Molecular Structure
Chemistry 120
Both atoms and molecules are quantum systems
We need a method of describing molecules in a
quantum mechanical way so that we can predict
structure and properties
The method we use is the
Linear Combination of Atomic Orbitals
where we can use the properties of atoms to predict
the properties of molecules.
Molecular Structure
Chemistry 120
We combine atoms to form molecules by
considering the phase of the atomic orbitals we are
using
We represent the phase via the shading we give the
orbital.
The phase represents the sign of the wavefunction
Molecular Structure
Chemistry 120
We combine atoms to form molecules by
considering the phase of the atomic orbitals we are
using
The phase represents the sign of the wavefunction
We represent the phase via the shading we give the
orbital.
Molecular Structure
Chemistry 120
For an s orbital, the orbital has the same phase
everywhere:
1s orbital, n = 1, l = 0
For a p orbital, there is a
change in the sign of the
wavefunction across the
nodal plane:
2p orbital, n = 2, l = 1, ml = -1
Molecular Structure
Chemistry 120
Consider two H atoms (1s1) coming together from
infinite separation.
There are two possibilities:
1 The wavefunctions are in phase
2 The wavefunctions are not in phase
Molecular Structure
Chemistry 120
Case 1: The wavefunctions are in phase
The atoms move together and the electron waves
overlap with the same phase, producing
constructive interference and a build up of electron
density between the nuclei
The energy of the system drops and we form a
bond
Chemistry 120
r=8Å
r=3Å
r=7Å
r=2Å
r=6Å
r=1Å
r=5Å
r = 0.75 Å
Molecular Structure
Chemistry 120
Case 2: The wavefunctions are out of phase
The atoms move together and the electron waves
have opposite phase.
The electron waves overlap producing destructive
interference and electron density between the
nuclei is reduced.
The energy of the system rises and we have an
antibonding situation
Chemistry 120
r=8Å
r=3Å
r=7Å
r=2Å
r=5Å
r=1Å
r=4Å
r = 0.75 Å
Two atoms with wavefunctions in
phase overlap with constructive
interference. Electron density
increases between the nuclei and the
overall energy decreases.
Chemistry 120
When the wavefunctions are of
opposite phase, the electron density
between the nuclei decreases due to
destructive interference. The energy
of the system rises and we have an
antibonding situation
Bonding
Antibonding
Chemistry 120
Bonding
Antibonding
Organic Structure and Bonding
Review of diatomic bonding
There are two types of bond that are
important in this part of the
Periodic Table
s bonds and p bonds
Chemistry 120
B
5
C
6
N
7
O
8
F
9
Si
14
P
15
S
16
Cl
17
Se
34
Br
35
I
53
Organic Structure and Bonding
Chemistry 120
s bonds and p bonds
s bonds are in general stronger than p bonds and
can be formed from either s or p orbitals:
Organic Structure and Bonding
Chemistry 120
s bonds and p bonds
s bonds have no nodal plane that contains the two
nuclei.
The s* antibonding orbital has a nodal plane
between the two nuclei
Organic Structure and Bonding
s bonds and p bonds
p bonds have a nodal
plane that contains
both nuclei,
The p* antibonding
orbital also has a
plane between the
nuclei
Chemistry 120
Organic Structure and Bonding
Chemistry 120
s bonds and p bonds
These s, p bonding orbitals and s*, p* antibonding
orbitals are the orbitals that are used to bind all
simple organic molecules together.
We can also describe the bonding in diatomic
molecules
important models for larger organic systems
Organic Structure and Bonding
Chemistry 120
s bonds and p bonds
To describe the bonding in the diatomic molecules
such as O2, N2 and X2 (X = F, Cl, Br and I), we use
both the s orbitals and the p orbitals on the two
atoms as a basis set - the palette of atomic orbitals
from which we will build the molecular orbitals.
The energies of the two different l states, s and p,
are slightly different in polyelectronic atoms.
Organic Structure and Bonding
Chemistry 120
s bonds and p bonds
The s orbitals and the p orbitals appear as follows
Organic Structure and Bonding
s bonds and p bonds
Chemistry 120
We arrange the atoms along one of the axes for
convenience and so the first pair of orbitals we
construct are the ss and ss* orbitals from the s
orbitals on the atoms.
Organic Structure and Bonding
s bonds and p
bonds
We now us the
higher energy p
orbitals to
construct ps
and pp orbitals
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
The complete molecular
orbital diagram for all the
diatomic molecules from Li2
to N2
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
The complete molecular
orbital diagram for all the
diatomic molecules from Li2
to N2
As each molecule has a
different number of
electrons,
Li2 2 Be2 4 B2 6 C2 8
N2 10 O2 12 F2 14 Ne2 16
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
Li2 2 Be2 4 B2 6 C2 8
N2 10 O2 12 F2 14 Ne2 16
We can write the electronic
structure of each molecule
by placing electron pairs
into the orbitals.
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
Li2 2 Be2 4 B2 6 C2 8
N2 10 O2 12 F2 14 Ne2 16
Something peculiar happens
after N2
Recall that as the charge on
the nucleus increases, the
orbitals become more
stabilized and the electrons
become more strongly
bound.
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
Li2 2 Be2 4 B2 6 C2 8
N2 10 O2 12 F2 14 Ne2 16
This happens by different
amounts, depending on the
orbital.
After N2 (10 e-), the
ordering of the orbitals
derived from p change their
order in the molecule
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
For N2 (10 e-), the ordering is this
For O2 (12 e-),
the ordering is
this
Chemistry 120
Organic Structure and Bonding
s bonds and p bonds
Chemistry 120
This is an example of configurational interaction
Each electron moves in the field of the other
electrons. If the energies of the two molecular
orbitals are sufficiently close and the nodal
properties are correct, molecular orbitals will
interact and shuffle their energies in the molecule.
This causes the s orbitals to change their energetic
ordering but only when the nuclear charge is high
enough to force the electrons close in energy.
Organic Structure and Bonding
s bonds and p bonds
Chemistry 120
Configurational interaction
Each electron moves in the field of the other
electrons. If the energies of the two molecular
orbitals are sufficiently close and the nodal
properties are correct, molecular orbitals will
interact and shuffle their energies in the molecule.
This causes the s orbitals to change their energetic
ordering but only when the nuclear charge is high
enough to force the electrons close in energy.
Chemistry 120
Molecular Structure
A full description of the structure of a molecule
requires the solution of the Schrödinger equation for
the entire molecule.
Hˆ   E
Tˆ  Vˆ r, ,   E
The potential term Vˆ r ,  ,  is far too complicated
to be solved analytically and so we need an
empirical approach to molecular structure.
Molecular Structure
There are two common approaches
Chemistry 120
- Lewis description
- Valence Shell Electron Pair Repulsion (VSEPR)
theory
and both are based on the electron count at the
central atom of the molecule or fragment of the
molecule.
Molecular Structure
Lewis description
The covalent chemical bond can be
thought of as a pair of electrons
shared between atoms;
By considering the number of
electrons in the valence shell and the
number of electrons in the outer
atoms, we can explain the presence of
lone pairs and the gross structure of
the molecule.
Chemistry 120
G. N. Lewis
Molecular Structure
Lewis description
Chemistry 120
The covalent chemical bond can
be thought of as a pair of
electrons shared between atoms;
By considering the number of
electrons in the valence shell and the
number of electrons in the outer
atoms, we can explain the presence of
lone pairs and the gross structure of
the molecule in simple cases.
G. N. Lewis
Molecular Structure
Lewis description
Chemistry 120
The Lewis description arose form
an attempt to cram the observed
properties of atoms in
combination into a mechanically
classical picture of the physical
world then prevalent; in fact even
classically, the structure of the
atom was not explicable.
G. N. Lewis
http://www.chem.yale.edu/~chem125/125/history99/7BondTheory/LewisOctet/
cubicoctet.html
Molecular Structure
Lewis description
Chemistry 120
The Lewis description is based on the observed
requirement that the atom achieves the valence shell
octet associated with the noble gases - a noble gas
configuration.
Consider the formation of MgCl2
Mg: 1s22s22p63s2 or [Ne]3s2
Cl:
1s22s22p63s23p5 or [Ne]3s23p5
Molecular Structure
Lewis description
Chemistry 120
We know that MgCl2 is ionic and so the changes in
the valence shell configurations are
Mg:
1s22s22p63s2 or [Ne]3s2
Mg2+:
1s22s22p6
Cl:
1s22s22p63s23p5 or [Ne]3s23p5
Cl-:
1s22s22p63s23p6 or [Ne]3s23p6 (i.e. [Ar])
or [Ne]
Molecular Structure
Lewis description
Chemistry 120
We therefore account for the stability of MgCl2
through the formation of closed shell ions with
noble gas configurations, namely
Mg2+:
1s22s22p63s2 or [Ne]
and
Cl-:
1s22s22p63s23p5 or [Ne]3s23p6 (i.e. [Ar])
Molecular Structure
Lewis description
Chemistry 120
In this respect, the Lewis description of bonding is
accurate, but there are major failures with
molecules.
Lewis described molecular structure through the
idea that the atom had some inherent tetrahedral
quality and that the electrons were distributed in
static manner at the vertices of the tetrahedron
Molecular Structure
Lewis description
Chemistry 120
Lewis described molecular structure through the
idea that the atom had some inherent tetrahedral
structure and that the electrons were distributed in
static manner at the vertices of the tetrahedron.
Molecular Structure
Lewis description
Chemistry 120
Molecular species therefore take structures via
sharing electrons through the vertices of the
tetrahedron. This naturally implies that all
molecules are tetrahedral, which causes major
problems for those that are not……….
Molecular Structure
Lewis description
Chemistry 120
Examples: BH3, CH4, NH3, OH2 and FH
All these structures are based on the tetrahedron and
the sharing of electrons in bonds or the presence of
lone pairs at the corner of the tetrahedron.
Molecular Structure
Lewis description
Chemistry 120
Examples: BH3, CH4, NH3, OH2 and FH
We can depict the valence shell (i.e. the shell with
the highest principle quantum number) as
C
B
2
2
1
1s 2s 2p
2
O
N
2
2
1s 2s 2p
2
2
3
1s 2s 2p
2
F
2
4
1s 2s 2p
2
2
5
1s 2s 2p
Molecular Structure
Lewis description
Chemistry 120
Examples: BH3, CH4, NH3, OH2 and FH
C
B
2
2
1
1s 2s 2p
2
N
2
2
1s 2s 2p
1s22s22p3
O
1s22s22p4
F
1s22s22p5
We satisfy the open valences of these atoms with H
atoms : H
Molecular Structure
Examples: BH3, CH4, NH3, OH2 and FH
1s22s22p1
1s22s22p2
3H
4H
H
H
B
H
1s22s22p4
1s22s22p3
1s22s22p4
1s22s22p5
N
O
F
C
B
Chemistry 120
3H
1H
2H
H
H
C
H
H
1s22s22p6
H
N
H
H
1s22s22p6
H
O
F
H
1s22s22p6
1s22s22p6
H
Molecular Structure
Examples: BH3, CH4, NH3, OH2 and FH
H
H
B
H
H
C
H
H
H
H
H
H
N
O
F
B
H
H
H
H
H
H
H
H
H
H
Chemistry 120
C
H
H
H
N
H
H
H
O
F
Molecular Structure
Examples: BH3, CH4, NH3, OH2 and FH
H
H
B
H
H
C
H
H
H
H
H
H
N
O
F
B
H
H
H
H
H
H
H
H
H
H
Chemistry 120
C
H
H
H
N
H
H
H
O
F
Molecular Structure
Examples: BH3, CH4, NH3, OH2 and FH
Chemistry 120
The structures of the first row hydrides are not
accurately predicted by the Lewis Theory of
structure and bonding.
H
H
H
H
B
H
H
H
C
H
H
H
N
H
H
H
O
F
Molecular Structure
VSEPR
Chemistry 120
The other model for molecular structure is VSEPR.
We consider a closed shell atom and we also assume
that it is spherical. The structure is then determined
by the number of “stereochemically active units”
present in the outer shell.
These stereochemically active units are the ‘lone
pairs’ and the bond pairs that are formally assumed
to exist in a molecule from a Lewis picture of
structure and bonding.
Molecular Structure
VSEPR
Chemistry 120
Once we assume that bond and lone pairs exist, we
introduce some other assumptions, one about
structure and one about energies of interactions.
The structural types that we use are based on the
distribution of points on the surface of a sphere such
that the distance between them is a maximum.
Molecular Structure
VSEPR
Chemistry 120
For two stereochemically active units, the obvious
geometry is linear:
Molecular Structure
VSEPR
Chemistry 120
For three stereochemically active units, we form a
triangular arrangement of atoms around the central
atom:
This geometry is termed
Trigonal Planar
Molecular Structure
VSEPR
Chemistry 120
Four stereochemically active units are arranged in
the form of a tetrahedron
Molecular Structure
VSEPR
Chemistry 120
For five stereochemically active units, there are two
choices. The one most commonly encountered is
the trigonal bipyramid
Molecular Structure
VSEPR
Chemistry 120
Six stereochemically active units have only one
choice for the base geometry – the octahedron
Molecular Structure
VSEPR
Chemistry 120
In each of these geometries, the sites that are
predicted are occupied either by an atom or by an
‘electron pair’
The final requirement is to detail the interaction
between these various pairs of electrons – bond and
lone – in the atom.
Molecular Structure
VSEPR
Chemistry 120
The interaction energies that we are interested in are
the repulsions between these pairs of electrons.
As bond pairs are more tightly confined, the
reulsions due to bond pairs are less. Lone pairs,
assumed to be more diffuse, suffer from higher
repulsions and thus the energy ordering is:
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Molecular Structure
VSEPR
Chemistry 120
With these rules in hand, and a knowledge of the
possible geometries, we can now predict with some
certainty the expected molecular geometry of any
main group species.
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
Step 1:
atom.
S
6
Count the electrons on each central metal
C
4
Xe
8
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
Step 1:
atom.
S
6
Count the electrons on each central metal
C
4
Xe
8
Step 2:
Determine the number of bond pairs that
each atom has.
In this case it is 4 each.
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
Step 3:
Determine the number of ‘lone pairs’
that each atom has. As each bond has 1 e- from the
central atom, then the number of electrons in lone
pairs is just
S
6-4=2
C
4-4=0
Xe
8-4=4
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
Step 4:
Determine the number of
sterochemically active species by type
SF4
4 bond pairs, 1 lone pair
CF4
4 bond pairs, 0 lone pair
XeF4
4 bond pairs, 2 lone pairs
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
Step 5:
Apply the energy rules to each structure
type and determine the structure type
SF4
trigonal bipyramid
CF4
tetrahedron
XeF4
octahedron
Molecular Structure
VSEPR
Chemistry 120
Example: What are the structures of SF4, CF4 and
XeF4?
And so the structures are……..
F
F
F
S
F
F
F
F
Xe
F
F
F
C
F
F
Molecular Structure Review
• Lewis electron dot structures
Chemistry 120
• Valence Shell Electron Pair Repulsion (VSEPR)
theory
Both are based on the electron count at the central
atom, A, of the molecule or fragment of the
molecule AXn.
Chemistry 120
Molecular Structure Review
We observe that the atoms achieve a noble gas
configuration in the valence shell, often (but not
always) an octet.
The first three noble gas configurations:
[He]: 1s2
2 valence electrons
[Ne]: 1s22s22p6
8 valence electrons
[Ar]: 1s22s22p63s23p6
8 valence electrons
Molecular Structure Review
Chemistry 120
He
H
N O
Li Be
B C
Na Mg
Al Si P S Cl Ar
F Ne
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y
Zr Nb Mo Tc Os Rh Pd Ag Cd In Sn Sb Te
I Xe
Cs Ba La Hf Ta W Re Ru Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
Molecular Structure Review
AHn Examples:
Chemistry 120
BH3, CH4, NH3, OH2 and FH
C
B
2
1
[He] 2s 2p
[He] 2s22p2
N
[He] 2s22p3
O
[He] 2s22p4
F
[He] 2s22p5
We satisfy the open valences of these atoms with H
atoms :
H
Molecular Structure Review
Chemistry 120
He
H
Li Be
B C
N O
Na Mg
Al Si P S Cl Ar
F Ne
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y
Zr Nb Mo Tc Os Rh Pd Ag Cd In Sn Sb Te
I Xe
Cs Ba La Hf Ta W Re Ru Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
Molecular Structure Review
AHn: BH3, CH4, NH3, OH2 and FH
[He]2s22p1
[He]2s22p2
[He]2s22p3
C
N
B
4H
3H
B
H
[He]2s22p4
[He]2s22p4
[He]2s22p5
O
3H
F
1H
2H
H
H
H
Chemistry 120
H
C
H
H
[He]2s22p6
or
[Ne]
H
N
H
H
O
H
H
[He]2s22p6
or
[Ne]
[He]2s22p6
or
[Ne]
F
H
[He]2s22p6
or
[Ne]
Chemistry 120
Molecular Structure Review
VSEPR: What are the structures of BH3, CH4,
NH3, OH2, and FH?
Step 1:
Count the electrons on each central atom.
B
3
[He]2s22p1
C
4
[He]2s22p2
N
5
[He]2s22p3
O
6
[He]2s22p4
F
7
[He]2s22p5
Chemistry 120
Molecular Structure Review
Step 2:
Determine the number of bond pairs for
each central atom.
H
H
B
H
H
H
C
H
H
H
N
H
Number of bond pairs:
3
4
3
H
H
O
F
H
2
1
H
Chemistry 120
Molecular Structure Review
Step 3:
Determine the number of ‘lone pairs’
remaining on the central atom.
H
H
B
H
H
H
C
H
H
Number of bond pairs:
3
4
Number of lone pairs:
0
0
H
N
H
H
H
O
F
H
3
2
1
1
2
3
H
Chemistry 120
Molecular Structure Review
Step 4:
Determine the total number of
stereochemically active units (bond pairs and lone
pairs of electrons).
BH3
3 bond pairs + 0 lone pairs =
3
CH4
4 bond pairs + 0 lone pairs =
4
NH3
3 bond pairs + 1 lone pair =
4
OH2
2 bond pairs + 2 lone pairs =
4
FH
1 bond pairs + 3 lone pairs =
4
Chemistry 120
Molecular Structure Review
Step 5:
Determine the structure type by applying
the energy rules to each structure type.
The most favorable structure minimizes steric
interactions among the stereochemically active
units.
For a molecule with a central atom (B), imagine the
stereochemically active units (A) on the surface of a
sphere as far apart in space as possible.
Chemistry 120
Molecular Structure Review
The number of stereochemically active units
determines the geometry around the central atom.
BA2
BA3
BA4
linear
trigonal planar
tetrahedral
Molecular Structure Review
BA5
BA6
trigonal bipyramid
octahedral
Chemistry 120
Chemistry 120
Molecular Structure Review
Which VSEPR geometry is appropriate for these
compounds?
BH3
3 bond pairs + 0 lone pairs =
3
CH4
4 bond pairs + 0 lone pairs =
4
NH3
3 bond pairs + 1 lone pair =
4
OH2
2 bond pairs + 2 lone pairs =
4
FH
1 bond pairs + 3 lone pairs =
4
Molecular Structure Review
BH3
3 bond pairs + 0 lone pairs =
Chemistry 120
3
Molecular Structure Review
BH3
Chemistry 120
3 bond pairs + 0 lone pairs =
3
trigonal planar
Chemistry 120
Molecular Structure Review
CH4, NH3, OH2, FH: All have 4 stereochemically
active units.
Chemistry 120
Molecular Structure Review
CH4, NH3, OH2, FH: All have 4 stereochemically
active units.
tetrahedral
Molecular Structure Review
Chemistry 120
CH4, NH3, OH2 and FH
H
H
H
H
C
H
H
H
N
H
H
H
O
F
All have a tetrahedral geometry at the central atom.
Chemistry 120
Molecular Structure Review
Q: If there are a mix of bonding pairs and lone pairs
around the central atom, what determines their
positions?
Chemistry 120
Molecular Structure
Q: If there are a mix of bonding pairs and lone pairs
around the central atom, what determines their
positions?
A: As bond pairs are more tightly confined, the
repulsions due to bond pairs are less. Lone pairs,
assumed to be more diffuse, suffer from higher
repulsions and thus the energy ordering is:
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
PCl5
PCl3
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
PCl5
PCl3
Cl
Cl
P
Cl
Cl
Cl
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
PCl5
PCl3
Cl
Cl
P
Cl
Cl
Cl
AB5
Cl
Cl
P
Cl
AB4
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
SF2
SF4
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
SF2
F
F
SF4
S
Molecular Structure Review
Trigonal Bipyramid
Aa
120
Ae
B
90
Aee
Ae
Aa
Chemistry 120
Molecular Structure
Chemistry 120
B.P.-B.P. < B.P.-L.P. < L.P.-L.P
Examples :
SF2
F
F
SF4
S
F
F
S
F
F
Chemistry 120
Molecular Structure Review
Q: What if the species is charged, e.g. PH4+ ?
Chemistry 120
Molecular Structure Review
Q: What if the species is charged, e.g. PH4+ ?
A: Take the charge into account in the valence
electrons of the central atom.
P0
5 valence electrons
P+
4 valence electrons
Chemistry 120
Molecular Structure Review
Q: What if the species is charged, e.g. PH4+ ?
A: Take the charge into account in the valence
electrons of the central atom.
P0
5 valence electrons
P+
4 valence electrons
H
H
H
P
H
Chemistry 120
Molecular Structure Review
Q: If there are three different atoms in a molecule,
which one is the central one?
Chemistry 120
Molecular Structure Review
Q: If there are three different atoms in a molecule,
which one is the central one?
Chemistry 120
Molecular Structure Review
Q: If there are three different atoms in a molecule,
which one is the central one?
A: The least electronegative atom (most
electropositive atom) is in the center.
Chemistry 120
Molecular Structure Review
Q: If there are three different atoms in a molecule,
which one is the central one?
A: The least electronegative atom (most
electropositive atom) is in the center.
Example:
COCl2
Chemistry 120
Molecular Structure Review
Q: If there are three different atoms in a molecule,
which one is the central one?
A: The least electronegative atom (most
electropositive atom) is in the center.
Example:
O
COCl2
C
Cl
Cl
Organic Structure and Bonding
Organic chemistry is the chemistry
of the top right corner of the
Periodic table.
It is the branch of chemistry that is
most closely connected to biology
and health-related science:
Medicine and Pharmaceuticals
The key elements are carbon,
hydrogen, oxygen, nitrogen and
phosphorous.
Chemistry 120
Organic Structure and Bonding
Chemistry 120
s and p review
s bonds: No nodal plane that contains both nuclei
p bonds:
1 nodal plane that contains both nuclei
Organic Structure and Bonding
Bonding in organic molecule almost
always only contains these two
types of bond.
In organic molecules, neutral
carbon atoms always have 4 and
only 4 bonds:
They can be single, double or triple
Chemistry 120
Chemistry 120
Organic Structure and Bonding
Bonding in organic molecule
almost always only contains
Alkyne: 2-butyne single and triple bonds
these two types of bond.
H
H
In organic molecules, neutral
carbon atoms always have 4
and only 4 bonds:
They can be single, double
or triple
H
H
H
H
H
H
H
H
H H
H H
Alkene: 2-butene single and double bonds
H
H
H
H
H
H
H H
H H
Alkane: butane all single bonds
Chemistry 120
Organic Structure and Bonding
In methane, all the bonds are identical and methane
is tetrahedral.
How do we describe the bonding in methane?
Two methods:
1.
Full molecular orbital theory
needs group theory and quantum mechanics
2.
Hybridization of atomic orbitals
an approximation that works well in organic
chemistry
Chemistry 120
Hybridization
This is a method for describing the bonding in
organic molecules by adding all of the
wavefunctions together on a single carbon atom,
using the three 2p orbitals and the 2s orbital.
We can form three different combinations:
sp3
sp2
sp
where the superscripts show the number of p
orbitals that we are adding to the s orbital
Chemistry 120
sp3 hybridization
In terms of energy, sp3 hybridization looks like:
sp3 hybridization
The four sp3 hybrid orbitals
point naturally at the corners
of the tetrahedron.
Carbon is tetrahedrally
coordinated
All are equal in length and the
angle between the orbitals and
therefore the bonds is ~ 109.5°
Carbon atoms with four single
bonds are sp3 hybridized
Chemistry 120
Chemistry 120
sp2 hybridization
We can also add two p orbitals to the s orbital to
form an sp2 hybrid, leaving one p orbital unused
2p
Hybridize
2s
Atom
three sp2 hybrids
Hybridized atom
sp2 hybridization
The three sp2 hybrid orbitals
point naturally at the corners
of a triangle – the coordination
at carbon is trigonal planar.
All are equal in length and the
angle between the orbitals and
therefore the bonds is 120°
Carbon atoms with two single
bonds and one double bond
are sp2 hybridized
Chemistry 120
sp2 hybridization
The double bond can occur
due to the p orbital that we
have not used on the carbon
atom:
Chemistry 120
sp2 hybridization
Imagine two sp2 hybridized
carbon atoms forming a s
bond using one sp2 hybrid:
The ‘spare’ p orbitals can then
form the p bond.
Chemistry 120
sp2 hybridization
Alkenes, ketones, aldehydes,
and any double bonded atom
are all sp2 hybridized. The s
bonds are formed from the
hybrids and the p bond from
the p orbital left over on each
atom.
Chemistry 120
Chemistry 120
sp hybridization
The last possible hybrid is the sp hybrid. We use one
p orbital and 1 s orbital:
sp hybridization
The two sp hybrids point at
180° to each other.
The two p orbitals can form
two p bonds
Chemistry 120
sp hybridization
Any triple bonded atom is sp
hybridized
Alkynes, CO and CN- are all
triply bonded.
Chemistry 120
Chemistry 120
Double and Triple bond structures
In a alkene, or other double bonded structure, the
sp2 hybrids from the s framework
The bond between carbon
atoms contains 4 electrons,
two in an sp2 s bond and
two in the pp bond
Chemistry 120
Double and Triple bond structures
In a alkyne, or other triple bonded structure, the sp
hybrids from the s framework
The bond between carbon
atoms contains 6 electrons,
two in an sp s bond and
four in the two pp bonds
Chemistry 120
Polar and non-polar bonds
So far, we have considered bonds between identical
atoms – homoatomic bonds.
The electron distributions are equal as the orbitals
on the atoms have identical energies and sizes.
In a heteroatomic system, this is not true and the
atomic orbitals that make up the molecular orbitals
have different energies.
Chemistry 120
Polar and non-polar bonds
The orbital energies on an electronegative atom are
lower in energy and therefore stabilize an electron
more effectively.
In a heteroatomic bond, this causes a small change
in the distribution of the bonding electron density
and thus a small, permanent charge difference
Polar and non-polar bonds
For a non-polar molecule, the
molecular orbital diagram is
the standard diagram for a
diatomic, shown here for O2F2
Polar molecules are skewed
in energy.
Chemistry 120
Polar and non-polar bonds
Polar molecules are skewed
in energy.
The bonding molecular
orbitals are more similar to
the lower energy atomic
orbitals – those on the
electronegative element.
The opposite is true for the
antibonding orbitals.
Chemistry 120
Phase Changes
Chemistry 120
Matter exists primarily in three phases:
solid
liquid
gas
Both elements and compounds are found in
these three phases, and are denoted by
subscripts, e.g. N2 (g), H2O(l), and Au(s).
Phase Changes
Chemistry 120
The three phases interconvert with one another.
gas
liquid
solid
Phase Changes
Chemistry 120
Interconversions of gases and liquids are called:
gas
vaporization
condensation
liquid
solid
Phase Changes
Chemistry 120
Interconversions of liquids and solids are called:
gas
liquid
freezing
melting
solid
Phase Changes
Chemistry 120
Interconversions of gases and solids are called:
gas
sublimation
liquid
solid
deposition
Phase Changes
Chemistry 120
AB – sublimation/deposition AD – melting/freezing
AC – vaporization/condensation
Phase Changes
Chemistry 120
Changing from a less dense phase to a more
dense phase (e.g. condensation) is exothermic.
Changing from a more dense phase to a less
dense (e.g. vaporization) one is endothermic.
For any two phases, the energy changes in both
directions are equal in magnitude, but opposite
in sign.
DHvap = -DHcon
Recall that enthalpy is a state function.
Phase Changes
H2O enthalpy of fusion (melting)
Chemistry 120
Phase Changes
Chemistry 120
Liquids and their vapors are in equilibrium.
Pressure of a vapors (gas phase)
= vapor pressure of liquid
Pressure and temperature are directly proportional.
Recall
PV = nRT
Phase Changes
SCF = supercritical fluid
Chemistry 120
Phase Changes
Chemistry 120
AB – sublimation/deposition AD – melting/freezing
AC – vaporization/condensation
Phase Changes
Chemistry 120
The temperature at which a liquid boils is called
its boiling point (bp). Boiling point is a function
atmospheric pressure, or the pressure above the
solution.
Normal boiling point is the boiling temperature
of a liquid at 1 atmosphere (atm) pressure.
Phase Changes
Chemistry 120
Critical temperature, Tc, is the highest
temperature at which liquid and vapor exist in
equilibrium.
Critical pressure, Pc, is the vapor pressure at the
critical temperature.
Critical point is reached at Tc and Pc.
Triple point is the temperature and pressure at
which all three phases coexist.
Phase Changes
H2O phase diagram
A = triple point
C = Tc, Pc
Chemistry 120
Phase Changes
Chemistry 120
Phase diagram for HgI2 (mercuric iodide)
HgI2 (a) and HgI2 (b) are both solids but
different phases.
Intermolecular Forces
Chemistry 120
The phase changes from solid to liquid to gas are
governed by intermolecular forces.
Intramolecular forces are the chemical bonding
forces discussed previously.
These intermolecular forces have both attractive
and repulsive components. Collectively they are
called van der Waals forces after the Dutch
Nobel laureate (physics) who described them.
Intermolecular Forces
Johannes Diderik van der Waals
Van der Waals forces describe
the behavior of a non-ideal
gas, which includes both
attractive and repulsive
components.
[P + a(n/V)2](V-bn) = nRT
Chemistry 120
Intermolecular Forces
Chemistry 120
All molecules exert weak attractions on one
another due to the mutual attraction of nuclei
and electrons. These attractive forces are only
significant at very short distances.
At such small distances the intermolecular
repulsion of the electrons on different atoms is
also significant.
Intermolecular Forces
Chemistry 120
The electrons orbiting all atoms and molecules
can be perturbed by an electric field, with
greater or lesser ease. This property is called
polarizability.
The electron cloud around an atom or molecule
can give an instantaneous dipole any time that
the electrons are not distributed perfectly
symmetrically.
Such a dipole can induce dipoles in other
species nearby.
Intermolecular Forces
Chemistry 120
Intermolecular Forces
The attractive forces
between an
instantaneous dipole
and an induced dipole
are called London
dispersion forces after
the physicist Fritz
London.
These forces are
stronger for more
polarizable species.
Chemistry 120
Intermolecular Forces
Chemistry 120
Some molecules have a permanent dipole
because of differences in electronegativities
among the atoms.
Such molecules experience dipole-dipole forces.
All molecules experience dispersion forces and
induced dipoles, and polar molecules also
experience dipole-dipole forces.
Intermolecular Forces
Dipole-dipole interactions.
Chemistry 120
Intermolecular Forces
Chemistry 120
The intermolecular forces that we have seen
London dispersion forces
induced dipole interactions
dipole-dipole interactions
have a strong effect on the boiling points of
liquids, along with the molecular weight of a
compound, and hydrogen bonding.
Intermolecular Forces
(a) CS2
(b) CH3OH
(c) CH3CH2OH
(d) H2O
(e) C6H5NH2
NH2
Chemistry 120
Intermolecular Forces
Chemistry 120
Hydrogen bonding is an additional type of
bonding interaction that requires a hydrogen atom
on one molecule and a source of electron density
on another molecule, usually a lone pair.
Hydrogen bonding can be intramolecular as well
as intermolecular.
Intermolecular Forces
Chemistry 120
Intermolecular Forces
Chemistry 120
Intermolecular Forces
Chemistry 120
Intermolecular Forces
Boiling points of covalent hydrides
Chemistry 120
Intermolecular Forces
Boiling points of covalent hydrides
Chemistry 120
Intermolecular Forces
Chemistry 120
The second-row hydrides NH3, H2O, and HF
exhibit much higher boiling points that would be
expected based on their molecular weights.
Strong hydrogen bonding between the molecules
is responsible for the large liquid phase range of
these compounds.
CH4 has a low boiling point because it has no lone
pairs to form strong hydrogen bonds.
Intermolecular Forces
Hydrogen bonding is crucial
for the double-helical
structure of DNA.
An understanding of
hydrogen bonding between
base pairs made the
structural solution possible
for James Watson and
Francis Crick.
Chemistry 120
Intermolecular Forces
Chemistry 120
DNA base pairs hydrogen bond between
G-C and
A-T
Structure and Bonding
Chemistry 120
s bonds are in general stronger than p bonds and
can be formed from either s or p orbitals:
s bonds have no nodal planes along the line
containing the two nuclei.
Structure and Bonding
Chemistry 120
The s* antibonding orbital has one nodal plane
between the two nuclei
Structure and Bonding
p bonds have one
nodal plane that
contains both nuclei.
The p* antibonding
orbital also has one
nodal plane between
the nuclei.
Chemistry 120
Hybridization Review
Chemistry 120
Recall:
sp3 hybridization
4 s bonds
sp2 hybridization
3 s bonds, 1 p
bond
H
C
H
sp3
sp hybridization 2 s bonds, 2 p bonds
H
H
H
C
H
H
C
H
sp2
H C
C H
sp
Hybridization Review
methanol
H
H
H
C
OH
Chemistry 120
water
ethanol
H
HH
OH
C C
H
H
H
O
H
Hybridization Review
methanol
H
H
H
C
OH
3
sp
Chemistry 120
water
ethanol
H
H
H
OH
C C
H
HH
O
H
Hybridization Review
methanol
H
H
H
C
OH
3
sp
Chemistry 120
water
ethanol
H
H
H
OH
C C
H
H
H
3
3
sp sp
O
H
Hybridization Review
methanol
H
H
H
C
OH
3
sp
Chemistry 120
ethanol
water
HH
OH
C C
H
H
H
H
3
3
sp sp
O
H
3
sp
Hybridization Review
carbon disulfide
S
C
Chemistry 120
aniline
S
NH2
Hybridization Review
carbon disulfide
S
C
sp
Chemistry 120
aniline
S
NH2
Hybridization Review
carbon disulfide
S
C
Chemistry 120
aniline
NH2
S
sp2
sp
sp2
Hybridization Review
carbon disulfide
S
C
Chemistry 120
aniline
sp3
NH2
S
sp2
sp
sp2