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CHAPTER SEVEN
Molecular Geometry
7.13
Molecular geometry may be defined as the three-dimensional arrangement of atoms in a molecule.
The study of molecular geometry is important in that a molecule’s geometry affects its physical
and chemical properties, such as melting point, boiling point, density, and the types of reactions it
undergoes.
7.14
Linear triatomic: A-----B-----C. Bond angles = 180 degrees.
Trigonal planar: Bond angles = 120 degrees.
B
< B – A – B = 120°
‫׀‬
A
/
\
B
B
Tetrahedral: Bond angles = 109.5 degrees
B
< B – A – B = 109.5°
‫׀‬
A ----- B
/
\
B
B
Trigonal bipyramidal: Bond angles = 90 or 120 degrees (depends on angle in question)
C B
< C – A – B = 90°
‌ /
< B – A – B = 120°
B — A
/
\
B
B
Octahedral: Bond angles = 90 degrees.
B B
‌ /
B — A — B
/ ‌
B B
< B – A – B = 90°
7.15
Tetrahedral: 4 atoms
Trigonal bipyramidal: 5 atoms
Octahedral: 6 atoms
7.16
The VESPR model, or the valence-shell electron pair repulsion model, accounts for the geometric
arrangements of electron pairs around a central atom in terms of the electrostatic repulsion
between electron pairs. Using this model we can predict the geometry of molecules (and ions) in a
systematic way. With this model double bonds and triple bonds can be treated like single bonds,
which is good for qualitative purposes. Also, if a molecule has two or more resonance structures,
we can apply the VESPR model to any one of them. Formal charges are usually not shown.
The magnitude of repulsion decreases in that order because electrons in a bond are held by the
attractive forces exerted by the nuclei of the two bonded atoms. These electrons have less “spatial
distribution” than lone pairs; that is, they take up less space than lone-pair electrons, which are
associated with only one particular atom. Because lone-pair electrons in a molecule occupy more
space, they experience greater repulsion from neighboring lone pairs and bonding pairs.
7.17
By the VSEPR theory, lone pair – bonding pair repulsion is stronger than bonding pair – bonding
pair repulsion. Therefore, the lone pair occupies a spot where it has two bonds 120 degrees away
and two bonds 90 degrees away, instead of the spot which has three bonds 90 degrees away and
one bond 180 degrees away.
7.18
In the proposed arrangement each C-H bond has two C-H bonds 90 degrees away and one C-H
bond 180 degrees away. This is not as stable as the arrangement which puts three C-H bonds 109.5
degrees away, which is the observed geometry.
H
<H – C – H = 90˚
H
‌
|
H — C — H
C
H
‌
/
\
H
H
H
Square Planar (proposed)
Tetrahedral (actual)
7.19
(a)
The Lewis structure of PCl3 is shown below. Since in the VSEPR method the number of
bonding pairs and lone pairs of electrons around the central atom (phosphorus, in this case)
is important in determining the structure, the lone pairs of electrons around the chlorine
atoms have been omitted for simplicity. There are three bonds and one lone electron pair
around the central atom, phosphorus, which makes this an AB3E case. The information in
Table 10.2 shows that the structure is a trigonal pyramid like ammonia.
Cl
P
Cl
Cl
What would be the structure of the molecule if there were no lone pairs and only three bonds?
(b)
The Lewis structure of CHCl3 is shown below. There are four bonds and no lone pairs
around carbon which makes this an AB4 case. The molecule should be tetrahedral like
methane (Table 10.1).
H
Cl
C
Cl
Cl
(c)
The Lewis structure of SiH4 is shown below. Like part (b), it is a tetrahedral AB4 molecule.
H
H
Si
H
H
(d)
The Lewis structure of TeCl4 is shown below. There are four bonds and one lone pair which
make this an AB4E case. Consulting Table 10.2 shows that the structure should be that of a
distorted tetrahedron like SF4.
Cl
Cl
Te
Cl
Cl
Are TeCl4 and SF4 isoelectronic? Should isoelectronic molecules have similar VSEPR
structures?
7.20
In each case, you should use the following approach.
Step 1: Write the Lewis structure of the molecule. The lone pairs on the terminal atoms have
been omitted.
Step 2: Count the number of electron pairs around the central atom.
Step 3: Build a model, or consult Table 10.1 of your text to predict the geometry of the molecule.
Lewis structure
Geometry
Electron pairs
Electron
Lone pairs
on central atom
arrangement
3
trigonal planar
2
linear
4
tetrahedral
Cl
(a)
(b)
Cl
Al
Cl
Cl
Zn Cl
Cl
Cl
7.21
7.23
linear, AB2
Zn Cl
0tetrahedral, AB4
The lone pairs of electrons on the bromine atoms have been omitted for simplicity.
Br
7.22
0
2−
Cl
(c)
0 trigonal planar, AB3
Hg Br
linear
N
+
−
N
O
linear
S
N−
C
linear
(a)
AB4
tetrahedral
(f)
AB4
tetrahedral
(b)
AB2E2
bent
(g)
AB5
trigonal bipyramid
(c)
AB3
trigonal planar
(h)
AB3E
trigonal pyramid
(d)
AB2E3
linear
(i)
AB4
tetrahedral
(e)
AB4E2
square planar
The Lewis structure is:
H
H
C
O
C
O
H
H
O
H
H
C
AB3 trigonal planar
C
AB4 tetrahedral
H
H AB2E2 bent
O
7.24
Only molecules with four bonds to the central atom and no lone pairs are tetrahedral (AB4).
2−
Cl
I
Cl
Cl
Si
Cl
Cl
I
C
Cl
I
Cd
Cl
Cl
I
What are the Lewis structures and shapes for XeF4 and SeF4?
7.25
Only (c) will not be tetrahedral. All the others have AB4−type Lewis structures and will therefore
be tetrahedral. For SF4 the Lewis structure is of the AB4E type which gives rise to a distorted
tetrahedral geometry (Table 10.2 of the text).
7.26
Step 1: Write the Lewis structure of the molecule.
Br
Hg
Br
Step 2: Count the number of electron pairs around the central atom. There are two electron pairs
around Hg.
Step 3: Since there are two electron pairs around Hg, the electron-pair arrangement that
minimizes electron-pair repulsion is linear.
In addition, since there are no lone pairs around the central atom, the geometry is also linear
(AB2).
You could establish the geometry of HgBr2 by measuring its dipole moment. If mercury(II)
bromide were bent, it would have a measurable dipole moment. Experimentally, it has no dipole
moment and therefore must be linear.
7.27
Geometry: bent; hybridization: sp3.
7.28
The Lewis structures and VSEPR geometries of these species are shown below. The three
nonbonding pairs of electrons on each fluorine atom have been omitted for simplicity.
−
F
+
+
F
F
F
F
F
Xe
F
Xe
Sb
F
F
F
F
F
F
F
AB3E2
T-shaped
7.29
(a)
AB5E
Square Pyramid
AB6
Octahedral
The Lewis structure is:
F
B
F
F
The shape will be trigonal planar (AB3)
(b)
The Lewis structure is:
O
Cl
O
-
O
The molecule will be a trigonal pyramid (nonplanar).
(e)
The Lewis structure is:
N
O
O
The nitrogen atom is of the AB2E type, but there is only one unshared electron rather than
the usual pair. As a result, the repulsion will not be as great and the O−N−O angle will be
greater than 120° expected for AB2E geometry. Experiment shows the angle to be around
135°.
Which of the species in this problem has resonance structures?
7.30
To predict the bond angles for the molecules, you would have to draw the Lewis structure and
determine the geometry using the VSEPR model. From the geometry, you can predict the bond
angles.
(a)
BeCl2:
AB2 type, 180° (linear).
(b)
BCl3:
AB3 type, 120° (trigonal planar).
(c)
CCl4:
AB4 type, 109.5° (tetrahedral).
(d)
CH3Cl:
AB4 type, 109.5° (tetrahedral with a possible slight distortion resulting from the
different sizes of the chlorine and hydrogen atoms).
7.31
(e)
Hg2Cl2: Each mercury atom is of the AB2 type. The entire molecule is linear, 180° bond
angles.
(f)
SnCl2:
AB2E type, roughly 120° (bent).
(g)
H2O2:
The atom arrangement is HOOH. Each oxygen atom is of the AB2E2 type and the
H−O−O angles will be roughly 109.5°.
(h)
SnH4:
AB4 type, 109.5° (tetrahedral).
(d)
The Lewis structure is:
−
H
H
Si
H
The ion has a trigonal pyramidal geometry (AB3E).
7.32
Only ICl2− and CdBr2 will be linear. The rest are bent.
7.33
The Lewis structure is shown below.
2−
Cl
Cl
Be
Cl
Cl
The molecule is of the AB4 type and should therefore be tetrahedral. The hybridization of the Be
atom should be sp3.
7.34
For an octahedral AX4Y2 molecule only two different structures are possible: one with the two Y’s
next to each other like (b) and (d), and one with the two Y’s on opposite sides of the molecule like
(a) and (c). The different looking drawings simply depict the same molecule seen from a different
angle or side.
It would help to develop your power of spatial visualization to make some simple models and
convince yourself of the validity of these answers. How many different structures are possible for
octahedral AX5Y or AX3Y3 molecules? Would an octahedral AX2Y4 molecule have a different
number of structures from AX4Y2? Ask your instructor if you aren’t sure.