Intermolecular forces
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Some exam "Dos and Don'ts": Intermolecular forces
Advice on answering questions on topics often highlighted by examiners as being poorly answered in exams. Make sure that the detail covered by the question is included in your syllabus specification.
INTERMOLECULAR FORCES
Exam questions involving intermolecular forces usually either ask directly about the forces between given molecules, or indirectly, by asking you to compare the melting or boiling points or solubilities of two or more molecules and explain the differences or trends.
Boiling and melting points depend on the strength of the intermolecular forces, so make sure that in your answer you make the connection between the strength of the forces and the effect on these properties.
Remember that volatility is a measure of the ease with which a substance evaporates: the lower the boiling point, the higher the volatility is likely to be.
It should go without saying that the strength of covalent bonds inside a molecule are not responsible for the strength of attraction between molecules, but some candidates still persist in, for example, attributing the high boiling point of water to the strong O-H bonds in the molecule, rather than the strong H-bonding between the molecules.
van der Waals' forces
These are the only forces between non-polar molecules, but are also present between polar molecules - and even ions. van der Waals' forces are usually regarded as weak, but this is only compared to the other types of intermolecular forces between molecules of similar size. In very large non-polar molecules, such as in polymers like poly(ethene) or poly(propene), the van der Waals' forces can be strong - as a mountaineer hanging on poly(propene) rope should know!
When other, stronger intermolecular forces are present between small molecules, van der Waals' forces are often ignored. However, if the question appears to be asking for all the forces present between polar molecules, include van der Waals' forces together with dipole-dipole attraction - and H-bonding, if applicable.
van der Waals' forces depend on two main factors:
a) the number of electrons in the molecule - the more, the stronger the forces.
This should be used for comparing non-polar molecules of a similar type, but with increasing molecular mass, e.g. straight-chain alkanes.
Some examiners may accept increasing RMM (relative molecular mass ) instead of no. of electrons, but avoid vague terms like 'size'.
b) the surface contact area between the molecules.
Use this reason when comparing molecules of very similar or identical RMM, such as the isomers of C5H12. A sketch using outline shapes of two straight-chain or two branched-chain molecules to show the difference in contact area would improve your answer.
Dipole-dipole forces - attraction between molecules with permanent dipoles
Remember that C-H bonds are not regarded as polar at this level.
Comparing the dipole-dipole forces between polar molecules is difficult without more information than is usually supplied in exam questions, so comparisons often involve non-polar vs polar molecules, or two polar molecules, one of which can form H-bonds as well, e.g. an aldehyde and an alcohol of similar RMM.
Remember that molecules of polar substances, such as propanone, may only have dipole-dipole attraction between them (and hence are relatively volatile) whereas with solvents such as water they can form strong H-bonds. Propanone is thus completely miscible with water.
Hydrogen bonding (H-bonding)
This is the strongest type of intermolecular attraction - between molecules of similar molecular mass.
A H-atom attached to O, N, or F but not C (normally) must be present.
Candidates are often asked for a diagram showing the H-bonding between two identical or two different molecules. To be sure of maximum marks (which can be up to 4) be sure to show:
>the H-bond as a dashed or dotted line and label it.
>the polarity of the X-H (X = O,N or F) bond involved the H-bonding, using the d+/d- notation. Also show the polarity of the bond containing the O, N or F atom with which the H-bond is formed, and a lone pair of electrons on this atom as part of the H-bond.
>the angle between the H-bond and the X-H covalent bond as 180 o - the orientation when H-bonding is strongest.
When questions involve molecules containing H-atoms attached to electronegative atoms other than O,N or F, such as S or Cl, explain the lack of H-bonding in terms of the lower polarity of the X-H bond (due to a smaller difference in electronegativity between X and H than between H and O,N or F).
Solubility and miscibility
When asked to explain the miscibility, or lack thereof, of two liquids, A & B, remember to consider:
a) the force(s) of attraction between molecules of A before mixing.
b) the force(s) of attraction between molecules of B before mixing.
c) the new force(s) of attraction that could arise between molecules of A and B, if mixed.
In general, when there are strong forces (e.g. H-bonding) between molecules of one type before mixing, but no similarly strong forces of attraction likely between A and B, then A and B are unlikely to be miscible, e.g. hexane and water.
When both A and B have only weak forces between their respective molecules, and similar forces between A and B, then they are likely to be miscible, e.g. hexane and tetrachloromethane.
When there are strong intermolecular forces between both molecules of A and between molecules of B, and between A and B, then A and B are likely to be miscible, e.g. water and ethanoic acid.
A trickier question can involve the decreasing solubility of, for example, an alcohol in water as the chain length increases. H-bonding is the most important factor in making the small alcohols soluble in water. As the non-polar hydrocarbon chain of the alcohols increases in length, more and more H-bonding between water molecules would be disrupted if the alcohol dissolved, without any new strong attraction between the hydrocarbon chains and the water molecules replacing it.
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updated: 01 June 2007
