0.7 Molecular structure and physical properties  (Page 5/5)

Observation 3: dipole moments in polyatomic molecules

We might reasonably expect from our analysis to observe a dipole moment in any molecule formed from atoms withdifferent electronegativities. Although this must be the case for a diatomic molecule, this is not necessarily true for a polyatomicmolecule, i.e. one with more than two atoms. For example, carbon is more electronegative than hydrogen. However, the simplest moleculeformed from carbon and hydrogen ( e.g. $C{H}_4$ ) does not possess a dipole moment, as we see in . Similarly, oxygen is significantly more electronegative than carbon, yet $C{O}_2$ is a non-polar molecule. An analysis of molecular dipole moments in polyatomic molecules requires us to apply our understanding ofmolecular geometry.

Note that each $CO$ bond is expected to be polar, due to the unequal sharing of the electron pairs between the carbon and the oxygen. Thus, the carbonatom should have a slight positive charge and the oxygen atom a slight negative charge in each $CO$ bond. However, since each oxygen atom should have the same net negative charge, neither end of the molecule would display agreater affinity for an electric field. Moreover, because $C{O}_2$ is linear, the dipole in one $CO$ bond is exactly offset by the dipole in the opposite direction due to the other $CO$ bond. As measured by an electric field from a distance, the $C{O}_2$ molecule does not appear to have separated positive and negative charges and therefore does not display polarity. Thus, inpredicting molecular dipoles we must take into account both differences in electronegativity, which affect bond polarity, andoverall molecular geometry, which can produce cancellation of bond polarities.

Using this same argument, we can rationalize the zero molecular dipole moments observed for other molecules,such as methane, ethene and acetylene. In each of these molecules, the individual $CH$ bonds are polar. However, the symmetry of the molecule produces a cancellation of these bond dipoles overall, and none of thesemolecules have a molecular dipole moment.

As an example of how a molecular property like the dipole moment can affect the macroscopic property of asubstance, we can examine the boiling points of various compounds. The boiling point of a compound is determined by the strength ofthe forces between molecules of the compound: the stronger the force, the more energy is required to separate the molecules, thehigher the temperature required to provide this energy. Therefore, molecules with strong intermolecular forces have high boilingpoints.

We begin by comparing molecules which are similar in size, such as the hydrides $\mathrm{Si}{H}_4$ , $P{H}_3$ , and $S{H}_2$ from the third period. The boiling points at standard pressure for these molecules are, respectively, -111.8°C, -87.7°C, and-60.7°C. All three compounds are thus gases at room temperature and well below. These molecules have very similarmasses and have exactly the same number of electrons. However, the dipole moments of these molecules are very different. The dipolemoment of $\mathrm{Si}{H}_4$ , is 0.0D, the dipole moment of $P{H}_3$ is 0.58D, and the dipole moment of $S{H}_2$ is 0.97D. Note that, for these similar molecules, the higher the dipole moment, the higher the boiling point. Thus, molecules withlarger dipole moments generally have stronger intermolecular forces than similar molecules with smaller dipole moments. This is becausethe positive end of the dipole in one molecule can interact electrostatically with the negative end of the dipole in anothermolecules, and vice versa.

We note, however, that one cannot generally predict from dipole moment information only the relative boilingpoints of compounds of very dissimilar molecules

Review and discussion questions