Chapters 13 and 14

Dipole Moments and Polarity

When there is a net dipole moment for a polar molecule it can be measured. The strength of a dipole is a measurement of its dipole moment. A quantity of charge (\(q\)) is separated by a distance (\(r\)) and the dipole moment is easily shown mathematically to be:

$$ \mu = q \cdot r $$

This is all explained in section 13.3 of your book. If you need the conversion from C·m to debye units I would give you that (you won't need it, OK). You DO need to know this though: ANY polar molecule MUST have a non-zero net dipole moment. All non-polar molecules will have zero net dipole moments. Remember that symmetry is the biggest factor you must consider when assigning whether a molecule is polar or not. Symmetry will mathematically cancel out polar bond vectors within a molecule.

VSEPR and VB Theories

Know what each of these theories “brings to the table” as far as chemical bonding and shapes goes. While I'm at it - you DO know what VSEPR and VB stand for - right? You’ll need to know all the shapes that are covered in the book – linear to octahedral which matches up with 2 electron regions through 6 electron regions. Know the VB hybrids that are necessary to get each shape that we covered (Section 14.1 and summarized in Figure 14.24) VSEPR will help you arrive at the right shape but it says nothing about HOW you get there. VB theory steps in and helps us visualize how this gets done. By allowing atomic orbitals to hybridize, we can get all the necessary geometries and we can “see” orbitals on neighboring atoms overlapping to give bonding orbitals (\(\sigma\) and \(\pi\) bonds). Know your shapes, angles, angle tweaks, and hybridization. Be sure and check out my other help sheet on VSEPR and VB Theory - and read the book.

One more thing...If someone asks what the "geometry" is of a molecule they are referring to the molecular geometry. One has to specify electronic geometry if that is what they mean - molecular geometry is always implied.

Sigma (\(\sigma\)) and Pi (\(\pi\)) Bonding

VB theory (LE or Localized Electron theory in your book) also brings us \(\sigma\) and \(\pi\) bonds. Know the key difference in these two types of bonds. What kind of overlap does each bond use? Which one is generally stronger? A double bond in VB theory is always a \(\sigma\) bond reinforced with a \(\pi\) bond. A triple bond is a \(\sigma\) bond reinforced with two \(\pi\) bonds. Unfortunately VB theory doesn’t always work out – especially for certain non-bonding electrons and radical species (O₂ is a good example).

MO Theory

In MO theory we use the atomic orbital wavefunctions (\(\psi\)) to generate new molecular orbital wavefunctions for the electrons in the complete molecule. For each atomic orbital we will get a new molecular orbital. MO theory still uses the σ and π designations for bonding orbitals. MO also introduces antibonding orbitals \(\sigma\)* and \(\pi\)* (sigma-star and pi-star). Antibonding orbitals are the result of destructive interference when combining the wavefunctions for the atomic orbitals (\(\psi_1\) and \(\psi_2\) are out of phase). Know the relative energies of bonding vs antibonding orbitals. Know the ordering of the molecular orbitals for all 2nd row homonuclear diatomic molecules (Figure 14.41). Know HOW to fill this diagram properly with electrons. For heteronuclear diatomics know how to fill in a diagram if one is given to you. You should also realize that the MO diagram for heteronuclear molecules with atomic orbitals of similar energy (such as NO) are essentially the same as the homonuclear diatomics. In heteronuclear diatomics the most electronegative element will have the lower orbital energies. As a result the bonding MOs will more closely resemble the atomic orbitals from those elements. The anti-bonding will be more similar to the atom with the smaller electronegativity (aka: the more electropositive one).

And finally, there are times in which atomic orbitals (AO) on one atom don’t overlap at all with the AOs on the other atom. This results in non-bonding orbitals which is where lone pair electrons tend to reside. These orbitals are essentially identical to the AO on that element. They don’t lower or raise the energy compared to the AO. Therefore they don’t participate in the bonding (or bond order calculation).

Orbitals in Polyatomic Molecules

Know what the overall concept is. We often COMBINE VB theory and MO theory taking the best parts out of each. Don’t fret the details but do know that atomic orbitals will combine in all sorts of ways to get you various levels of energies within a molecule. Check out how the \(\pi\)-system in benzene is really represented with MO’s even though the 6-carbon ring is best pictured as 6 carbon atoms with sp² hybridization. Realize that the idea of bond “delocalization” is best understood from the molecular orbital model. I DID spend plenty of class time on the concept of \(\pi\)-systems. Know what that is and how it is manifested in a molecule.

Chapter 16 - Liquids and Solids

Intramolecular forces

Intramolecular forces are the forces of attraction within molecules and are simply the bonds that hold the atoms together to make the molecule in the first place. These are considerably large forces. Looking back to Chapter 13 you can see how strong an intramolecular force (covalent bonding) really is. Check out Table 13.6 in your book to see numerous bond dissociation energies. Those are the amounts of energy that would be required to pull those bonds apart. The lowest value on that table is 149 kJ/mol and goes up to 1072 kJ/mol for a carbon-oxygen triple bond. Realize that is just a sampling of all the possible covalent bond strengths, but you need to have an idea of where all those strengths are on the energy scale and what their average is. Let’s put the average at around 400 kJ/mol. I also like to include ionic bonding here which is what holds ionic compounds (salts) together. They too are very strong forces reaching as high as 15,000 kJ/mol for some 3+ to 3- salts. So all of those were BIG FORCES. Now let’s get back to the main topic here...

What’s in a name? When its Van der Waals, its a lot.

Intermolecular Forces have historically also been called Van der Waal forces which is really the correct interpretation - meaning it covers all the forces between molecules. However, be warned, many books (and questions) equate Van der Waal forces with ONLY dispersion forces. Check out Wikipedia for Van der Waals forces - you’ll see there under the definition section that the term can be meant one way or another. Best to know the authors intent as they say. Unfortunately our history is steeped in tradition and the term on Quest usually refers to just dispersion forces. We’ll try to stick with this concept on the exam. In other classes (biology, etc...) you might have to adjust your thinking on this term.

Dipole-dipole interaction

We now knock down the strength of the positive/negative interaction by a considerable amount. Now we use partial positive and partial negative as our “sticking points”. Dipoles truly deserve the word partial in partial charge. Only a little bit of charge is there - a little positive (\(\delta^+\)) and a little negative (\(\delta^-\)). Although there is variability in the amounts of these charges (think electronegativity scale), they are still piddly little compared to full-blown charge found in cations and anions. This interaction in equation form is shown as

$$E_{\rm P} \propto -{{\mu_1\mu_2}\over r^3}$$

Where \(\mu \) is the dipole strength (dipole moment). Notice that the distance dependence is \(1/r^3\). From distance alone this puts the force at about 1% of ion-ion attraction. Most "regular" dipole-dipole potential energies have averages around 4 to 5 kJ/mol of interactions. I say "regular" because I must contrast this to H-bonding which is just a stronger version of the same thing. Also remember that these dipoles are permanent dipoles and are at the heart of polar substances.

(London) Dispersion Forces

Sometimes the word "dispersion" is not there, and sometimes the word "London" is not there. Sometimes the term Van der Waal’s Forces are (mistakingly) used to describe only these particular forces. The fact is that dispersion forces are present in ALL molecules, polar AND non-polar. The roll they play (whether major or minor) is what we must consider. These forces are the ONLY force for all non-polar molecules. All substances CAN be liquefied and therefore they ALL have some degree of attractive forces within them. The potential energy for London dispersion forces is

$$E_{\rm P} \propto -{{\alpha_1\alpha_2}\over r^6}$$

Where \(\alpha \) is the polarizability of the molecules. The polarizability quantifies the tendency of a collection of charges (like the electrons in a molecule) to become distorted by an applied electric field (like a nearby charge or dipole). The larger the molecule, the more polarizable and therefore, the greater the London forces. For atomic systems, the higher the atomic number the larger the polarizability and thus the London force. This is because the more electrons there are the more they start to populate the larger atomic orbitals. The further out an electron is from the nucleus the easier it can be influenced and moved by another source. Once the charge is shifted you get a temporary dipole - there one minute, gone the next (or should I say femtosecond?). Temporary dipoles are the heart of dispersion forces - these are also known as instantaneous dipoles in some books.

Once you have the temporary dipole, you can have it participate in a dipole-induced dipole. This is the result when a non-polar (no permanent dipole) molecule can be induced to have a dipole. It doesn’t matter if it is induced by a real (permanent) dipole from a polar molecule or whether by a temporary dipole from a non-polar molecule. Once set-up, the partial charges in the dipoles do cause attractions. These attractions are reinforced, the bigger the molecule happens to be.

Now if we only look at that the \(1/r^6\) part of that equation we can certainly see why dispersion forces are the smallest of all intermolecular forces. At the same distances, a dispersion interaction can be 1000 to 10000 times less than ion-ion interactions. That is so small, that if there is no appreciable polarizability (\(\alpha\)), then the substance is most likely a gas at normal temperatures (case in point : all the noble gases). So by taking into account distance (\(r\)) and typical \(\alpha\)’s we find the typical molecule to have about 2 kJ/mol for this interaction - certainly the smallest of our three intermolecular forces. But hey! That 2 kJ/mol average is only for a "typical" little molecule. What exactly IS a typical molecule? That’s a loaded question if ever there was one. Let’s try this in a different way, read on...

Relatively small molecules are going to have very small amounts of dispersion forces present. As a matter of fact, very small molecules will almost always be gases because of this. Water is one of those rare small molecules that has enough "pull" to become a liquid at room temperature thanks to H-bonding between the molecules. As molecules get larger (careful, larger in size doesn’t always mean mass - remember your trends), polarizability increases and dispersion forces start playing a larger role in the cohesiveness (see cohesive forces later in this review) of a substance. Two molecules could have the same mass and make-up (isomers) and the one with the largest surface area will have the largest dispersion forces. Increasing surface area will increase the number of electrons available for polarization and therefore increase the forces in between.

Get that Intermolecular Feeling...

Now that we’ve got all those different types of interactions down lets look at the figure at the end of this review sheet and get a feel for the actual amounts of energy involved. Look at it and realize that those are approximate values. Take a look at my figure and try to get a good mental picture of all these interactions and their relative values. Are you feeling it? I knew you could. I must remind you that although dispersion forces are the weakest of all the forces, that is just per single interaction. Real molecules have real shapes, volumes, and sizes. Those little interactions can certainly add up to a substantial amount of force. It is for this reason that most substances are solids once you get a big enough molecule. The dispersion forces are quite impressive in great numbers.

Put it all together now

You now know the forces that hold substances together. You also know the relative strengths of those forces. So now you should be able to predict some relative outcomes of physical properties DUE to those forces.

So read the book and your notes (and any other reference you want) and KNOW how the following properties are affected by intermolecular forces.

viscosity
surface tension

capillary action
meniscus (shape?)

vapor pressure
melting and boiling point

Think about what the property does when forces get bigger and smaller. How does temperature affect these forces? BTW, nobody should miss how temperature affects these forces. You DO know right? The hotter you get something, the more likely you are to evaporate or melt it, right? The end result of constant heating is to vaporize everything into gas state – which happens to be the state where ALL these forces of attraction are so RELATIVELY weak that they can be ignored. The thermal motion brought on by heat will eventually overcome any “sticking power” that is present. More heat, less stick, less stick, less likely to be a solid or liquid and MORE likely to become a gas. See, that wasn’t too hard was it. Hopefully this makes perfectly logical sense to you. And the truer that previous sentence becomes, the better your understanding will be of the material.

Standard Read the Book Part

Once again I’m asking you to READ. Yes, read your book. You must read over and over to get things straight. Read everything in context. Of course, only read the sections that we are covering. If our book is not getting through to you, go to the chemistry library and read another one.

Class Matters

Attending class does matter. Anything I told you in class is fair game on an exam. Turns out that about 97% of what I tell you is in the book - so if you know the book and what it preaches, then you’ll do fine. If I ever tell you in class about a good exam question, you should wake up and take note. I generally don’t kid about this stuff. Those who are alert and pay attention can reap the rewards.