### Topic 10

```Topic 10
Molecular Geometry
and Bonding Theory
1
Molecular Geometry
The subject of molecular geometry deals with shapes of
molecules. Recall that a Lewis structure is a twodimensional representation of a molecule or ion; its
intent is to show how the valence electrons of atoms are
distributed in the molecule. A Lewis structure does not
necessarily have to depict the actual shape of the
molecule or ion, but we can use it deduce what the
actual shape is.
Since atoms are always in motion, when we talk about
the shapes of molecules or ion, we mean the shape
based on the average location of the atoms.
2
Molecular Geometry
A description of the geometry of a molecule or ion
involves the positions of all the atoms in the molecule
relative to each other which includes the parameters of
bond angles and bond lengths.
In a Lewis structure, two atoms are joined by a line.
When more than two atoms are joined together, these
lines intersect at a given central atom and forms angles
with each other. These angles are called bond angles.
The line or bond joining two atoms connects the nuclei
of the two atoms with the length or distance between
those two nuclei being called the bond length.
3
Molecular Geometry
The bond angles and bond lengths of a molecule or ion
dictate the particular geometry of the species.
Lewis structures are two-dimensional representations of a
molecule but because molecules are three dimensional, a
picture of the molecular geometry must convey
information about the third dimension as well.
Our drawing surface is flat; therefore, we use a projection
which includes dashed line-wedges for bonds to indicate
the three-dimensional molecule.
4
Molecular Geometry
Bonds are represented as ordinary lines if they are in the
plane of the paper; solid wedges if they are pointing out of the
paper; dashed wedges or dashed lines if they are pointing into
the paper.
:
Ammonia has a geometry that is triangular pyramidal (looks
like a pyramid with N at the top and H at the base of the three
corners). If the N atom is in the plane of the paper, one 3-D
arrangement could have two H atoms pointing back into the
paper with the third pointing out of the paper or vice versa.
N
H
H
H
5
Molecular Geometry
To determine the shape of a molecule based on the bond angles
and bond length, we use a simple theory called VSEPR (Valence
Shell Electron Pair Repulsion Theory).
The basic idea behind VSEPR is that regions of high electron
density (electron groups) around each atom in the molecule tend
to be oriented as far away from one another as possible.
Depending on the number of electron groups around the central
atom, there will be a particular geometry available that minimizes
the electron repelling forces in the molecule.
Once we know how the electron groups around an atom are
oriented, it becomes easy to imagine the shape of the molecule.
6
Molecular Geometry
----
90
----
Let’s take a look at the following example:
Which orientation gives the most space between the two
negative charges around the central positive charge; the
90o or the 180o?
180o
(-) ----(+)
(-) ----(+)
(-)
o
(-)
Obviously, the 180o orientation gives the best use of
space and allows the two negative charges to get as far
apart as possible.
This means that a central atom that has two electron
groups around it will adapt a geometry that is linear in
shape; electron groups will be 180o apart.
7
Molecular Geometry
----
----
90
90
----
Let’s look at another example:
Which orientation gives the most space between the three
negative charges around the central positive charge?
(-)
(-)
120o
180o
(+)120o
(-) ----(+)
(-)
o
120
o
o
(-)
(-)
Obviously, the shape that has all three negative charges
120o apart gives the best use of space and allows the
three negative charges to get as far apart as possible.
This means that a central atom that has three electron
groups around it will adapt a geometry that is trigonal
planar in shape; electron groups will be 120o apart.
8
Molecular Geometry
Bottom line: Depending on the number of electron groups
around the central atom, there will be a particular geometry
that allows all the electron groups to maximize their
distance from each other.
The first step in figuring out the shape of a molecule is to
determine the steric number of atoms that are bonded to at
least two other atoms.
Steric number (SN) refers to the number of electron groups
around an atom and gives us information on how these
groups will orient relative to each other.
9
Steric Number
To determine the steric number, we must account for the
following:
-
a single bond is counted as one group
a double bond is counted as one group
a triple bond is counted as one group
a lone pair is counted as one group
Basically, each electron density region (group of electrons
between atoms) around an atom counts as one group, and
a lone pair counts as one group as well.
10
Steric Number
To determine the steric number (SN), we must first draw
the Lewis structure and then count the number of electron
groups around an atom.
What are the steric numbers for the C atoms in the Lewis
structures below?
Both carbons have three electron groups
H
H
around them (2 single bonds and one
C C
double bond);
therefore, both have a SN of 3.
H
H
H
C
C
H
Both carbons have two electron groups
around them (one single and one triple
bond;
11
therefore, both have a SN of 2.
Steric Number
.. ..
The O atom has four electron groups
around it (2 single bonds and two lone
pairs);
therefore, O has a SN of 4 in this
structure.
..
What are the steric numbers for the O and N atoms in the
Lewis structures below?
The nitrogen atom has four electron
groups around it (3 single bonds and one
lone pair);
therefore, N has a SN of 4 in this
structure.
12
H - O-H
-
H - N-H
H
Bond Angles
Once we have determined the steric number of an atom, we
can predict the measure of the angle formed by the electron
groups sticking out of that atom.
However, when the groups on the atom are not identical,
we must realize that the bond angles will differ slightly.
Bond angles are affected by the type of electron density
regions (electron groups) around the atom.
Lone pairs (LP) repel other groups more strongly than
bonded pairs (BP) and multiple bonds repel more strongly
than single bonds. Repelling force interactions are
greater in LPLP > LPBP > BPBP and cause the bond
angles to decrease.
13
Polarity
Many physical properties of molecules are directly related to
whether they are nonpolar or polar. The overall polarity of a
molecule depends not only on the polarity of its bonds but
also on its geometry.
Each polar bond has a bond moment, but each molecule
also has an overall molecular dipole moment that is
approximately the sum of all its individual bond moments.
This vector sum of bond moments must include not only the
magnitude of the bond moments but also their direction.
To determine if a molecule is polar or nonpolar requires the
examination of the bond moments and their orientations
within the geometry of the molecule. If bond moments are
oriented in such a way that it cancels each other out, 14
the molecule is nonpolar.
an Atom Based on Steric Number
2 groups: steric number, SN = 2
bond angles: 180o
molecular geometry: linear
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 2 atoms bonded (X) to central atom (A), AX2
i.e CO2
3 groups: SN = 3
molecular geometry: trigonal planar
bond angles: 120o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 3 atoms bonded (X) to central atom (A), AX3
i.e NO3
15
an Atom Based on Steric Number
3 groups: SN = 3
molecular geometry: bent or angular
bond angles: <120o
polarity (assuming all atoms the same around central atom): polar
geometry involves: 2 atoms bonded to central atom which has one lone pair (E), AX2E
i.e NO2
4 groups: SN = 4
molecular geometry: tetrahedral
bond angles: 109.5o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 4 atoms bonded to central atom, AX4
i.e CH4
16
an Atom Based on Steric Number
4 groups: SN = 4
molecular geometry: triangular pyramidal
bond angles: <109.5o
polarity (assuming all atoms the same around central atom): polar
geometry involves: 3 atoms bonded to central atom which has one lone pair (E), AX3E
i.e NH3
4 groups: SN = 4
molecular geometry: bent or angular
bond angles: <109.5o
polarity (assuming all atoms the same around central atom): polar
geometry involves: 2 atoms bonded to central atom which has two lone pairs, AX2E2
i.e H2O
17
an Atom Based on Steric Number
5 groups: SN = 5
molecular geometry: triangular bipyramidal
bond angles: 90o, 120o, 180o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 5 atoms bonded to central atom, AX5
i.e PF5
5 groups: SN = 5
molecular geometry: see-saw
bond angles: <90o, <120o, <180o polarity (assuming all atoms the same around central atom): polar
geometry involves: 4 atoms bonded to central atom which has one lone pairs, AX4E
i.e SF4
18
an Atom Based on Steric Number
5 groups: SN = 5
molecular geometry: T-shaped
bond angles: <90o, <180o
polarity (assuming all atoms the same around central atom): polar
geometry involves: 3 atoms bonded to central atom which has two lone pairs, AX3E2
i.e ClF3
5 groups: SN = 5
molecular geometry: linear
bond angles: 180o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 2 atoms bonded to central atom which has three lone pairs, AX2E3
i.e I3
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an Atom Based on Steric Number
6 groups: SN = 6
molecular geometry: octahedral
bond angles: 90o, 180o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 6 atoms bonded to central atom, AX6
i.e SF6
6 groups: SN = 6
molecular geometry: square pyramidal
bond angles: <90o
polarity (assuming all atoms the same around central atom): polar
geometry involves: 5 atoms bonded to central atom which has one lone pairs, AX5E
i.e BrF5
20
an Atom Based on Steric Number
6 groups: SN = 6
molecular geometry: square planar
bond angles: 90o, 180o
polarity (assuming all atoms the same around central atom): nonpolar
geometry involves: 4 atoms bonded to central atom which has two lone pairs, AX4E2
i.e XeF4
Note:
There are other molecular geometries with SN =6 (AX3E3, T-shaped;
AX2E4, linear) as well as molecular geometries with SN>6, but we
will not cover these in this course.
21
Valence Bond Model
The VSEPR model provides a way to predict
molecular shapes but says nothing about the
electronic nature of covalent bonds. To describe
bonding, the valence bond (VB) model was
developed.
The VB model provides an orbital picture of how
electron pairs are shared in a covalent bond.
A covalent bond results when two atoms approach
each other closely enough so that a singly
occupied valence orbital on one atom overlaps a
22
singly occupied valence orbital on another atom.
Valence Bond Model
These paired electrons located within the
overlapping orbitals are attracted to the nuclei of
both atoms forming a covalent bond between the
atoms.
For example, the H-H bond in H2 results from the
overlap of two singly occupied hydrogen 1s
orbitals:
H
+
H
H
H
23
Hybridization Model
According to this model, it would seem that for an atom to
form a covalent bond, it must have an unpaired electron.
However, experience tells us that this is not true.
To adapt the valence bond model to include bonding
between atoms with no unpaired electrons, the
hybridization model was developed.
This model accounts for the experimental evidence about
molecular structure much better than the simple overlap VB
model.
The hybridization model modifies the description of atomic
orbitals when they overlap to form bonds.
24
Hybridization Model
In this model, we do not describe the orbitals by using simple s and p
orbitals, but instead, we use hybrid orbitals which are combination
orbitals formed by the mixing of simple atomic orbitals. The number of
hybrid orbitals formed is equal to the number of atomic orbitals mixed.
Although it appears that hybrid orbitals are nothing more than a different
way of describing electrons, there is an underlying physical reality that
we are actually describing. The new orbitals are more suited for
forming bonds.
A bond is the result of the overlap of two orbitals, one from each atom of
the bond. The better the overlap, the stronger the bond. Pure atomic
orbitals are quite symmetrical resulting in relatively little overlap before
their nuclei approach close enough to begin repelling each other.
Hybrid orbitals are less symmetrical than unhybridized ones resulting in
more effective overlap and stronger bonding.
25
Hybridization Model
The formation of the BeH2 molecule can be explained using
the hybridization model.
It is assumed that as two hydrogen atoms approach Be, the
atomic orbitals of the Be atom undergo a significant
change:
a 2s orbital is mixed or hybridized with a 2p orbital to form
two new sp hybrid orbitals that will have better overlap
(bonding) with the 1s orbitals of H:
one s atomic orbital + one p atomic orbital  two sp hybrid orbitals
The next slide describes the process using an orbital
diagram.
26
Hybridization Model
H : Be : H
BeH2
s sp sp s
Be
____ ____
1s22s2
1s
____
2s

“ground state”
-
Be does not have unpaired e ;
therefore, it must promote an electron
to a p orbital to form the valence state
before bonding will occur.
2s
____
____
____
____ valence state
2s
2px
2py
2pz “excited state”
____ ____
sp
sp
Hybridization process: 2s and 2p
orbitals are mixed so as to form two
hybrid (mixed) orbitals (sp) that are
better suited for overlaping with the 1s
orbitals of H.
27
Hybridization Model
A similar process occurs for all atoms bonded in molecules.
They undergo hybridization to form new orbitals that allow
for better overlapping and bonding.
The bonding in carbon might be explained as follows:
CH4
C: 2s22p2
___
2s
___ ___ ___
2px
2py
2pz
ground state
Four unpaired electrons are formed as an electron
from the 2s orbital is promoted (excited) to the vacant
2p orbital to form the valence state.
28
Energy
2p
2p
2s
2s
1s
1s
C atom (ground state)
C atom (promoted)
valence “excited” state
29
Hybridization Model
This promotion would give four unpaired electrons available
for bonding: one bond on carbon would form using the 2s
orbital while the other three bonds would use the 2p orbitals.
However, this does not explain the fact that the four bonds
in CH4 appear to be identical and not a single 2s and three
2p orbitals.
The hybridization model assumes that the four available
atomic orbitals in carbon combine to make four equivalent
“hybrid” orbitals.
The 2s, 2px, 2py, 2pz mix to form 4 new equivalent sp3
orbitals that are better suited for overlapping in bonding
and containing an unpaired electron.
____ ____ ____ ____
30
sp3
sp3
sp3
sp3
Hybrid Orbitals
Hybrid orbitals are orbitals used to describe
bonding that are obtained by taking
combinations of atomic orbitals of an isolated
atom.
In the carbon case, a set of hybrids are
constructed from one “s” orbital and three “p”
orbitals, so they are called sp3 hybrid orbitals.
The four sp3 hybrid orbitals take the shape of a
tetrahedron. Each C-H covalent bond results
from an overlap between a singly occupied
hydrogen 1s orbital with a singly occupied carbon31
sp3 hybrid orbital.
s
sp3
s
sp3 s
sp3
sp3
s
32
You can represent the hybridization
of carbon in CH4 as follows.
Energy
2p
sp3
sp3
C - H bonds
2s
1s
C atom
(promoted or
excited state)
1s
C atom
(hybridized state)
1s
C atom
(in CH4)
33
Hybridization Correlated to Geometry
In general, we can correlate the steric number to the
hybridization of the atom. There is a relationship
between the type of hybrid orbitals and the
geometric arrangement of those orbitals.
•
•
•
•
•
If the steric number is 2, the hybridization is sp.
If the steric number is 3, the hybridization is sp2.
If the steric number is 4, the hybridization is sp3.
If the steric number is 5, the hybridization is sp3d.
If the steric number is 6, the hybridization is sp3d2.
34
:
What are the molecular geometry, angles between
bonded atoms, hybridization, and polarity of XeF2?
-
:F:
-
First, you will have to draw it’s Lewis structure (22e available):
Xe
-
Next, we see that there are 5 electron groups around the
central atom giving this structure a steric number of 5.
The molecular geometry for SN=5 with two bonded atoms and
3 lone pairs is linear.
:
: F:
The angle between bonded pairs is 180o.
Xe
Since this is a SN=5, the hybridization is sp3d.
All dipoles cancel each other out (all 3 lone pairs pull out from
the center while the two bonded atoms pull in opposite
directions); therefore, this is a nonpolar molecule.
35
What are the molecular geometry, angles between
bonded atoms, hybridization, and polarity of NH3?
-
First, you will have to draw it’s Lewis structure (8e available):
:
Next, we see that there are 4 electron groups around the
central atom giving this structure a steric number of 4.
H N H
H
The molecular geometry for SN=4 with three bonded atoms
:
and 1 lone pairs is triangular pyramidal.
The angles between bonded pairs is <109.5o. The lone pair
causes the angles of the bonded pairs to decrease.
Since this is a SN=4, the hybridization is sp3.
N
H
All dipoles are additive (lone pair pulls out and all three
bonded atoms pull toward the center resulting in an overall
dipole moment for the molecule; therefore, this is a polar
molecule.
H
H
36
1Steric
No.
2Angles,
XAX
3Hybrid-
ization
Lone Pairs
4Molecular
Geometry
5Sketch
6Polarity
2
180o
sp
0
linear
no
3
120o
sp2
0
trigonal planar
no
<120o
sp2
1
angular or bent
yes
109.5o
sp3
0
tetrahedral
no
<109.5o
sp3
1
triangular pyramidal
yes
<109.5o
sp3
2
angular or bent
yes
4
37
1Steric
5
6
No.
2Angles,
XAX
3Hybrid-
ization
Lone Pairs
4Molecular
Geometry
5Sketch
6Polarity
90o 120o 180o
sp3d
0
triangular
bipyramidal
no
<90o <120o
<180o
sp3d
1
see-saw
yes
<90o <180o
sp3d
2
T- shaped
yes
180o
sp3d
3
linear
no
90o, 180o
sp3d2
0
octahedral
no
<90o
sp3d2
1
square pyramidal
yes
90o, 180o
sp3d2
2
square planar
no
38
Molecular Geometry Table Footnotes
•
•
•
•
•
•
1Steric
number equals the number of electron density
regions. Single bond, double bond, triple bond, and lone
pair all equal one electron density region.
2Angles are affected by the type of electron density regions.
Repelling forces are greater in LPLP>LPBP>BPBP (LP
– lone pair, BP – bonding pair) and multiple bonds are
greater than single bonds.
3Hybridization can be easily determined by the Steric
number and remembering the combined s, p, and d orbitals
4Molecular Geometry gives the shape of the atoms in the
molecule.
5Example sketch of electronic geometry around of central atom.
6Polarity is given for case when all the ligands are the same.
39
Multiple Bonding
Insofar as geometry is concerned, a multiple bond acts as if
it were a single bond. In other words, the extra electron
pairs in a multiple bond (double, triple) have no effect
on the geometry of the molecule.
Insofar as hybridization is concerned, the extra electron
pairs in the multiple bond are not located in hybrid
orbitals.
According to the valence bond model, only one electron
pair occupies one needed hybrid orbital for each bond
(whether a single or multiple bond).
hybrid
unhybrid
Triple bond
hybrid
Double bond
40
sp2
sp2
unhybridized
2p
sp2
sp2
sp2
For example, consider the molecule ethene:
Each carbon atom is bonded to three other atoms and
no lone pairs, which indicates the need for three sp2
hybrid orbitals with one electron pair in each orbital.
The third 2p orbital is left unhybridized and lies
perpendicular to the plane of the trigonal sp2 hybrids.
The following slide represents the sp2 hybridization of the
41
carbon atoms.
(unhybridized)
2p
2p
sp2
Energy
2s
1-2s and 2-2p orbitals
form 3-sp2 hybrid
orbitals leaving 1-2p
orbital unchanged
1s
C atom (ground state)
1s
C atom (hybridized)
42
Multiple Bonding
To describe the multiple bonding in
ethene, we must first distinguish between
two kinds of bonds.
with a cylindrical shape about the bond axis. This occurs
when two “s” orbitals overlap or “p” orbitals overlap along
their axis (hybridized).
C : C
axis
– A p (pi) bond is a “side-to-side” overlap of parallel “p”
orbitals, creating an electron distribution above and
below the bond axis (unhybridized).
.. ..
axis
C : C
..
..
43
axis
overlap, s
axis
overlap, s
axis
side-to-side
overlap, p
44
Multiple Bonding
Two of the sp2 hybrid orbitals of each carbon
overlap with the 1s orbitals of the hydrogens and
the remaining sp2 hybrid orbital on each carbon
overlap to form s bonds.
There are 5 s bonds in ethene.
s
s
s
s
s
45
Multiple Bonding
The remaining “unhybridized” 2p orbitals on
each of the carbon atoms overlap side-to-side
p
forming a p bond.
s
s
s
s
s
You therefore describe the carbon-carbon double
bond as one s bond and one p bond.
Ethene has 5 s bonds and 1 p bond.
Bottom line:
single bond – s
double bond – 1s and 1p
triple bond – 1s and 2p
46
Multiple Bonding
How many s and p bonds does acetylene, C2H2, have?
First, you must draw a Lewis structure:
H–C
s
p
p
s
C–H
s
Since the C’s have a SN of 2, we know that we have a
hybridization of sp on the carbons.
This gives us 1 sp-sp head-on overlap (C-C) and 2 s-sp headon overlaps (H-C) giving us 3 s bonds.
We also have 2 unhybridized side-on overlaps (C-C) giving us
2 p bonds.
This gives us a total of 3s and 2p bonds.
47
HW 68
code: geometry
```