I recently encountered a question on a test:

Substances whose molecules have a high degree of symmetry will have:

A. a low specific heat
B. a high melting point
C. a high heat of fusion
D. a low heat of vaporization
E. strong Van der Waals forces

Looking through a few compounds, I found that $\ce{CO2}$ is symmetric (it's shape is a straight line), $\ce{CH4}$ is symmetric (tetrahedrons seem symmetric). Through these examples, I was able to eliminate A, B, and C, and have a suspicious feeling about E, so I guessed D and was right.

I was wondering what exactly "high degree of symmetry" meant. Through my math courses, I knew about rotational symmetry (where if you turn any 2-D shape in a plane by n degrees, it will look identical to the shape before the rotation). Is a high degree of symmetry when the lowest possible value of n for a compound is over some constant, lets say 90 degrees, or is it something else?

  • $\begingroup$ Answer B also can be defended durene has one more plane of symmetry in comparison to its isomers and much higher melting point. $\endgroup$
    – Mithoron
    Mar 9 '17 at 0:56
  • $\begingroup$ Symmetry in 3D is much more complicated than in 2D. So "high degree of symmetry" is an informal expression with hardly any rigorous meaning. If the question asked about central symmetry in particular, then B would be the right answer indeed. $\endgroup$ Mar 9 '17 at 11:31

I think that the question meant to test your knowledge about the polarity of simple molecules.

The fact that symmetric molecules contain identical bond polarities between the central atom and surrounding atoms, the net dipole moment usually tends to be zero, creating a nonpolar molecule.

In other words, I think symmetry in this case refers to the fact that all of the surrounding atoms are identical.


Although the term 'high degree of symmetry' is somewhat subjective the only molecules with dipoles are those that belong to the point groups $C_1$ (no symmetry elements other than rotation by 360 degrees), the group $C_s$ which has only one mirror plane and the groups $C_n$ ($n \ge 2$ ) which have rotation axes of $2\pi/n$ degrees and $C_{nv}$ which in addition have a (vertical) mirror plane along a rotation axis. An example of a $C_{2v}$ molecule is $\ce{H2O}$, and $\ce{NH3}$ is $C_{3v}$. All these point groups do perhaps have 'lower' symmetry than say an octahedral group and so the question suggests that molecules with low symmetry have a dipole moment and the rest don't. (You can see 3D models of many molecules of all common point groups at molecule-viewer.com and add/remove their symmetry elements)

As @Dan Burden suggest the question is therefore most probably to do with the presence or absence of a dipole moment. One might suppose that a dipole will cause the molecule to have a larger van der waals interaction energy due to dipole-dipole interaction, than a similar molecule without such a dipole and so have higher boiling and melting points. Isomers for example, could be used to test this.

Three such could be 1,2-, 1,3- and 1,4-dichlorobenzene; I chose at random. The 1,2- and 1,3- have a dipole ($C_{2v}$ point group) but the 1,4- does not ($D_{2h}$ point group). The data (from Wikipedia pages) are below

boiling point (centigrade) $1,2 = 180.5, ~~~~ 1,3 = 172,~~~~ 1,4 = 174 $ so not much difference

melting point(centigrade) $1,2 = -17, ~~~~1,3= -22, ~~~~ 1,4 = 53$ not what one would expect if dipole interaction is important, quite the opposite in fact.

vapour pressure $1,2= 1.0, ~~~~ 1,4 = 1.3$ mmHg so again, not much difference here

To me the question does not have clear answers and it seems possible to choose molecules to support whatever the 'official' answers are or to show the opposite.


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