Intermolecular Forces and Hydrogen Bonding

Covalent bonds hold the atoms together inside a molecule, but something much weaker holds separate molecules together in a liquid or a solid. Those attractions between molecules, the intermolecular forces, decide boiling points, volatility and solubility, and they explain why water, with a tiny molecule, is a liquid at room temperature. This note covers bond polarity and dipole moments, the umbrella term van der Waals' forces, the two types you must name (instantaneous dipole–induced dipole and permanent dipole–permanent dipole), and hydrogen bonding with the anomalous properties of water. "Explain the difference in boiling points" questions built on these ideas appear in nearly every Paper 2.

Bonds versus intermolecular forces

There are two very different kinds of attraction in a molecular substance:

• intramolecular forces: the covalent bonds within each molecule;
• intermolecular forces: the attractions between neighbouring molecules.

When a molecular substance such as water boils, only the intermolecular forces are overcome. The molecules separate, but each $\ce{H2O}$ stays intact: no O–H bonds break. Steam is still $\ce{H2O}$.

So molecular substances have low melting and boiling points compared with ionic, metallic and giant covalent substances, where strong bonds must be broken.

Bond polarity and dipole moments

When two atoms with different electronegativities share a pair of electrons, the more electronegative atom pulls the pair towards itself. It gains a small negative charge, $\delta-$, and the other atom a small positive charge, $\delta+$. The bond is polar, and it has a dipole: a separation of positive and negative charge.

The bigger the difference in electronegativity, the more polar the bond. Using Pauling values: C–H ($2.5$ and $2.1$, difference $0.4$) is almost non-polar; O–H ($3.5$ and $2.1$, difference $1.4$) is strongly polar.

A molecule's dipole moment measures its overall separation of charge. It depends on two things:

1. the polarity of each bond;
2. the shape of the molecule, which decides whether the bond dipoles cancel.

A quick test in the laboratory: a charged rod held next to a thin stream of a polar liquid (water, propanone) deflects the stream; a non-polar liquid such as hexane is barely affected.

Van der Waals' forces

Under this umbrella the syllabus names two types: instantaneous dipole–induced dipole forces (present between all molecules) and permanent dipole–permanent dipole forces (only between polar molecules), of which hydrogen bonding is a special, strong case.

Instantaneous dipole–induced dipole forces

Even in a non-polar molecule such as $\ce{Cl2}$ or an atom of argon, the electrons are constantly moving. At any instant the electron cloud may be slightly more on one side than the other, giving a temporary, instantaneous dipole. This instantaneous dipole repels or attracts the electrons of a neighbouring molecule, inducing a dipole in it, lined up so that the two attract. A moment later the electrons have moved, the dipoles have changed, and new ones form. The net result is a weak attraction between all molecules all the time.

These forces exist between all molecules and atoms, polar or not. Their strength depends on:

• the number of electrons: more electrons give a larger, more easily distorted electron cloud, so larger instantaneous dipoles. Boiling points rise with molecular size (and so with $M_r$).
• the area of contact between molecules: long, unbranched molecules can lie alongside each other with more points of contact than compact, branched ones.

substance	electrons per molecule	boiling point / °C
$\ce{F2}$	18	$-188$
$\ce{Cl2}$	34	$-34$
$\ce{Br2}$	70	$59$
$\ce{I2}$	106	$184$

Down Group 17 the halogens change from gases ($\ce{F2}$, $\ce{Cl2}$) to a liquid ($\ce{Br2}$) to a solid ($\ce{I2}$): volatility decreases because id-id forces get stronger as the number of electrons increases.

The isomers of pentane ($\ce{C5H12}$, all with 42 electrons) show the effect of shape:

isomer	shape	boiling point / °C
pentane	unbranched chain	$36$
2-methylbutane	one branch	$28$
2,2-dimethylpropane	almost spherical	$10$

Permanent dipole–permanent dipole forces

Polar molecules have a permanent dipole. Neighbouring molecules line up so that the $\delta+$ end of one is near the $\delta-$ end of the next, and they attract. These permanent dipole–permanent dipole (pd-pd) forces act in addition to the id-id forces that all molecules have.

Compare two molecules with almost the same number of electrons:

substance	$M_r$	polar?	forces	boiling point / °C
butane, $\ce{CH3CH2CH2CH3}$	58	no	id-id only	$-1$
propanone, $\ce{CH3COCH3}$	58	yes (C=O)	id-id and pd-pd	$56$

Hydrogen bonding

What a hydrogen bond is

Why is it so much stronger than an ordinary pd-pd force? Three features combine:

1. N, O and F are the most electronegative elements, so the N–H, O–H and F–H bonds are very polar and the hydrogen carries a large $\delta+$ charge.
2. Hydrogen has no inner electrons. When its bonding pair is pulled away, its nucleus (a bare proton) is almost exposed, and it is very small, so it can get very close to a lone pair on a neighbouring molecule.
3. N, O and F are small atoms with concentrated lone pairs for the hydrogen to be attracted to.

The syllabus limits hydrogen bonding to molecules containing N–H and O–H groups, with ammonia and water as the simple examples (hydrogen fluoride also forms them, and you may see it in data questions).

Ammonia and water

• In ammonia, each molecule has three N–H bonds but only one lone pair, so on average each molecule forms only one hydrogen bond as donor and one as acceptor.
• In water, each molecule has two O–H bonds and two lone pairs, so each molecule can form up to four hydrogen bonds (two through its own hydrogens, two through its lone pairs). This is why water's boiling point ($100\ ^\circ\text{C}$) is far above ammonia's ($-33\ ^\circ\text{C}$) and above hydrogen fluoride's ($20\ ^\circ\text{C}$): fluorine has three lone pairs but only one H to donate, so HF averages fewer hydrogen bonds per molecule than water.

Boiling points of the hydrides

For the hydrides of Groups 14 to 17, boiling point normally increases down each group because the number of electrons, and so the id-id forces, increase. The Period 2 hydrides of Groups 15, 16 and 17 break the pattern.

period	Group 14	Group 15	Group 16	Group 17
2	$\ce{CH4}$ $-162$	$\ce{NH3}$ $-33$	$\ce{H2O}$ $100$	$\ce{HF}$ $20$
3	$\ce{SiH4}$ $-112$	$\ce{PH3}$ $-88$	$\ce{H2S}$ $-60$	$\ce{HCl}$ $-85$
4	$\ce{GeH4}$ $-88$	$\ce{AsH3}$ $-62$	$\ce{H2Se}$ $-41$	$\ce{HBr}$ $-67$
5	$\ce{SnH4}$ $-52$	$\ce{SbH3}$ $-17$	$\ce{H2Te}$ $-2$	$\ce{HI}$ $-35$

(boiling points in °C)

$\ce{NH3}$, $\ce{H2O}$ and $\ce{HF}$ boil far higher than the trend predicts because they form hydrogen bonds, which need much more energy to overcome. Methane follows the trend: carbon is not electronegative enough for hydrogen bonding, and $\ce{CH4}$ is non-polar.

The anomalous properties of water

Water has three properties that are unusual for such a small molecule, and all three come from its extensive hydrogen bonding.

1. Relatively high melting and boiling points. By comparison with $\ce{H2S}$, $\ce{H2Se}$ and $\ce{H2Te}$, water should boil at about $-80\ ^\circ\text{C}$. Instead it boils at $100\ ^\circ\text{C}$, because a large amount of energy is needed to overcome the many hydrogen bonds between molecules (up to four per molecule), which are much stronger than the id-id forces in the other hydrides.

2. Relatively high surface tension. A molecule inside the liquid is pulled equally in all directions by hydrogen bonds to its neighbours. A molecule at the surface has neighbours only beside and below it, so it is pulled inwards. Because the hydrogen bonds are strong, the surface behaves like a stretched skin: water forms near-spherical droplets and can support small insects or a carefully placed needle.

3. Ice is less dense than liquid water. In ice, every water molecule forms four hydrogen bonds arranged tetrahedrally around the oxygen (two through its hydrogens, two through its lone pairs). This holds the molecules in a rigid, open lattice with a lot of empty space. When ice melts, some hydrogen bonds break, the open structure collapses and the molecules pack more closely. So ice floats on water. (Liquid water is densest at about $4\ ^\circ\text{C}$.)

The density effect matters: ponds and lakes freeze from the top down, and the floating ice insulates the water beneath, so aquatic life survives the winter.

Hydrogen bonding and solubility

Substances that can hydrogen bond with water tend to dissolve in it. Ammonia is extremely soluble in water, and ethanol mixes with water in all proportions, because new hydrogen bonds form between their molecules and water molecules, releasing energy that compensates for breaking the hydrogen bonds in pure water. Non-polar substances such as hexane cannot form these bonds and do not dissolve.

Choosing the strongest force: a method

Worked examples

Practice

Answers

1. The intermolecular forces between molecular entities other than those due to bond formation (a generic term for all intermolecular forces). Types: instantaneous dipole–induced dipole (id-id, London dispersion) forces; permanent dipole–permanent dipole (pd-pd) forces, including hydrogen bonding.
2. The electrons in an argon atom move randomly, so at any instant the electron cloud may be unevenly distributed, giving an instantaneous dipole. This induces a dipole in a neighbouring atom by attracting or repelling its electrons. The instantaneous dipole and the induced dipole attract. The dipoles constantly disappear and re-form, giving a weak, continuous attraction.
3. (a) Non-polar: four polar C–Cl bonds arranged symmetrically (tetrahedral) cancel. (b) Polar: tetrahedral but with one H and three Cl, so the dipoles do not cancel. (c) Polar: non-linear (lone pair on S), so the S=O dipoles do not cancel. (d) Non-polar: trigonal planar and symmetrical, so the B–F dipoles cancel.
4. Two $\ce{NH3}$ molecules. On the first, an N–H bond with $\delta+$ on H and $\delta-$ on N. On the second, N labelled $\delta-$ with its lone pair drawn. A dashed line from the H of the first molecule to the lone pair on N of the second, with N–H···N in a straight line.
5. Both are $\ce{C5H12}$ with the same number of electrons and only id-id forces. Pentane is an unbranched chain, so molecules can lie close alongside one another with a large area of contact; 2,2-dimethylpropane is compact and nearly spherical, with a smaller area of contact. The id-id forces between pentane molecules are stronger, so more energy is needed to separate them.
6. Propane ($-42\ ^\circ\text{C}$) < ethanal ($20\ ^\circ\text{C}$) < ethanol ($78\ ^\circ\text{C}$). Propane is non-polar: id-id only. Ethanal has a polar C=O bond: pd-pd as well as id-id. Ethanol has an O–H group: hydrogen bonding, the strongest intermolecular force. Similar numbers of electrons means similar id-id forces, so the type of additional force decides the order.
7. Each water molecule has two H atoms and two lone pairs, so it can form up to two hydrogen bonds as a donor and two as an acceptor: an average of two hydrogen bonds per molecule in the liquid. An HF molecule has only one H atom, so although its hydrogen bonds are individually stronger, it averages only one hydrogen bond per molecule. More energy is needed to overcome the larger number of hydrogen bonds in water.
8. Ammonia has N–H bonds and a lone pair on N, so it forms hydrogen bonds with water molecules (both as donor and acceptor). The energy released in forming these compensates for breaking hydrogen bonds between water molecules. Methane is non-polar and cannot hydrogen bond with water, so it does not dissolve to any great extent.
9. Both molecules are polar and have both pd-pd and id-id forces. HCl is more polar, so its pd-pd forces are slightly stronger. But HBr has many more electrons (36 against 18), so its id-id forces are considerably stronger. The increase in id-id forces outweighs the decrease in pd-pd forces, so overall the intermolecular forces in HBr are stronger and its boiling point is higher. The id-id forces are the dominant contribution in both.
10. Two molecules face each other: the O–H of each molecule hydrogen bonds to the C=O oxygen of the other, forming a ring (a cyclic dimer) held by two hydrogen bonds. In the liquid, ethanoic acid forms strong hydrogen bonds through both its O–H and C=O groups, and the dimers have twice as many electrons, giving stronger id-id forces between them; more energy is needed to boil it than propan-1-ol, which forms only O–H···O hydrogen bonds between single molecules. Because some molecules remain paired as $\ce{(CH3COOH)2}$ ($M_r = 120$) in the vapour, the average $M_r$ measured is between 60 and 120, so greater than 60.