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Chapter 10
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Chemical Bonding II:
Molecular Geometry and
Hybridization of Atomic Orbitals
2
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding)
pairs.
AB2 2 0
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
linear linear
B B
3
Cl ClBe
2 atoms bonded to central atom
0 lone pairs on central atom
2
4
AB2 2 0 linear linear
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0trigonal
planar
trigonal
planar
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Boron Trifluoride
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AB2 2 0 linear linear
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0 tetrahedral tetrahedral
3
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Methane
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AB2 2 0 linear linear
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0 tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
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Phosphorus Pentachloride
4
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AB2 2 0 linear linear
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0trigonal
planar
trigonal
planar
AB4 4 0 tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB6 6 0 octahedraloctahedral
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Sulfur Hexafluoride
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5
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bonding-pair vs. bonding-
pair repulsion
lone-pair vs. lone-pair
repulsion
lone-pair vs. bonding-
pair repulsion> >
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0trigonal
planar
trigonal
planar
AB2E 2 1trigonal
planarbent
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3E 3 1
AB4 4 0 tetrahedral tetrahedral
tetrahedraltrigonal
pyramidal
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB4 4 0 tetrahedral tetrahedral
AB3E 3 1 tetrahedraltrigonal
pyramidal
AB2E2 2 2 tetrahedral bent
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
AB3E2 3 2trigonal
bipyramidalT-shaped
7
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB5 5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
AB3E2 3 2trigonal
bipyramidalT-shaped
AB2E3 2 3trigonal
bipyramidallinear
20
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB6 6 0 octahedraloctahedral
AB5E 5 1 octahedralsquare
pyramidal
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Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB6 6 0 octahedraloctahedral
AB5E 5 1 octahedralsquare
pyramidal
AB4E2 4 2 octahedralsquare
planar
8
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Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
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 10.1
Use the VSEPR model to predict the geometry of the following
molecules and ions:
(a) AsH3
(b) OF2
(c)
(d)
(e) C2H4
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Example 10.1
Strategy The sequence of steps in determining molecular
geometry is as follows:
Solution
(a) The Lewis structure of AsH3 is
There are four electron pairs around the central atom;
therefore, the electron pair arrangement is tetrahedral
(see Table 10.1).
Example 10.1
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 NH3.
We cannot predict the HAsH angle accurately, but we know
that it is less than 109.5° because the repulsion of the
bonding electron pairs in the As—H bonds by the lone pair
on As is greater than the repulsion between the bonding
pairs.
(b) The Lewis structure of OF2 is
There are four electron pairs around the central atom;
therefore, the electron pair arrangement is tetrahedral.
Example 10.1
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 H2O. We
cannot predict the FOF angle accurately, but we know that it
must be less than 109.5° because the repulsion of the
bonding electron pairs in the O−F bonds by the lone pairs
on O is greater than the repulsion between the bonding
pairs.
(c) The Lewis structure of is
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Example 10.1
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 ClAlCl angles are all 109.5°.
(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.
Example 10.1
Recall that the lone pairs preferentially occupy the
equatorial positions in a trigonal bipyramid (see Table 10.2).
Thus, removing the lone pairs leaves us with a linear
geometry for , that is, all three I atoms lie in a straight line.
(e) The Lewis structure of C2H4 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 BF3, discussed earlier.
Example 10.1
Thus, the predicted bond angles in C2H4 are all 120°.
Comment
(1) The ion is one of the few structures for which the bond
angle (180°) can be predicted accurately even though the
central atom contains lone pairs.
(2) In C2H4, all six atoms lie in the same plane. The overall
planar geometry is not predicted by the VSEPR model, but we
will see why the molecule prefers to be planar later. In reality,
the angles are close, but not equal, to 120° because the bonds
are not all equivalent.
11
31
Dipole Moments and Polar Molecules
H F
electron rich
regionelectron poor
region
d+ d-
m = Q x r
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
32
Behavior of Polar Molecules
field off field on
33
Bond moments and resultant dipole moments in NH3 and NF3.
12
34
Example 10.2
Predict whether each of the following molecules has a dipole
moment:
(a) BrCl
(b) BF3 (trigonal planar)
(c) CH2Cl2 (tetrahedral)
Example 10.2
Strategy
Keep in mind that the dipole moment of a molecule depends on
both the difference in electronegativities of the elements
present and its geometry.
A molecule can have polar bonds (if the bonded atoms have
different electronegativities), but it may not possess a dipole
moment if it has a highly symmetrical geometry.
13
Example 10.2Solution
(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 10.2
(b) Because fluorine is more electronegative than boron, each
B−F bond in BF3 (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:
An analogy is an object that is pulled in the directions shown
by the three bond moments. If the forces are equal, the
object will not move. Consequently, BF3 has no dipole
moment; it is a nonpolar molecule.
Example 10.2
(c) The Lewis structure of CH2Cl2 (methylene chloride) is
This molecule is similar to CH4 in that it has an overall
tetrahedral shape. However, because not all the bonds are
identical, there are three different bond angles: HCH, HCCl,
and ClCCl. These bond angles are close to, but not equal
to, 109.5°.
14
Example 10.2
Because chlorine is more electronegative than carbon,
which is more electronegative than hydrogen, the bond
moments do not cancel and the molecule possesses a
dipole moment:
Thus, CH2Cl2 is a polar molecule.
41
Change in Potential Energy of Two Hydrogen Atoms
as a Function of Their Distance of Separation
42
Change in electron density as two hydrogen atoms
approach each other.
15
43
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s
and p). Hybrid orbitals have very different shape
from original atomic orbitals.
2. Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
b. Overlap of hybrid orbitals with other hybrid
orbitals
44
Formation of sp3 Hybrid Orbitals
45
Formation of Covalent Bonds in CH4
16
46
Predict correct
bond angle
sp3-Hybridized N Atom in NH3
47
Formation of sp Hybrid Orbitals
48
Formation of sp2 Hybrid Orbitals
17
49
# of Lone Pairs+
# of Bonded Atoms Hybridization Examples
2
3
4
5
6
sp
sp2
sp3
sp3d
sp3d2
BeCl2
BF3
CH4, NH3, H2O
PCl5
SF6
How do I predict the hybridization of the central atom?
1. Draw the Lewis structure of the molecule.
2. Count the number of lone pairs AND the number of
atoms bonded to the central atom
50
Example 10.3
Determine the hybridization state of the central (underlined)
atom in each of the following molecules:
(a) BeH2
(b) AlI3
(c) PF3
Describe the hybridization process and determine the molecular
geometry in each case.
18
Example 10.3
Strategy The steps for determining the hybridization of the
central atom in a molecule are:
Solution
(a) The ground-state electron configuration of Be is 1s22s2 and
the Be atom has two valence electrons. The Lewis structure
of BeH2 is
H—Be—H
Example 10.3
There are two bonding pairs around Be; therefore, the electron
pair arrangement is linear. We conclude that Be uses sp hybrid
orbitals in bonding with H, because sp orbitals have a linear
arrangement (see Table 10.4). The hybridization process can
be imagined as follows. First, we draw the orbital diagram for
the ground state of Be:
By promoting a 2s electron to the 2p orbital, we get the excited
state:
Example 10.3
The 2s and 2p orbitals then mix to form two hybrid orbitals:
The two Be−H bonds are formed by the overlap of the Be sp
orbitals with the 1s orbitals of the H atoms. Thus, BeH2 is a
linear molecule.
19
Example 10.3
(b) The ground-state electron configuration of Al is [Ne]3s23p1.
Therefore, the Al atom has three valence electrons. The
Lewis structure of AlI3 is
There are three pairs of electrons around Al; therefore,
the electron pair arrangement is trigonal planar. We
conclude that Al uses sp2 hybrid orbitals in bonding with I
because sp2 orbitals have a trigonal planar arrangement.
The orbital diagram of the ground-state Al atom is
Example 10.3
By promoting a 3s electron into the 3p orbital we obtain the
following excited state:
The 3s and two 3p orbitals then mix to form three sp2 hybrid
orbitals:
The sp2 hybrid orbitals overlap with the 5p orbitals of I to form
three covalent Al−I bonds. We predict that the AlI3 molecule is
trigonal planar and all the IAlI angles are 120°.
Example 10.3
(c) The ground-state electron configuration of P is [Ne]3s23p3.
Therefore, P atom has five valence electrons. The Lewis
structure of PF3 is
There are four pairs of electrons around P; therefore, the
electron pair arrangement is tetrahedral. We conclude that P
uses sp3 hybrid orbitals in bonding to F, because sp3 orbitals
have a tetrahedral arrangement. The hybridization process can
be imagined to take place as follows. The orbital diagram of the
ground-state P atom is
20
Example 10.3
By mixing the 3s and 3p orbitals, we obtain four sp3 hybrid
orbitals.
As in the case of NH3, one of the sp3 hybrid orbitals is used to
accommodate the lone pair on P. The other three sp3 hybrid
orbitals form covalent P−F bonds with the 2p orbitals of F. We
predict the geometry of the molecule to be trigonal pyramidal;
the F−F angle should be somewhat less than 109.5°.
Example 10.4
Describe the hybridization state of phosphorus in phosphorus
pentabromide (PBr5).
Example 10.4
Strategy Follow the same procedure shown in Example 10.3.
Solution The ground-state electron configuration of P is
[Ne]3s23p3. Therefore, the P atom has five valence electrons.
The Lewis structure of PBr5 is
There are five pairs of electrons around P; therefore, the
electron pair arrangement is trigonal bipyramidal. We conclude
that P uses sp3d hybrid orbitals in bonding to Br, because sp3d
hybrid orbitals have a trigonal bipyramidal arrangement.
21
Example 10.4
The hybridization process can be imagined as follows. The
orbital diagram of the ground-state P atom is
Promoting a 3s electron into a 3d orbital results in the following
excited state:
Example 10.4
Mixing the one 3s, three 3p, and one 3d orbitals generates five
sp3d hybrid orbitals:
These hybrid orbitals overlap the 4p orbitals of Br to form five
covalent P−Br bonds. Because there are no lone pairs on the P
atom, the geometry of PBr5 is trigonal bipyramidal.
63
sp2 Hybridization of Carbon
22
64
Unhybridized 2pz orbital (gray), which is perpendicular
to the plane of the hybrid (green) orbitals.
65
sp Hybridization of Carbon
