FORCES BETWEEN COVALENT MOLECULES 77
Hydrogen bonding
Hydrogen bonding is a special type of dipole–dipole attraction in which the hydrogen
atom acts as a bridge between two electronegative atoms. A hydrogen bond follows the
general formula
A——H· · · · · B
Examples of hydrogen bonding are F–H··· O, where A is F and B is O, and
F–H··· F, where A and B are F (see Example 5.6). These three atoms are
usually bonded in a straight line. A and B are the electronegative atoms, such as F, O
or N. Such atoms possess one or more lone pairs of electrons.
BOX 5.5
London dispersion forces and the shapes of molecules
Molecules of pentane (C 5 H 12 ) exist with the
same number of electrons, but very different
shapes. One form of C 5 H 12 has molecules
which have a cylindrical shape, whereas the
other has molecules which are spherical. The
cylindrical form has a boiling point of 36 °C,
while the spherical form boils at 10 °C,
suggesting that weaker London interactions
exist between the spherical molecules. The
valence electrons in the molecules with a
spherical shape are ‘hidden’ deeper within
the molecule than if the molecule is
cylindrical, and cannot interact as strongly
with neighbouring molecules (Fig. 5.14).
Fig. 5.14Different shapes of pentane
molecules.
Example 5.6
A molecule of hydrogen fluoride has a large dipole moment. The partial positive
charge on the hydrogen is attracted to the lone pair of electrons on the partially
negatively charged fluorine atom of a neighbouring molecule – it forms a hydrogen
bondwith the fluorine. Liquid hydrogen fluoride contains zigzag chains of
molecules of HF joined together by hydrogen bonds (Fig. 5.15).
Fig. 5.15Hydrogen bonding between molecules of
hydrogen fluoride (hydrogen bonds are usually represented
by dotted lines).
London forces
(i)Arrange the following in order of decreasing
boiling points. Explain your answer.
Inert gas Diameter of atom/pm
He 100
Ne 130
Ar 190
Kr 220
(ii)At room temperature, F 2 and Cl 2 are
gases, whereas Br 2 is a liquid and I 2 a solid.
Explain these observations in terms of the
strength of intermolecular forces within
these substances.
Exercise 5J