CHEMISTRY 161
HW CH#9
23, 30, 36, 38, 50, 52, 56, 60, 68, 76, 82, 86,88, 90, 100, 108
9-23
B
Removing an atom from one of the axial
positions gives a pentagonal pyramidal
geometry
B
B
A
B
B
B
B
Removing an atom from one of the equatorial
positions gives a distorted octahedral
geometry
B
B
A
B
B
B
9-30
(a)
SN = 4
Electron-pair geometry = tetrahedral
One lone pair + one unpaired
electron
Molecular geometry = bent
(b)
SN = 4
Electron-pair geometry = tetrahedral
One unpaired electron
Molecular geometry = trigonal
pyramidal
(c)
F
F
F
SN = 5
Electron-pair geometry = trigonal
bipyramidal
Two lone pairs
Molecular geometry = T-shaped
I
F
(d)
S
F
I
F
F
F
F
F
F
SN = 5
Electron-pair geometry = trigonal
bipyramidal
One lone pair
Molecular geometry = seesaw
S
F
F
F
9-38
H
Xe
S
SN = 5
Electron-pair geometry = trigonal
bipyramidal
Three lone pairs
Molecular geometry = linear around Xe
H
9-50
(a) Nitrogen is more electronegative than P so NH 3 has the most polar bonds and, therefore, is more
polar than PH3.
(b) Bromine is less electronegative than Cl. In CBr2F2 the Br atoms do not counteract the electron pull
from the F atoms as well as Cl does, so CBr2F2 is more polar than CCl2F2.
9-52
From the periodic table we would estimate the atoms increase in electronegativity in the order Si < S
<
C
<
N < O.
The greater the electronegativity difference between the atoms, the more polar the diatomic molecule.
Compounds made up of elements in the list that are far from each other (e.g., Si and O) have the
largest dipole moment because they are the most polar. Molecules made up of elements closest
together in the series have small differences in electronegativity and so are less polar with lower
dipole moments. We might guess an order of polarities, then, as SiS < CS < NO < CO < SO < SiO.
Using the electronegativity values from Figure 8.5, we obtain the molecules in order of increasing
dipole moment of CS < NO < SiS < SO = CO < SiO.
9-56
+
(a)
O
N
O
(b)
O
N
O
SN = 2
sp hybridized
SN = 3
sp2 hybridized
SN = 2 for both N atoms
sp hybridized
SN = 3 for both N atoms
sp2 hybridized
SN = 3 for both N atoms
sp2 hybridized
–
(c)
O
(d)
O
O
N
N
O
O
O
(e)
N
N
O
O
9-60
F
F
S
SN = 4
sp3
F
F
F
F
S
F
SN = 5
sp3d
F
F
S
F
F
F
SN = 6
sp3d 2
9-68
Because the steric number, and thus the geometry, around each central atom in a larger molecule must
be defined separately and may be different for adjacent atoms, the overall molecular geometry is
sometimes hard to name. For example, if a tetrahedral atom is bonded to a trigonal planar atom, just
one term cannot describe the geometry of the molecule
X
Y
A
X
X
B
Y
9-76
The S atoms have SN = 4 and are tetrahedral. The central O atom also has SN = 4 and is bent with sp3
hybridization.
9-82
Molecular orbital theory better explains magnetic properties because the orbitals are placed in terms
of energy, like atomic orbitals. If there are unpaired electrons in any molecular orbital energy level,
the molecule will be attracted to a magnetic field.
9-86
Overlap of an s orbital with either p or d orbitals side-on will result in no net overlap because of the
different phases of the lobes on the p and d orbitals.
Overlap of a p orbital with a d orbital gives constructive interference, but the overlap is not too
effective because the p and d orbitals are of different energies.
Overlap of a 3p orbital with a 3p orbital, or 3s with 3s, or 3d with 3d, gives constructive interference
with efficient overlap because of their similar energies and shapes.
9-88
9-90
He2 would have the MOs
 1s 1*s
Ne2 would have the MOs
 2s 2*s 2 p 2 p 2* p 2* p
The bond order (BO) is calculated from
BO = 12 (number of e– in bonding MOs– – number of e– in antibonding MOs)
Solve
+
He 2
Total number of electrons = 3 e 
( 1s ) 2 ( 1*s )1
BO = 12 (2  1) = 0.5
Ne 2
+
total number of electrons = 15 e
( 2 s ) 2 ( 2*s ) 2 ( 2 p ) 2 ( 2 p ) 4 ( 2* p ) 4 ( 2* p )1
BO = 12 (8  7) = 0.5
Each of these species has a bond order of
and Ne2.
1
2
, so each would be predicted to be more stable than He2
9-100
In this Lewis structure all the negative formal
charges are on the most electronegative atoms
(oxygen)
SN = 4
Electron-pair geometry = tetrahedral
Molecular geometry = tetrahedral
Bond angles = 109.5˚
9-108
F
Al
SN = 3 with 3 bonds to Al
Molecular geometry = trigonal planar
sp2 hybridization
F
F
–
F
F
Al
SN = 4 with 4 bonds to Al
Molecular geometry = tetrahedral
sp3 hybridization
F
F
2–
F
F Al
F
F
SN = 5 with 5 bonds to Al
Molecular geometry = trigonal bipyramidal
sp3d hybridization
F
SN = 6 with 6 bonds to Al
Molecular geometry = octahedral
sp3d 2 hybridization