**X. FUN**

Now let's look at some more examples and see if we can apply what we have been studying.

(The polarizability is represented by the symbol α )Let's consider very small at first. Consider He and Ne. Neither has a permanent dipole moment nor charge so the only difference is size and therefore polarizability. The potential energy of interaction between two He atoms would be proportional to .202 and between two Ne atoms would be proportional to .392 (just the polarizability squared). The ratio for Ne to He potential interactions would then be 3.80 considering all other factors equal (ionization potential and distance). Of course we would predict that Ne would boil higher than He due to the difference in polarizability. This rough calculation says that the ratio would be 3.80 and the actual ratio is 27.3/4.216 = 6.47 so we are considerably off quantitatively.

Now let's compare molecules which are a little bigger but still have something in common so that we don't have too many factors to consider. Let's consider C_{2}H_{6} and C_{3}H_{8} where neither has a dipole moment, but
one is larger than the other. This is straightforward, right? The larger molecule should have the higher boiling point. Experimentally we see the following:

α(Å^{3}/molecule) | µ(D) | TB(K) | |
---|---|---|---|

C_{2}H_{6} | 5.0 | 0 | 184.5 |

C_{3}H_{8} | 7.1 | 0 | 231 |

So we were correct in our intelligent estimation.

What about two molecules with about the same polarizability and different dipole moments? We can consider isobutane and isobutylene where the relevant data is:

α(Å^{3}/molecule) | µ(D) | |
---|---|---|

isobutane | 9.30 | 0.440 |

isobutylene | 9.30 | 1.63 |

They have the same polarizability so the dipole moment difference should make isobutylene boil considerably above isobutane, right? We didn't forget anything did we? No hydrogen bonds and no free charges, so we considered everything. The isobutane boils at 263 K and the isobutylene boils at 267 K. Qualitatively we were ok, except that we expected more of a difference due to the increase in dipole moment. Let's remember that.

Now increase the complexity a little bit by considering the molecules H_{2} and HCl (why did I skip HF at this point?). Of course we know that H_{2} has no dipole moment and it is smaller than HCl so we know that HCl will boil much
higher than H_{2}. There also will be a little hydrogen bonding between the two HCl molecules. Based upon all of these factors we certainly expect HCl to boil at a temperature considerably above H_{2}. The experimental data are the
following:

α(Å^{3}/molecule) | µ(D) | TB(K) | |
---|---|---|---|

H_{2} | .819 | 0 | 20.39 |

HCl | 2.63 | 1.08 | 191.1 |

We see that our qualitative deduction was very true.

So let's go a little bigger. Which would you expect to have the higher boiling point, CCl_{4} or CHCl_{3}? Always ask "Are there hydrogen bonds present?" Here the answer is no. Then we ask if there are any full charges present and we
see again that the answer is no. So what are the basic differences between the two molecules? We see that one is somewhat larger than the other and we see that one is a polar molecule. The size affect will provide a polarization difference and the
polar molecule will have dipole moment contributions. Looking in our table in a previous section we see that

α(Å^{3}/molecule) | µ(D) | |
---|---|---|

CCl_{4} | 10.5 | 0 |

CHCl_{3} | 8.5 | 1.01 |

This is a little tough to intelligently guess at, isn't it? One has larger polarizability and the other has a dipole moment. So we would expect the boiling points to be similar. Intuitively, I would think most people would think that the dipole moment
difference would shift the answer toward the chloroform for the guess at the higher boiling substance. However CCl_{4} boils at 349.9 K and chloroform boils at 334.9 K. Why? Remember the isobutane and isobutylene above? Evidently the
polarizability factor is more important than the difference in dipole moments and this leads us to an important principle which we will see illustrated further as we go along: when a molecule becomes "large", the dominant interaction is the dispersion
interaction. Remember this as we go along. So the dipole difference in isobutylene indeed made it boil higher, but the similarity in polarizability and the fact that we are close to "large" made the difference not as significant as we had guessed

We can see a similar result by comparing propane and dimethyl ether from the following data:

α(Å^{3}/molecule) | µ(D) | TB(K) | |
---|---|---|---|

propane | 7.1 | 0.28 | 231 |

dimethyl ether | 6.7 | 4.34 | 248 |

There is a huge increase in dipole moment and indeed the ether boils at a higher temperature (by 7.3%), but not as high as we would guess from the dipole moment difference.

Now get bigger. Consider the isomers of dinitrobenzene:

α(Å^{3}/molecule) | µ(D) | TB(K) | |
---|---|---|---|

para | 21 | 0 | 572(sublimes) |

meta | 21 | 13.0 | 576 |

ortho | 21 | 20 | 592 |

Now we see a huge increase in dipole moment, but very little increase in boiling point (3.5%). So with a much higher dipole moment increase than that for the case of propane and dimethyl ether, we have half the percentage increase in boiling point.

At this point, you probably should be able to do some ordering of dipole moments based upon the basic nature of intermolecular interactions. So consider CH3F and CH3OH. Which has the higher boiling point?

α(Å^{3}/molecule) | µ(D) | |
---|---|---|

CH_{3}F | 4.27 | 6.04 |

CH_{3}OH | 3.3 | 5.67 |

Well methyl fluoride has the higher polarizability and the higher dipole moment, so this is a no-brainer, right? If you then said that methyl fluoride boils at the higher temperature, you were ..........wrong! Methanol boils at 338 K and methyl fluoride boils at 195 K. What did you forget? Hydrogen bonding. Hydrogen bonding. Hydrogen bonding. Always remember to consider if you have hydrogen bonding for such is very strong.

Next we will return to the second virial coefficient to consider some applications of it. If you are ready now, then go now to this page.

*Web Author: Dr. Leon L. Combs*
*Copyright ©2012 by Dr. Leon L. Combs - ALL RIGHTS RESERVED*