Cambridge International AS Level Chemistry
weak permanent dipole–dipole force CH3 δ+ C
δ– O
CH3
δ– O
δ+ C CH3
CH3
Figure 4.39 Dipole–dipole forces in propanone.
For small molecules with the same number of electrons, permanent dipole–dipole forces are often stronger than van der Waals’ forces. For example, propanone (CH3COCH3, Mr = 58) has a higher boiling point than butane (CH3CH2CH2CH3, Mr = 58) (Figure 4.40). This means that more energy is needed to break the intermolecular forces between propanone molecules than between butane molecules.
CH3
CH2
CH2
CH3
butane, boiling point 0 °C 64
CH3 δ+ C
δ– O
When a hydrogen atom is covalently bonded to a very electronegative atom, the bond is very highly polarised. The δ+ charge on the hydrogen atom is high enough for a bond to be formed with a lone pair of electrons on the F, O or N atom of a neighbouring molecule (Figure 4.41). The force of attraction is about one-tenth of the strength of a normal covalent bond. For maximum bond strength, the angle between the covalent bond to the hydrogen atom and the hydrogen bond is usually 180°. H H
N
H
Figure 4.41 Hydrogen bonding between two ammonia molecules. A hydrogen bond is represented by a line of dots.
The average number of hydrogen bonds formed per molecule depends on: ■
the number of hydrogen atoms attached to F, O or N in the molecule the number of lone pairs present on the F, O or N.
■
propanone, boiling point 56 °C
Water has two hydrogen atoms and two lone pairs per molecule (Figure 4.42). So water is extensively hydrogen bonded with other water molecules. It has an average of two hydrogen bonds per molecule. H
The permanent dipole–dipole forces between propanone molecules are strong enough to make this substance a liquid at room temperature. There are only van der Waals’ forces between butane molecules. These forces are comparatively weak, so butane is a gas at room temperature. QUESTION 11 Bromine, Br2, and iodine monochloride, ICl, have the same number of electrons. But the boiling point of iodine monochloride is nearly 40 °C higher than the boiling point of bromine. Explain this difference.
Hydrogen bonding Hydrogen bonding is the strongest type of intermolecular force. For hydrogen bonding to occur between two molecules we need:
■
N
H
CH3
Figure 4.40 The difference in the boiling points of propanone and butane can be explained by the different types of intermolecular force between the molecules.
■
H H
one molecule having a hydrogen atom covalently bonded to F, O or N (the three most electronegative atoms) a second molecule having a F, O or N atom with an available lone pair of electrons.
H
δ+
H
δ+ H O δ–
H
O δ–
δ+
H
O δ–
Figure 4.42 Water can form, on average, two hydrogen bonds per molecule.
Ammonia is less extensively hydrogen bonded than water (see Figure 4.41). It can form, on average, only one hydrogen bond per molecule. Although each ammonia molecule has three hydrogen atoms attached to the nitrogen atom, it has only one lone pair of electrons that can be involved in hydrogen bond formation. QUESTION 12 Draw diagrams to show hydrogen bonding between the following molecules: a ethanol, C2H5OH, and water b ammonia and water c
two hydrogen fluoride molecules.