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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