2.
Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3.
Use VSEPR to predict the geometry of the molecule.

Example
8.13
Use the VSEPR model to predict the geometry of the following
molecules and ions:
(a)AsH
3
(b)OF
2
(c)
(d)
(e)C
2
H
4

Example
8.13
Solution
(a)The Lewis structure of AsH
3
is
There are four electron pairs around the central atom;
therefore, the electron pair arrangement is tetrahedral.
Recall that the geometry of a molecule is determined only by
the arrangement of atoms (in this case the As and H atoms).
Thus, removing the lone pair leaves us with three bonding pairs
and a trigonal pyramidal geometry, like NH
3
.

Example
8.13
(b) The Lewis structure of OF
2
is
There are four electron pairs around the central atom;
therefore, the electron pair arrangement is tetrahedral.
Recall that the geometry of a molecule is determined only by
the arrangement of atoms (in this case the O and F atoms).
Thus, removing the two lone pairs leaves us with two bonding
pairs and a bent geometry, like H
2
O.

Example
8.13
(c) The Lewis structure of
is:
There are four electron pairs around the central atom;
therefore, the electron pair arrangement is tetrahedral.
Because there are no lone pairs present, the arrangement of
the bonding pairs is the same as the electron pair arrangement.
Therefore, has a tetrahedral geometry and the Cl-Al-Cl angles
are all 109.5°.

Example
8.13
(d) The Lewis structure of
is
There are five electron pairs around the central I atom;
therefore, the electron pair arrangement is trigonal bipyramidal.
Of the five electron pairs, three are lone pairs and two are
bonding pairs.
Recall that the lone pairs preferentially occupy the
equatorial positions in a trigonal bipyramid (see Slide 76).
Thus, removing the lone pairs leaves us with a linear geometry
for
, that is, all three I atoms lie in a straight line.

Example
8.13
(e) The Lewis structure of C
2
H
4
is
The C=C bond is treated as though it were a single bond in the
VSEPR model.
Because there are three electron pairs
around each C atom and there are no lone pairs present,
the
arrangement around each C atom has a trigonal planar
shape like BF
3
, discussed earlier.
Thus, the predicted bond angles in C
2
H
4
are all 120°.

84
Dipole Moments and Polar Molecules
H
F
electron rich
region
electron poor
region
=
Q
x
r
Q
is the charge
r
is the distance between charges
1 D = 3.36 x 10
-30
C m

85
How Polar Molecules React to Magnetic Fields
field off
field on

86
Bond moments and resultant dipole moments in NH
3
and NF
3
.

87
Dipole Moments of Common Polar Molecules

Example
8.14
Predict whether each of the following molecules has a dipole
moment:
(a)BrCl
(b)BF
3
(trigonal planar)
(c)CH
2
Cl
2
(tetrahedral)

Example
8.14
Solution
(a) Because bromine chloride is diatomic, it has a linear
geometry.
Chlorine is more electronegative than bromine
(see
Figure 9.5), so BrCl is polar with chlorine at the negative end
Thus, the molecule does have a dipole moment.
In fact,
all diatomic molecules containing different elements possess a
dipole moment.

Example
8.14
(b) Because fluorine is more electronegative than boron, each
B−F bond in BF
3
(boron trifluoride) is polar and the three
bond moments are equal.
However, the symmetry of a
trigonal planar shape means that the three bond moments
exactly cancel one another:

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