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CHAPTER 9 | Molecular Geometry and Bonding Theories
9.18. Collect and Organize
For the seesaw, tetrahedral, and square pyramidal geometries, we are to rank these in order of increasing bond
angles for the smallest bond angle for each geometry.
Analyze
In seesaw molecules (based on the trigonal bipyramidal geometry) the smallest bond angle is 90˚ (there are
also bond angles of 120˚ and 180˚). In tetrahedral molecules, all the bond angles are 109.5˚. In square
pyramidal molecules, the smallest bond angle is 90˚ (there are also 180˚ bond angles).
Solve
In order of increasing bond angle, (a) seesaw = (c) square pyramidal < (b) tetrahedral.
Think about It
In general, as steric number increases, the smallest bond angle for the geometry decreases.
9.20. Collect and Organize
Of the geometries for molecules with ABx formulas where x = 2–6, we are to list those that retain the same
bond angles even if one or more of the atoms were removed and replaced with a lone pair.
Analyze
The molecular geometries based on steric number (SN) and electron-pair geometry are as follows:
SN
Electron-Pair Geometry Molecular Geometry
Ideal Bond Angles
2
Linear
Linear
180˚
3
Trigonal planar
Trigonal planar
120˚
Bent
120˚
4
Tetrahedral
Tetrahedral
109.5˚
Trigonal pyramidal
109.5˚
Bent
109.5˚
5
Trigonal bipyramidal
Trigonal bipyramidal
90˚, 120˚, 180˚
Seesaw
90˚, 120˚, 180˚
T-Shaped
90˚, 180˚
Linear
180˚
6
Octahedral
Octahedral
90˚, 180˚
Square pyramidal
90˚, 180˚
Square planar
90˚, 180˚
Solve
From the table we can see that replacing one atom in all the structures with a lone pair does not change any of
the bond angles (ideally). When we remove two atoms from a trigonal bipyramidal structure we lose the 180˚
bond angle, and when we replace three atoms on the trigonal bipyramidal structure with lone pairs, we lose the
90˚ and 120˚ bond angles between atoms. Thus, the molecular geometries that have the same bond angles
when lone pairs replace one or more atoms are linear, trigonal planar, tetrahedral, trigonal pyramidal, trigonal
bipyramidal (replacing one atom only), octahedral, and square pyramidal.
Think about It
There are slight changes in the bond angles in real molecules when an atom in a structure is replaced by a lone
pair. For example, CH4 and NH3 are both SN = 4 with a tetrahedral electron-pair geometry, but CH4 has bond
angles all at 109.5˚ while NH3 has bond angles of 107.3˚.
9.22. Collect and Organize
We are to determine if we can obtain linear triatomic molecules when we remove atoms from molecules with
trigonal bipyramidal, seesaw, and trigonal planar geometry.
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Molecular Geometry and Bonding Theories | 493
Analyze
For each geometry we take atoms away to obtain the VSEPR geometry until we obtain a triatomic molecule.
Solve
(a, b)
(c)
The trigonal planar geometry (c) does not lead to linear triatomic species.
Think about It
Since the seesaw geometry is derived from the trigonal bipyramidal electron pair geometry, both lead to the
same linear triatomic AB2E4 molecule.
9.24. Collect and Organize
We are asked to decide which atoms should be removed from a molecule with cubic geometry, AB 8, to obtain
an approximate octahedral geometry.
Analyze
All the positions on a cube are equal but we have the choice of removing adjacent atoms on the cube, ones
diagonally across the face of the cube, or ones from opposite corners of the cube.
Solve
Removing two adjacent edge atoms gives a
pyramidal shape
Removing two atoms across the diagonal of a
face gives a very distorted octahedron
Removing two atoms at opposite corners
gives an approximate octahedral geometry
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Think about It
To obtain the symmetrical octahedral geometry we had to remove atoms from the opposite ends of the cube;
this kept the resultant geometry symmetrical.
9.26. Collect and Organize
Using Lewis structures and VSEPR theory, we can determine the molecular geometries of NO 3–, NO43–, S2O,
and NF3.
Analyze
After drawing the Lewis structure for each molecule, we can determine the steric number for the central atom,
then locate the atoms about the central atom to see the bond angles, and finally determine the molecular shape
based on the location of the atoms.
Solve
(a)
O
SN = 3
Electron-pair geometry = trigonal planar
No lone pairs
Molecular geometry = trigonal planar
N
O
O
O
(b)
SN = 4
Electron-pair geometry = tetrahedral
No lone pairs
Molecular geometry = tetrahedral
N
O
O
O
(c)
S
S
O
SN = 3
Electron-pair geometry = trigonal planar
One lone pair
Molecular geometry = bent
(d)
SN = 4
Electron-pair geometry = tetrahedral
One lone pair
Molecular geometry = trigonal pyramidal
N
F
F
F
Think about It
Remember that lone pairs occupy positions in the electron-pair geometry, but the molecular geometry is
determined by the positions of the bonded atoms only.
9.27. Collect and Organize
Using Lewis structures and VSEPR theory, we can determine the molecular geometries of NH 4+, CO32–, NO2–,
and XeF5+.
Molecular Geometry and Bonding Theories | 495
Analyze
After drawing the Lewis structure for each molecule, we can determine the steric number for the central atom,
then locate the atoms about the central atom to see the bond angles, and finally determine the molecular shape
based on the location of the atoms.
Solve
(a)
+
H
H
N
H
SN = 4
Electron-pair geometry = tetrahedral
No lone pairs
Molecular geometry = tetrahedral
H
H
N
H
H
H
(b)
SN = 3
Electron-pair geometry = trigonal planar
No lone pairs
Molecular geometry = trigonal planar
O
C
O
O
(c)
SN = 3
Electron-pair geometry = trigonal planar
One lone pair
Molecular geometry = bent
N
O
O
(d)
SN = 6
Electron-pair geometry = octahedral
One lone pair
Molecular geometry = square pyramidal
Think about It
Notice how the other resonance structure of NO2– also has a bent geometry because the steric number remains
3 at the central N atom.
9.28. Collect and Organize
Using Lewis structures and VSEPR theory, we can determine the molecular geometries of SCN –, CH3PCl3+,
ICl2–, and PO33–.
Analyze
After drawing the Lewis structure for each molecule, we can determine the steric number for the central atom,
then locate the atoms about the central atom to see the bond angles, and finally determine the molecular shape
based on the location of the atoms.
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Solve
(a)
SN = 2
Electron-pair geometry = linear
No lone pairs
Molecular geometry = linear
(b)
S
C
N
+
H
H
Cl
C
P
H
Cl
Cl
SN = 4 (around P atom)
Electron-pair geometry = tetrahedral
No lone pairs
Molecular geometry = tetrahedral
(c)
Cl
SN = 5
Electron-pair geometry = trigonal
bipyramidal
Three lone pairs
Molecular geometry = linear
(d)
I
Cl
SN = 4
Electron-pair geometry = tetrahedral
One lone pair
Molecular geometry = trigonal pyramidal
P
O
O
O
Think about It
Even for the other resonance forms of SCN–,
the molecular geometry is linear because SN = 2 with no lone pairs on C.
9.36. Collect and Organize
We are given skeletal structures of NOCl, NO2Cl, and NO3Cl and asked to draw the Lewis structures and
predict the geometry about the nitrogen atom in each compound.
Analyze
The geometry around the N atom in each compound depends on the steric number for nitrogen, which may or
may not include lone pairs.
Solve
O
N
Cl
N
N
O
Cl
O
O
O
SN = 3
Electron-pair geometry = trigonal planar
One lone pair
Molecular geometry = bent
Cl
SN = 3
Electron-pair geometry = trigonal planar
No lone pairs
Molecular geometry = trigonal planar
N
O
Cl
Molecular Geometry and Bonding Theories | 497
O
O
O
N
O
Cl
SN = 3
Electron-pair geometry = trigonal planar
No lone pairs
Molecular geometry = trigonal planar
N
O
O
Cl
Think about It
The geometry about the O atom in NO3Cl will not be linear but will be bent because SN = 4 at the O atom.
9.39. Collect and Organize
Using the given skeletal structure of Sarin (Figure P9.39), we can complete the Lewis structure, assign formal
charges to the P and O atoms, and then predict the geometry around P using VSEPR theory.
Analyze
The molecular formula of Sarin is C4H10FOP, which has 44 e– and needs 76 e– to complete the octets (and
duets) on all the atoms. This gives a difference of 32 e – for 16 covalent bonds and leaves 12 e– in 6 lone pairs
to complete the Lewis structure.
Solve
Reducing formal charges
gives
The SN = 4 for the P atom in this molecule, and there are no lone pairs, so the geometry around the P atom in
Sarin is tetrahedral.
Think about It
Notice that we would predict the same molecular geometry around P for either resonance form.
9.45. Collect and Organize
We can look at the bond polarities and the molecular structure by VSEPR theory to determine which
molecules (CCl4, CHCl3, CO2, H2S, and SO2) are polar and which are nonpolar.
Analyze
All of the individual bonds in these molecules are polar, so the molecular geometry of each compound will
decide the overall molecular polarity. We can represent each bond polarity with a vector with the head of the
arrow pointed toward the more electronegative atom, which carries a partial negative charge. We then visually
inspect the molecule to see whether the individual bond dipoles add up or cancel out.
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Solve
(a)
Cl
C
Cl
Cl
Cl
All bond polarities are equal in magnitude and cancel
each other, so CCl4 is nonpolar.
H
(b)
C
Cl
Cl
Cl
The electronegativity of the atoms in CHCl3 are in order
Cl > C > H. Because the bond polarities do not cancel,
CHCl3 is polar.
(c)
O
C
(d)
O
The bond polarities in CO2 cancel, so it is nonpolar.
H
S
The molecular geometry of H2S is bent, so it is polar.
H
(e)
The molecular geometry of SO2 is bent, so it is polar.
Polar molecules are (b) CHCl3, (d) H2S, and (e) SO2. Nonpolar molecules are (a) CCl4 and (c) CO2.
Think about It
Molecules with polar bonds are nonpolar only for highly symmetrical geometries (linear, trigonal planar,
tetrahedral, trigonal bipyramidal, and octahedral).
9.48. Collect and Organize
By looking at the molecular structures and individual bond polarities present in C 4F8, C2ClF3, and Cl2HCClF2
we can determine which of these chlorofluorocarbons are polar and which are nonpolar.
Analyze
To determine bond polarities we need the electronegativity values for C (2.5), H (2.1), F (4.0), and Cl (3.0). In
each of these molecules the halogens are bonded to the carbon atoms. Because the electronegativities of the
halogens are higher than that of carbon, the bonds are polarized so that the halogen carries partial negative
charge. Because the electronegativity of carbon is higher than that of hydrogen, the C—H bonds are polarized
so that the carbon atom carries the partial negative charge.
Solve
(a)
(b)
(c)
F
Cl
C
F
Cl
Cl
C
F
C
F
H
C
Cl
F
Molecular Geometry and Bonding Theories | 499
Because of the different bond polarities of C—F, C—H, and C— Cl, only C4F8 has bond dipoles that cancel.
Therefore, C4F8 (a) is nonpolar whereas C2ClF3 (b) and Cl2HCCClF2 (c) are polar.
Think about It
Only molecules with completely symmetric geometries with all the same atoms attached to the central atoms
are nonpolar.
9.50. Collect and Organize
In the pairs of molecules given we can use differences in electronegativity between the bonded atoms to
predict which molecule is the most polar.
Analyze
The electronegativities of the atoms are N = 3.0, P = 2.1, H = 2.1, C = 2.5, Cl = 3.0, F = 4.0, Br = 2.8.
Solve
(a) Nitrogen is more electronegative than P so NH3 has the most polar bonds and, therefore, is more polar than
PH3.
(b) Bromine is less electronegative than Cl. In CBr 2F2 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.
Think about It
Notice that the direction of the bond polarity matters in determining molecular polarity. In part a the central
atom is – so if that atom is replaced by a less electronegative atom, the bond dipoles decrease and the
molecule is less polar. If the central atom, however, is + as in part b and is replaced by a less electronegative
atom, the molecule becomes more polar.
9.52. Collect and Organize
In this question we are to rank CO, CS, SiO, SiS, SO, and NO in order of increasing dipole moment first by
using periodic trends in electronegativity and then by using the actual values for electronegativities of the
atoms.
Analyze
Electronegativity decreases down a group and increases across a period in the periodic table. Bond polarity,
though, depends on the difference in electronegativity of the atoms so the relative position of the bonded atoms
as they appear in the periodic table is important.
Solve
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.
Think about It
The two rankings compare fairly well except it was difficult to judge the electronegativity difference for CS,
NO, and SiS from their positions in the periodic table.
9.107. Collect and Organize
Given the skeletal structure of phosphoric acid, we are to complete its Lewis structure and use VSEPR theory
to determine the molecular geometry around the phosphorus atom.
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Analyze
H3PO4 has 32 e– and needs 46 e–, giving a difference of 14 e– in seven covalent bonds. This leaves 18 e– in
nine lone pairs to complete the octets on the atoms in the structure. Phosphorus may expand its octet to reduce
the formal charges on the atoms.
Solve
H
O
O
P
O
O
H
All formal charges = 0
SN at phosphorus = 4
Electron-pair and molecular geometry around P = tetrahedral
H
Think about It
In this structure phosphorus forms three  bonds to OH, and a  plus a  bond to oxygen.
ures to miss the lone pair on the S
O sulfur atom.