enter image description here

I know that the highest boiling point has to do with which has the strongest intermolecular force. I also know that the strongest would be ionic, then hydrogen bonding, then dipole-dipole, then london dispersion.

Can someone show me how I am supposed to solve these questions step by step though?

Am I supposed to create the Lewis dot structure for all of them and then see which is nonpolar vs. polar?

Also, once I come up with the Lewis structure, is that all I need to determine polarity? I just look to see if it's symmetric or not?

I know the 3D VSEPR shapes, so am I supposed to come up with that and then see which is nonpolar or polar?

Can someone provide me with a step by step solution as to how I can solve any of these problems from start to finish so that I have a good methodology?

Thank you!

  • 4
    $\begingroup$ :/ It's impossible to do without some prior knowledge. If you have it, drawing anything would be usually a waste of time. Without it you might run out of time on exam and still put wrong answer. $\endgroup$ – Mithoron Feb 12 '19 at 19:15

Q1 and Q2 have the same approach. One notices that they are homologous alkanes and picks the biggest and the smallest, having the most London dispersion/van der Waals force and least, respectively. Other forces are known not to be present between alkanes.$^1$ (There is a small difference between branched and linear alkanes, but that is negligible compared to another $\ce{CH2}$ group.)

For Q3, one sees that they are combinations of hydrogen and halogen. While van der Waals forces are also present here, they are likely to be dominated by dipole-dipole interactions and hydrogen bonding (where applicable).$^2$ Which of the two is actually determining here, does not matter, since both point to a high boiling point in $\ce{HF}$.

In Q4, one notices that the question deals in homo-atomic molecules, so there can be neither hydrogen bonding nor dipole-dipole. Thus, the more electrons, the easier it is to polarize the molecule, and the heaviest molecule wins.

To summarize: To me, the questions are more about test-taking than actual chemistry. I cannot say whether the boiling point of $\ce{C4H10}$ or of $\ce{HF}$ is higher without looking up the values or knowing them and I do not expect most chemists to. Look for which forces may apply and reason about the trends. To do so, one may have to derive the molecular structure, but I would expect that in a test, one either does not need to or has the time to do so.

$^1$ This is where practice and knowledge are indispensable, which is unfortunate for the learner.

$^2$ In this case, this is trivial, but there are cases where exact quantitative calculations would be necessary. One can construct systems in which van der Waals forces outweigh ionic repulsion.

| improve this answer | |
  • 1
    $\begingroup$ Your rule of thumb that the smallest (lowest MW) alkane will have the lowest BP and the largest (highest MW) will have the largest is just wrong. It works in this case because the molecules are all very small, "essentially" linear. Branching in higher weight alkanes makes it impossible to predict BP solely on the basis of MW. For $\ce{C5H12}$, neopentane has a BP of 9.5 °C, but n-pentane has a BP 35.9 to 36.3 °C. N-butane has a BP of −1 to 1 °C, but isobutane has a BP of −11.7 °C. $\endgroup$ – MaxW Feb 12 '19 at 20:12
  • 1
    $\begingroup$ Yet the pentanes all boil higher than the butanes according to MaxW's list. $\endgroup$ – TAR86 Feb 12 '19 at 20:26
  • 1
    $\begingroup$ My point was that the "rule" doesn't work for higher alkanes because there is a significant spread in the BP due to branching. $\endgroup$ – MaxW Feb 12 '19 at 20:37

It is true that Q1 and Q2 can be addressed having the same approach since all of them are hydrocarbons, which do not have any other forces acting on them other than the London dispersion and van der Waals forces. Also, they are homologous alkanes, which increase those forces according to their size. Yet, these forces also depend on how branched their molecular structures are. Therefore, I have to agree with MaxW's concern that isomerization on boiling points has an upper hand over increasing $-\ce{CH2}-$ units. For example, compare boiling point of n-decane ($\ce{C10H22}$), which is $\pu{174.1 ^{\mathrm{o}}C}$ with boiling point of 2,2,4-trimethyl-3-isopropylpentane ($\ce{C11H24}$) that is $\pu{173.3 ^{\mathrm{o}}C}$ (Ref.1). Even lower molecular weight examples are given in Ref.1: Easily, if you look at $\ce{C10H20}$ vs $\ce{C9H20}$ isomers, you find the boiling points of 2,2,5,5-hexamethylhexane, 2,2,4,5-hexamethylhexane, and 2,2,3,5-hexamethylhexane, all of which are $\ce{C10H22}$ isomers, are $137.5, 147.9,$ and $\pu{148.4 ^{\mathrm{o}}C}$, respectively. Yet, the boiling point of n-nonane, which has one less $\ce{CH2}$ group ($\ce{C9H20}$ isomer), is $\pu{150.8 ^{\mathrm{o}}C}$. Another isomer of $\ce{C10H22}$, 2,4,6-trimethylheptane has boiling point of $\pu{144.8 ^{\mathrm{o}}C}$, which is also lower than that of n-nonane.

All of these examples means, in order to predict boiling point simply by molecular weights, we may need additional information such as molecules structural features, etc. We could have used the trend of lower atomic (molar) mass having a lower boiling point on Q1 and Q2, only if it states that answers (C) and (D) are n-$\ce{C5H12}$ and n-$\ce{C4H10}$, respectively (or i-$\ce{C5H12}$ and i-$\ce{C4H10}$, respectively). Simply, needs to say all have similar structural features.

The trend of lower atomic (molar) mass having a lower boiling point can be applied only to Q4 because all answers are homo diatomic molecules.

In the case of Q3, there is another force acting of $\ce{HF}$ other than London dispersion and van der Waals forces: H-bonding, which would overcome effects on molecular mass and other forces in this case. The simplest reason is $\ce{F}$ the highest electrnegative atom and capable of having strongest hydrogen bonding. If you Googgled, you may find boiling point of $\ce{HF}$ is $\pu{19.5 ^{\mathrm{o}}C}$, while that of $\ce{HI}$ and $\ce{HBr}$ are $\pu{−35.4 ^{\mathrm{o}}C}$ and $\pu{−66 ^{\mathrm{o}}C}$, respectively.


  1. S. W. Ferris, Handbook of Hydrocarbons; Academic Press, Inc. Publishers: New York, NY, 1955, pp. 187.
| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.