon atom
sp2
2p
2p
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.
9.29
Lewis structures are something like shorthand representations of VB descriptions of molecules.
9.30
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.
9.31
The ammonium ion has a tetrahedral geometry, with bond angles of 109.5°.
+
H
H
9.32
9.33
9.34
N
H
H
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.
See Figure 9.30.
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.
188
Chapter 9
9.35
Two resonance hybrids should be drawn.
sp2 hybrid orbitals are used for each carbon atom. Consequently, each bond angle is 120°. See Figure 9.42.
9.36
See Figure 9.33.
9.37
See Figures 9.35 and 9.36. It is important to note that, relative to the atomic orbitals from which they are
formed, the bonding orbital is lower in energy and the antibonding orbital is higher in energy.
9.38
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.
9.39
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 9.36. See Figure 9.36 for the MO diagram for H2.
9.40
See Figure 9.37.
9.41
(a)
9.42
As shown in Table 9.1, the highest energy electrons in dioxygen occupy the doubly degenerate π–
antibonding level:
2.5
(b)
↑
↑
π*2p x
π*2p y
1.5
(c)
1.5
Since their spins are unpaired, the molecule is paramagnetic.
9.43
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.
9.44
As bond order increases, bond energy (and strength) increases.
9.45
A delocalized MO is one that extends over more than two nuclei.
9.46
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.
9.47
9.48
Delocalization increases stability.
9.49
MO theory avoids the need in VB theory for the cumbersome and numerous resonance forms.
189
Chapter 9
Review Problems
9.50
(a)
(c)
(e)
tetrahedral
octahedral
linear
(b)
(d)
square planar
tetrahedral
9.51
(a)
(c)
(e)
linear
T–shaped
planar triangular
(b)
(d)
square planar
trigonal pyramidal
9.52
(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 a single bond)
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)
9.53
(a)
(c)
(e)
trigonal pyramidal
tetrahedral
linear
(b)
(d)
planar triangular
nonlinear (bent)
9.54
BrF4+
9.55
PF3
9.56
(a)
(c)
(e)
nonlinear
trigonal pyramidal
nonlinear
(b)
(d)
trigonal bipyramidal
trigonal pyramidal
9.57
(a)
(c)
(e)
distorted tetrahedral
nonlinear
tetrahedral
(b)
(d)
octahedral
tetrahedral
9.58
(a)
(c)
(e)
109.5°
120°
109.5°
(b)
(d)
109.5°
180°
9.59
(a)
(c)
(e)
109.5°
180°
120°
(b)
(d)
109.5°
120°
9.60
180°
9.61
all angles 120°
9.62
All are polar. (a), (b), (c), and (e) have asymmetrical structures, and (d) only has one bond, which is polar.
9.63
BeH2 is nonpolar since it is linear. H2S is bent and, therefore, polar. SCN– and CN– are linear, but have
dipoles that do not cancel out. BrCl3 is T-shaped and, therefore, polar.
9.64
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.
9.65
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.
190
Chapter 9
9.66
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.
9.67
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.
9.68
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
9.69
x
4p
This is shown in Figure 9.18. Another way to diagram it would be to show the orbitals of one of the
fluorine atoms:
Each F atom (x = an electron from the other F atom):
x
2s
9.70
(a)
2p
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
O
Two other resonance structures should also be drawn for SO3.
(c)
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
F
191
Chapter 9
9.71
(a)
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 Cl atoms are bonded to the Br atom, plus the Br atom has two lone pairs of electrons. This
requires the Br atom to be sp3d hybridized, and the geometry is T-shaped.
Cl
Cl
Br
Cl
(c)
The central Xe atom is bonded to four F atoms, plus it has two lone pairs of electrons. The
molecule has an square planar geometry. This requires sp3d2 hybridization of Xe.
F
F
Xe
F
F
9.72
(a)
There are three bonds to As and one lone pair at As, requiring As to be sp3 hybridized.
The Lewis diagram
As
Cl
Cl
Cl
The hybrid orbital diagram for As: (x = a Cl electron)
x
x
x
sp 3
(b)
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
192
Chapter 9
The hybrid orbital diagram for Cl: (x = a F electron)
x
x
3d
sp 3d
9.73
(a)
Sb has five bonds to Cl atoms and no lone pairs and is therefore sp3d hybridized.
The Lewis diagram
Cl
Cl
Cl
Sb
Cl
Cl
The hybrid orbital diagram for Sb: (x = a Cl electron)
x
x
x
x
5d
sp 3d 2
(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
The hybrid orbital diagram for Se: (x = a Cl electron)
x
x
sp
9.74
Atomic Be:
2s
2p
Hybridized Be: (x = a Cl electron)
x
x
2p
sp
9.75
(a)
Sn atom in SnCl4 (x = a Cl electron)
x
x
x
x
sp 3
(b)
Sb atom in SbCl5 (x = a Cl electron)
x
x
x
x
5d
sp 3d
193
Chapter 9
9.76
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
9.77
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
sp3
2.
sp
3.
sp2
4.
sp2
9.78
1.
9.79
1.
2.
3.
4.
9.80
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°.
9.81
The bonding in phosgene is the same as that diagrammed for H2CO in Figure 9.32, except that H atom 1s
orbitals are replaced by Cl atom sp3 hybrid orbitals.
9.82
(a)
one σ bond
one σ bond and two π bonds
one σ bond
one σ bond and one π bond
N in the C=N system:
2p
sp2
2p
sp2
sp2
sp2
(b)
sigma bond
pi bond
C
C
N
194
N
Chapter 9
(c)
120 o
H
120 o
H
9.83
(a)
C
N
120o
120 o H
sp– hybridized N atom:
2p
sp
(b)
The σ bonds:
C
N
The π bonds:
C
N
(c)
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.
(d)
The HCN bond angle should be 180°.
9.84
σ∗s
1s
2p
2p
σs
2s
2s
Hydrogen
Oxygen
There are two electrons in bonding MOs and three electrons in nonbonding MOs.
The net bond order is 1.
195
Chapter 9
9.85
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.
9.86
Here we pick the one with the higher bond order.
(a)
9.87
O2+
(b)
O2
(c)
N2
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:
(8 bonding e ) − (3 antibonding e ) = 5
−
Bond Order =
−
2
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.
N2+
9.88
(a)
9.89
All of the molecules or ions are paramagnetic except N2.
(b)
NO
(c)
O2–
Additional Exercises
9.90
Since the angles are nearer to 90° than to 109°, we must use atomic rather than hybrid orbitals on antimony.
9.91
(a)
(b)
(c)
9.92
planar triangular
9.93
Structure (d)
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.
196
Chapter 9
9.94
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
repulsions. 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
9.95
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.
9.96
Five valence electrons J Group VA
X
Cl
Cl
Cl
Three valence electrons J Group IIIA
Cl
X
Cl
Cl
Seven valence electrons J Group VIIA
Cl
X
Cl
Cl
It is unlikely that X would be in Group VI 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.
9.97
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.
9.98
(a)
(b)
(c)
(d)
(e)
planar triangular J tetrahedral
trigonal bipyramidal J octahedral
T-shaped J square planar
Trigonal pyramidal J trigonal bipyramidal
linear J planar
197
Chapter 9
9.99
(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.
(b)
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.
(c)
The situation described in part (b) is delocalized. We expect completely delocalized π type
bonding among the carbon atoms.
(d)
The bond is shorter because of the extra stability associated with the delocalization energy.
9.100
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.
9.101
(a)
(b)
(c)
9.102
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.
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.
O
S
Cl
strong repulsion
O
Cl
weak repulsion
9.103
Only the px orbital can form a π bond with dxz, if the internuclear axis is the z axis.
9.104
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.
9.105
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
198
Chapter 9
In BrF4–, the two lone pairs are located as far as possible from one another, giving a square geometry to the
ion:
_
F
F
Br
F
F
199
Bringing it Together: Chapters 7 – 9
1.
2.
Electrons, Protons, Neutrons
Electron
Proton
Neutron
5.48579903 × 10–4 u
1.007276470 u
1.008664905 u
Electron
Proton
Neutron
negative
positive
neutral
ν=
c 3.00 × 108 m s−1 ⎛ 109 nm ⎞
⎜⎜
⎟⎟ = 6.00 × 1014 s–1
=
1
m
500 nm
λ
⎝
⎠
Energy of one photon of 500 nm light = hν = (6.626 × 10–34 J⋅s)(6.00 × 1014 s–1) = 3.97 × 10–19 J
Energy of one mole of photons of 500 nm light = (3.97 × 10–19 J)(6.022 × 1023 photons) = 2.39 × 105 J
Blue light would have more energy per photon than the green light.
3.
Radio waves < microwaves < infrared light < red light < blue light < ultraviolet light < X-rays < gamma rays
4.
A continuous spectrum contains a continuous unbroken distribution of light of all colors.
An atomic spectrum is the line spectrum produced when energized or excited atoms emit electromagnetic
radiation.
5.
Matter can be diffracted.
6.
In a travelling wave the crests and troughs move in the direction the wave is moving. In a standing wave,
the crests and troughs do not change position. A node is a point of zero amplitude for a wave.
7.
As the number of nodes increase, the energy of the electron wave increases.
8.
A wave function is a mathematical function that describes the intensity of an electron wave at a specified
location in an atom.
The Greek letter: Ψ
An electron wave in an atom is an orbital.
9.
(a)
(b)
(c)
(d)
(e)
10.
For n = 4
l is s, p, d, f.
The maximum number of electrons is 32.
11.
(a)
(b)
(c)
(d)
(e)
Tin
Germanium
Silicon
Lead
Nickel
1s22s22p63s23p64s23d104p65s24d105p2
1s22s22p63s23p64s23d104p2
1s22s22p63s23p2
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
1s22s22p63s23p64s23d8
12.
(a)
(b)
(c)
(d)
(e)
Pb2+
Pb4+
S2–
Fe3+
Zn2+
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10
1s22s22p63s23p64s23d104p65s24d105p64f145d10
1s22s22p63s23p6
1s22s22p63s23p63d5
1s22s22p63s23p63d10
Sulfur
Strontium
Lead
Bromine
Boron
n=3
n=5
n=6
n=4
n=2
l = 0, 1
l=0
l = 0, 1
l = 0, 1
l = 0, 1
ml = 1, 0, –1, ms = 1/2 and –1/2
ml = 0, ms = 1/2 and –1/2
ml = 1, 0, or –1, ms = 1/2 and –1/2
ml = 1, 0, –1, ms = 1/2 and –1/2
ml = 1, 0, ms = 1/2 and –1/2
200
Bringing it Together: Chapters 7 – 9
13.
A paramagnetic atom or molecule has unpaired electrons
Atoms or molecules with all paired electrons are diamagnetic.
14.
(a)
(b)
(c)
(d)
(e)
15.
Ionization energy is the amount of energy required to remove an electron from an isolated gaseous atom,
ion or molecule.
Electron affinity is the energy change that occurs when an electron adds to an isolated gaseous atom or ion.
Metals tend to lose electrons and form cations since they have low ionization energies, and nonmetals tend
to gain electrons and form anions since they have high electron affinities. The cations and anions come
together to form ionic compounds.
16.
For the second ionization energy, an electron is removed from a negatively charged species. It requires
more energy to separate negative charges than to remove a negative charge from a neutral species.
17.
Be has the greatest difference between its second and third ionization energy since an electron is being
removed from a closed shell.
18.
(a)
(b)
(c)
(d)
19.
(a)
[Ar]184s23d8
[Ar]184s13d5
[Kr]364s2
[Kr]364s23d104p3
[Xe]546s24f145d106p4
Ni
Cr
Sr
Sb
Po
P–(g) + e– → P2–(g)
Fe3+(g) + e– → Fe2+(g)
Cl(g) + e– → Cl–(g)
S(g) + 2e– → S2–(g)
endothermic
exothermic
exothermic
endothermic
(b)
(c)
x
z
20.
Electron density is the concentration of the electron's charge within a given volume.
21.
Se
4s
4p
6s
6p
Tl
201
Bringing it Together: Chapters 7 – 9
22.
(a)
Fe2+
O–
23.
(c)
Ca and F will form an ionic compound.
(b)
24.
Ca
25.
Ca2+
S
+
S
(a)
H
Sb
H
H
(b)
F
I
F
F
(c)
H
O
Cl
O
(d)
2–
C
(e)
F
F
C
F
As
F
F
(f)
O
2–
O
(g)
H
O
C
O
O
202
2–
Bringing it Together: Chapters 7 – 9
(h)
F
F
F
Te
F
F
F
(i)
H
O
N
O
O
trigonal pyramidal
square pyramid
trigonal pyramidal
linear
bent
26.
(a)
(b)
(c)
(d)
(e)
SbCl3
IF5
AsH3
BrF2
OF2
27.
(a)
(b)
(c)
(d)
(e)
sp3
sp3d2
sp3
sp3d
sp3
28.
AsF5 is nonpolar.
29.
O
O
C
O
C
C
O
O
O
O
C
O
C
O
O
O
O
O
C
C
C
O
O
O
30.
Overlap of orbitals is the portion of two orbitals from different atoms that share the same space in a
molecule.
31.
Sigma bonds are bonds formed by the head-to-head overlap of two atomic orbitals and in which electron
density becomes concentrated along and around the imagniary line joining the two nuclei.
Pi bonds are formed by the sideways overlap of a pair of p orbitals and that concentrate electron density
into two separate regions that lie on opposite sides of a plane that contains an imagniary line joining the
nuclei.
The formation of π bonds allows double and triple bonds to form.
32.
0
O
0
C
0
O
+1
O
0
C
-1
O
-1
O
0
C
+1
O
The first structure is perferred because it has the least separation of charge, and oxygen does not have a
positive charge.
203
Bringing it Together: Chapters 7 – 9
33.
(a)
O
O
O
O
(b)
(c)
(d)
O
O
Nonlinear
O with the double bond is 0.
O with the single bond is –1
O in the middle is +1
Ozone is polar due to the separation of charge.
34.
The structure on the left is better since there is less separation of charge.
The best structure would have a double bond between the P and the O.
35.
(a)
(b)
(c)
(d)
Nonmetal
Group VIA
Any period below the second one, third or lower
Cl
Cl
Cl
Cl
X
X
X
Cl
Cl
XCl2
XCl4
XCl6
XO3
bent
distorted tetrahedral
octahedral
trigonal planar
polar
polar
nonpolar
nonpolar
(f)
X
O
O
(g)
XO2 see (f)
XO3
O
0
(h)
(i)
(j)
(k)
36.
X
O
0
0
O
Cl
Cl
Cl
Cl
Cl
SO3 has double bonds
(e)
O
Cl
0
sp3d
XCl4
sp3d2
XCl6
Al2X3
The more ionic bond would be in Na2X
5s25p4
The very reactive metals are in group 1.
The least reactive metals are in groups 10 and 11.
204
X
O
O
Bringing it Together: Chapters 7 – 9
37.
A bonding molecular orbital introduces a buildup of electron density between nuclei and stabilizes a
molecule when occupied by electrons.
An antibonding molecular orbital denies electron density to the space between nuclei and destabilizes a
molecule when occupied by electrons.
Bonding molecular orbitals have electron density in the region between the nuclei while antibonding
orbitals have nodes in that region.
Bonding molecular orbitals are lower in energy than antibonding molecular orbitals.
38.
A delocalized molecular orbital speads over more than two nuclei.
Molecular orbital thoery avoids the problem of resonance by recognizing that electron pairs can sometimes
be shared among overlapping orbitals from three or more atoms.
39.
(a)
(b)
(c)
(d)
40.
PF3 has a net dipole moment.
41.
(a)
PF3
PF4+
PF6–
PF5
42.
Trigonal pyramidal
Tetrahedral
Octahdehral
Trigonal bipyramid
σ2p*
π2p *
π2p
σ2p
σ2s*
σ2s
σ1s*
σ1s
MO theory explains the two unpaired electrons in O2 and allows for a double bond with a calculated bond
order of 2.0.
43.
Fe2+ is easily oxidized to Fe3+ because by removing the electron, a half filled set of d orbitals is formed.
Mn2+ is difficult ot oxidize to Mn3+ because Mn2+ has a half filled set of d orbitals, by removing that
electron, the d orbitals will not be half filled.
44.
(a)
(b)
(c)
(d)
MgO
BeO
NaF
MgO
Mg2+ and O2– have larger charges than Na+ and Cl–
Be2+ is smaller than Mg2+, thus the ions can come closer together.
F– is smaller than I–, thus the ions can come closer together.
The ions are smaller and closer together than CaS.
205
Bringing it Together: Chapters 7 – 9
45.
Al2O3 has higher charges, Al3+ and O2– than NaCl. The lattice energy for Al2O3 is higher than the lattice
energy for NaCl, therefore it is harder to separate the ions in Al2O3 and this gives a higher melting point.
46.
The change in atomic size is smaller among the transition elements than among the representative elements
because the electtron that is added to each successive element enters the d orbitals. These d orbitals are not
outer shell orbitals, but are inner shell orbitals. The amount of shielding provided by the addition of
electrons to this inner 3d level is greater than the amount of shielding that would occur if the electrons were
added to the outer shell, so the effective nuclear charge felt by the outer electrons increases more gradually.
47.
sp3d2
48.
(a)
(b)
σ bond
π bond
49.
(a)
1.
2.
3.
7.
4.
6.
5.
(b)
(c)
sp3
sp2
sp2
sp3
1 σ bond and 0 π bond
1 σ bond and 1 π bond
B.O. = 1.5
206