Download Intermolecular forces

Intermolecular forces
describe the interactions
between two or more molecules.
Dipole-Dipole Interactions vs
Induced Dipole Interactions
As a chemical phenomena, they
are electrostatic in nature, i.e.,
follows the formula of "like
charges repel, unlike charges
attract".
10-3 m
mm
10o m
m
10-6 m
μm
10-9 m
nm
10-10m
10-15m
fm
o
A
Gravitation
electromagnetism
mass
+
charge
-
strong
force
-
charge
mass
These can be splitted into different categories, and One of the
well-known classes are the Van der Waals' interactions.
I’M PROBABLY MISSING SOMEONE’S
FAVORITE INTERMOLECULAR FORCE FROM
THE VENN DIAGRAM...
π
π
anion/ cationinteractions
hydrogen
bonding
Van der Waal’s
interactions
halogen bonding
mechanical locking
intermolecular forces
which CAN BE AGAIN SPLITTED INTO 3 sub-classes:
debye FORCE
(PERMANENT DIPOLEinduced DIPOLE)
KEESOM FORCE
(PERMANENT DIPOLEPERMANENT DIPOLE)
Van der Waal’s
interactions
-
+
London Dispersion FORCE
(induced DIPOLE-induced DIPOLE)
-
+
Arises randomly,
and transiently
(10-11 sec?)
Many students reason that the Keesom force is the strongest and dominant in determining condensed phase properties (e.g., boiling point, binding constants). It’s not true and let’s look at why. In doing so, we’ll learn a valuable
subtlety about the mental image chemists ought to have.
Permanent >> Instantaneous,
because... more is better!!!
Nope.
To convince you, Let us start by reviewing some physical
evidence, easily obtained boiling points for some common
substances.
b.p. / K
400
300
309k
273k
200
185k
231k
Recall that liquid->gas requires
breaking all the intermolecular
interactions, and thus b.p. serves as a
proxy for measuring the total intermolecular forces.
100
0
b.p. / K
b.p. / K
400
400
300
300
306k
308k
271k
200
281k
200
249k
235k
195k
100
100
0
0
When the size increases (usually correlated with polarizability,
i.e., how "jiggly" the electron cloud is), the London dispersion
contribution increases; in these cases the boiling point
always increases.
b.p. / K
400
The shape of the molecule, which
is related to the available contact
surface area, also changes the
boiling point significantly. (With
exceptions like hydrogen-bonded
species.)
300
309k
200
301k
283k
100
0
b.p. / K
When the dipole moment is
increased and the size/shape stays
constant --- that is, increasing the
strength of the Keesom & Debye
components, the boiling points
sometimes increase and sometimes
decrease. (Whoops.)
400
300
309k
305k
308k
200
Moreover, even when we're adding
on some very polar groups like a
fluorine to a hydrocarbon, the
boiling point increase is often
only a minute 20K --- a difference
merely changing the shape of the
species can affect.
306k
301k
195k
100
185k
0
The Keesom/Debye interactions are simply weaker than London
dispersion. But why? Certainly a permanent attraction is stronger than a once-in-a-while attraction?
The key is that real molecules at
real temperatures tumble. The
dipole moments tumble with the molecules.
While strong attraction exists when
the dipoles are aligned, this is not
the only available configuration. If
we reverse one of the pair, now we
get a strong repulsion between the
like charges. (and I should stress
that this term is not present for
induced dipole interactions.)
attraction
θ
θ
This is a cosine function, passing through an
angle where the net attractive and repulsive
components of the 4 point charges is zero.
(What is this angle?)
repulsion
In a pair of freely rotating
molecules, there’s neither net
attraction nor repulsion. they
cancel out. The better question
is perhaps why should there be
any attraction at all.
+
=
The reason is that the attractive
and repulsive configurations are
not equally likely. The attractive
arrangement "sticks", and is present more often than the repulsive
ones. This minute asymmetry is
what results in the net observable effects of the Keesom
interactions.
How minute? This depends on the temperature. Intuitively, it
should: the higher temperature -> the higher kinetic energy ->
the less relative importance does the Keesom attraction holds.
Fulfilling my duty as a physical chemist, I present to you the
single expression that captures All of these insights quantitatively, describing the total energy:
strength of the dipoles
2 2
−2 μ1 μ2 1
Ε=
3kΤ (4πεο)2 r6
distance between
dipoles
boltzmann constant
temperature
permittivity
We’ve mentioned the temperature dependent kinetic energy
term already. The distance term buttress our earlier claim
that in gas phase these interactions are almost non-existent
-- a 6th power decay is a very sharp one. Not surprisingly the
larger the dipoles, the stronger their final interaction
energy will be; when the dielectric is higher (e.g., a more
polar solution), the interaction will be weaker.
One last mystery that may be nagging you at this point is when we
look back at an earlier graph with
empirical data. If increased dipole
moment always increases the intermolecular force, why should pentane have a higher b.p. than ether??
One possible reason is that we’re
changing more than just the dipole
moment of the whole molecule.
Remember that the London dispersion acts on polarizability, i.e.,
“sloshiness” of the electron
clouds.
The more electronegative fluorine
holds the electrons tighter, reducing the sloshiness and hence
the london dispersion component
while it raises the keesom component. The latter loses out (and we
know why), and the final consequence is that it has a lower net
intermolecular force, and hence
lower boiling point.
b.p. / K
400
300
200
309k
308k
306k
100
0
Bottom line: In real liquids, permanent dipole-dipole interactions actually play a less important role than induced dipole
interactions. The best case scenario simply don’t happen
often enough. An analogy is perhaps comparing playing the
lottery and working a regular job; winning the lottery is
better, but you don’t get to choose to win, but only to play.
Jon Chui / Feb 2012