Intermolecular forces, electronegativity and bond properties

1. Overview

Covalent bonds hold the atoms inside a molecule together. Much weaker intermolecular forces hold separate molecules to one another, and in simple molecular substances they decide the melting point, the boiling point and how easily the liquid evaporates. This section shows how electronegativity makes bonds, and some molecules, polar, then describes the types of van der Waals' forces: id-id, pd-pd and hydrogen bonding. It uses hydrogen bonding to explain three unusual properties of water.

Key Definitions

• Electronegativity: the power of an atom to attract electrons to itself; in a covalent bond, the more electronegative atom pulls the shared pair of electrons closer to itself.
• Polar bond: a covalent bond between atoms with different electronegativities, in which the shared pair is pulled towards the more electronegative atom, so that atom carries a partial negative charge (δ−) and the other atom a partial positive charge (δ+).
• Dipole moment: a measure of the separation of positive and negative charge in a bond or a molecule, which depends on the size of the partial charges and the distance between them; a molecule has an overall dipole moment, and is polar, when its bond dipoles do not cancel out.
• Van der Waals' forces: the general name for all intermolecular forces, that is, the attractions between molecular entities (molecules, or atoms of the noble gases) other than those caused by forming bonds.
• Instantaneous dipole-induced dipole (id-id) forces: attractions that arise when an uneven spread of electrons at one instant gives a molecule a temporary dipole, which induces a dipole in a neighbouring molecule; also called London dispersion forces, they act between all molecules and between noble gas atoms.
• Permanent dipole-permanent dipole (pd-pd) forces: attractions between the δ+ end of one polar molecule and the δ− end of a neighbouring polar molecule.
• Hydrogen bond: a special case of pd-pd force, in which a hydrogen atom covalently bonded to a highly electronegative atom (N, O or F; this course uses N-H and O-H groups) is attracted to a lone pair of electrons on an N, O or F atom of a neighbouring molecule.
• Surface tension: the tendency of the surface of a liquid to resist an increase in its area, caused by the intermolecular forces pulling the surface molecules inwards.

Content

Electronegativity and polar bonds

Electronegativity is the power of an atom to attract electrons to itself. In the Data Booklet's Pauling values, fluorine has the highest (4.0), then oxygen (3.5), then nitrogen and chlorine (3.0); carbon is 2.5 and hydrogen 2.1.

• Identical atoms (H-H, Cl-Cl) share their bonding pair equally. The bond is non-polar.
• Different atoms share it unequally. The pair sits closer to the more electronegative atom, which gains a partial negative charge, δ−, and leaves the other atom with a partial positive charge, δ+. The bond is polar: it has a bond dipole.

The larger the difference in electronegativity, the more polar the bond. A very large difference means the electrons are effectively transferred, and the bonding is ionic (section 3.1).

Bond	Electronegativities	Difference	Partial charges
C-H	C 2.5, H 2.1	0.4	only slightly polar; usually treated as non-polar
C-Cl	C 2.5, Cl 3.0	0.5	Cδ+, Clδ−
N-H	N 3.0, H 2.1	0.9	Nδ−, Hδ+
O-H	O 3.5, H 2.1	1.4	Oδ−, Hδ+
C-F	C 2.5, F 4.0	1.5	Cδ+, Fδ−

Dipole moments: the shape decides

Each bond dipole has a size and a direction. A molecule's dipole moment is the combined effect of all its bond dipoles, added the way forces are added, taking their directions into account. So the shape of the molecule (VSEPR, section 3.5) matters. If the bond dipoles cancel, the molecule is non-polar. If they do not, it is polar: it has a permanent dipole.

Molecule	Shape	Bond dipoles	Polar molecule?
CO₂	linear, 180°	two equal C=O dipoles, opposite directions: cancel	no
BF₃	trigonal planar, 120°	three equal B-F dipoles: cancel	no
HCN	linear, 180°	C-H and C≡N dipoles both point towards the N end: do not cancel	yes
NH₃	pyramidal, 107°	three N-H on one side of the N: do not cancel	yes
H₂O	non-linear, 104.5°	two O-H at an angle: do not cancel	yes

Worked example 1: polar bonds, but is the molecule polar?

Question: Boron trichloride, BCl₃, and phosphorus trichloride, PCl₃, each contain three bonds to chlorine. Pauling electronegativities: B 2.0, P 2.2, Cl 3.0. (a) Which molecule has the more polar bonds? Show the partial charges. (b) Which molecule has an overall dipole moment?

Step 1: electronegativity differences. $$\text{B-Cl: } 3.0 - 2.0 = 1.0 \qquad \text{P-Cl: } 3.0 - 2.2 = 0.8$$ Chlorine is the more electronegative atom in both, so in every bond Cl is δ−, and B or P is δ+.

Step 2: shapes. Boron (3 outer electrons) forms three bonding pairs and has no lone pair: trigonal planar, 120°, like BF₃. Phosphorus (5 outer electrons) has three bonding pairs and one lone pair: pyramidal, like NH₃.

Step 3: combine. In BCl₃ the three equal dipoles point outwards in one plane at 120° to each other and cancel exactly. In PCl₃ all three Cl atoms lie on the side of P away from the lone pair, so their dipoles do not cancel.

Answer: (a) BCl₃ has the more polar bonds (difference 1.0, against 0.8). (b) Only PCl₃ has an overall dipole moment. BCl₃ molecules attract one another by id-id forces only; PCl₃ molecules by id-id and pd-pd forces.

Van der Waals' forces

Van der Waals' forces is the general name for every kind of intermolecular force: the attractions between molecular entities (molecules, or atoms of a noble gas) that do not come from forming bonds. There are two types: id-id forces and pd-pd forces, and hydrogen bonding is a special case of pd-pd. When a question asks which forces act, "van der Waals' forces" is not specific enough: name the type.

Instantaneous dipole-induced dipole (id-id) forces

1. Electrons are always moving. At any instant, more of a molecule's electron cloud may be on one side than the other, giving an instantaneous dipole.
2. This dipole repels or attracts the electrons of a neighbouring molecule, creating an induced dipole lined up so that the two molecules attract.
3. These dipoles vanish a moment later, but new ones keep forming throughout the substance, so there is always an overall attraction.

id-id forces, also called London dispersion forces, act between all molecules, polar or not, and between noble gas atoms. Between non-polar molecules such as Cl₂, CH₄ and CO₂ they are the only intermolecular force. They are stronger when:

• there are more electrons: a larger electron cloud is distorted more easily, so larger dipoles form. Cl₂ (34 electrons) is a gas at room temperature, Br₂ (70) a liquid and I₂ (106) a solid;
• molecules have more contact: a long unbranched chain touches its neighbours at more points than a compact branched isomer with the same number of electrons, so it boils higher.

Permanent dipole-permanent dipole (pd-pd) forces

Polar molecules tend to line up so that the δ+ end of one sits near the δ− end of the next. That attraction is a pd-pd force. It acts as well as id-id forces, never instead of them.

For molecules with similar numbers of electrons, a polar substance usually boils higher than a non-polar one (Worked example 2). For molecules of different sizes, the id-id part often matters more: HCl, HBr and HI boil at higher and higher temperatures although the H-halogen bond becomes less polar, because the number of electrons rises (group 17, the dashed line in the diagram further down).

Hydrogen bonding

A hydrogen bond is a special case of a pd-pd force. It needs:

1. a hydrogen atom covalently bonded to a highly electronegative atom: N, O or F (this course uses N-H and O-H groups), and
2. a lone pair on an N, O or F atom of a neighbouring molecule.

A hydrogen bond is much stronger than a typical pd-pd force. The N-H and O-H bonds are very polar, so the H carries a large δ+. Hydrogen has no inner electrons, so with the bonding pair pulled away its nucleus is almost exposed. And the tiny H atom lets the lone pair get very close.

To draw one between two water molecules: put δ− on O and δ+ on H in the O-H bond; draw a lone pair on the O of the other molecule; join the H to that lone pair with a dashed line labelled "hydrogen bond"; keep O-H···O in a straight line, at about 180°.

Water, ammonia, alcohols, carboxylic acids and primary amines all form hydrogen bonds. Ammonia and small alcohols dissolve well in water because they form hydrogen bonds with water molecules. Hydrogen fluoride forms them too, through H-F.

No hydrogen bonds form when the H is bonded to carbon: fluoromethane, CH₃F, contains H and F, but its H atoms are on C. HCl and H₂S have id-id and pd-pd forces, but no hydrogen bonds. Chlorine has the same Pauling value as nitrogen (3.0), but it is a much larger atom, so its lone pairs are too spread out to attract the H closely.

Water and ammonia

Each hydrogen bond needs one δ+ H and one lone pair.

	Water, H₂O	Ammonia, NH₃
δ+ H atoms per molecule	2	3
Lone pairs per molecule	2	1
Bond to H (electronegativity difference)	O-H (1.4)	N-H (0.9)
Average hydrogen bonds per molecule, at most	2	1
Boiling point	100 °C	−33 °C

A water molecule can take part in up to four hydrogen bonds: two through its H atoms and two through its lone pairs. Each bond is shared by two molecules, so when every molecule forms all four (as in ice), the average is two per molecule. Ammonia's single lone pair limits it to an average of at most one. Water's hydrogen bonds are also individually stronger, because O is more electronegative than N. More hydrogen bonds, and stronger ones, give water the higher boiling point.

Worked example 2: three molecules with 18 electrons

Question: Ethane, C₂H₆, fluoromethane, CH₃F, and methylamine, CH₃NH₂, boil at −89 °C, −78 °C and −6 °C, in some order. Match each boiling point to its compound and explain the order. (Atomic numbers: H 1, C 6, N 7, F 9. Electronegativities: H 2.1, C 2.5, N 3.0, F 4.0.)

Step 1: count the electrons. $$\text{C}_2\text{H}_6:\ 2(6) + 6(1) = 18 \qquad \text{CH}_3\text{F}:\ 6 + 3(1) + 9 = 18 \qquad \text{CH}_3\text{NH}_2:\ 6 + 5(1) + 7 = 18$$ Equal numbers of electrons give similar id-id forces, so the differences come mainly from the other forces.

Step 2: polarity and hydrogen bonding.

• Ethane: only C-C (difference 0) and C-H (0.4) bonds, arranged symmetrically. Non-polar: id-id only.
• Fluoromethane: the C-F bond is very polar (1.5), and with one F and three H around the C, the dipoles cannot cancel. Polar: id-id and pd-pd. Its H atoms are on C, so no hydrogen bonds.
• Methylamine: N-H bonds (0.9) and a lone pair on N, so hydrogen bonds as well as id-id.

Step 3: rank. Stronger intermolecular forces need more energy to separate the molecules, so the boiling point rises from ethane to fluoromethane to methylamine.

Answer: ethane −89 °C (id-id only); fluoromethane −78 °C (id-id and pd-pd); methylamine −6 °C (id-id and hydrogen bonds).

Why water is unusual

Three properties of water are anomalous: they do not fit the pattern shown by similar substances. Hydrogen bonding explains all three.

1. Relatively high melting and boiling points.

For the hydrides of periods 3 to 5, boiling points rise steadily with the number of electrons, as id-id forces grow. H₂O has only 10 electrons, fewer than H₂S (18), yet boils at 373 K, about 160 K above H₂S (213 K). Boiling water means overcoming the hydrogen bonds between its molecules, which are much stronger than the id-id and pd-pd forces between H₂S molecules. Ice melts at a much higher temperature than solid H₂S for the same reason. NH₃ and HF sit well above their groups' trends too; CH₄, with no hydrogen bonds, does not. Boiling only separates the molecules: the O-H bonds inside them do not break, and steam is still H₂O molecules.

2. Relatively high surface tension.
A molecule inside the liquid is attracted by its neighbours equally in all directions, on average. A molecule at the surface has neighbours beside and below it but none above, so there is a net pull into the liquid, and the surface behaves like a stretched skin. Enlarging the surface means bringing molecules up from inside, overcoming some of their attractions. Because water's hydrogen bonds are strong compared with other intermolecular forces, its surface tension is high, about four times that of hexane, whose molecules attract only by id-id forces. Small insects can stand on a pond.

3. Ice is less dense than liquid water.
In ice, each molecule forms all four of its possible hydrogen bonds, so four neighbours sit tetrahedrally around each O atom. This regular lattice holds the molecules apart and leaves a lot of empty space: ice has an open structure. When ice melts, some hydrogen bonds break, the lattice partly collapses, and the molecules pack closer. The liquid is therefore denser, and ice floats. Most solids are denser than their liquids, so water is unusual.

Worked example 3: water expanding as it freezes

Question: A glass jar is filled with 375 cm³ of liquid water at 0 °C and placed in a freezer until all the water has frozen. At 0 °C the density of liquid water is 1.00 g cm⁻³ and the density of ice is 0.917 g cm⁻³. Assume no water is lost and the ice is still at 0 °C when its volume is found. (a) Calculate the volume of ice formed. (b) Calculate the percentage increase in volume. (c) Explain the expansion.

Step 1: mass of water. Density is mass per unit volume, $\rho = m/V$. Freezing does not change the mass. $$m = \rho V = 1.00 \times 375 = 375\ \text{g}$$

Step 2: volume of ice. Keep the unrounded value. $$V_{\text{ice}} = \frac{m}{\rho_{\text{ice}}} = \frac{375}{0.917} = 408.94\ \text{cm}^3$$ (Ice that cools below 0 °C contracts very slightly, changing only the last figure.)

Step 3: percentage increase. $$\frac{408.94 - 375}{375} \times 100 = 9.05\%$$ The increase itself, 409 − 375 = 34 cm³, is known to only 2 significant figures, so give the percentage to 2 s.f.

Answer: (a) 409 cm³ (3 s.f.). (b) 9.1% (2 s.f.), enough to crack a full, closed jar. (c) On freezing, every molecule forms four hydrogen bonds, arranged tetrahedrally. These hold the molecules apart in an open lattice with more empty space than in the liquid, so the same mass takes up more volume.

Bonds are stronger than intermolecular forces

In general, ionic, covalent and metallic bonding are all stronger than intermolecular forces.

Energy. Turning water at 100 °C into steam takes about 41 kJ mol⁻¹, most of it used to overcome the intermolecular forces. Breaking both O-H bonds in a mole of gaseous water molecules needs about $2 \times 460 = 920$ kJ mol⁻¹ (Data Booklet average): more than 20 times as much.

Melting points. Melting overcomes some of the attractions between the particles, so that they can move past one another. In a simple molecular substance these attractions are weak intermolecular forces. In a giant ionic or metallic structure they are strong ionic or metallic bonds acting throughout the lattice, so far more energy is needed.

Substance	Structure	Attractions between the particles	Melting point / °C
iodine, I₂	simple molecular	intermolecular (id-id)	114
calcium fluoride, CaF₂	giant ionic	ionic bonding	1418
nickel, Ni	giant metallic	metallic bonding	1455

A strong covalent bond inside a molecule does not by itself give a high boiling point. N≡N (944 kJ mol⁻¹) is one of the strongest bonds in the Data Booklet, yet nitrogen is a gas far below 0 °C, because only weak id-id forces act between N₂ molecules. "In general" matters: strengths vary within each type, so the statement compares typical values.

Key Equations

Electronegativity difference ($\chi$ = Pauling electronegativity from the Data Booklet; no units). The atom with the larger $\chi$ carries δ−; 0 means a non-polar bond, and a larger difference means a more polar bond: $$\text{difference} = \chi(\text{more electronegative atom}) - \chi(\text{less electronegative atom})$$

Electrons in a molecule ($N_{\text{e}}$ = number of electrons; $Z$ = atomic number of each atom). Similar $N_{\text{e}}$ means roughly similar id-id forces: $$N_{\text{e}} = \sum Z \qquad \text{e.g. CH}_3\text{F: } 6 + 3(1) + 9 = 18$$

Density ($\rho$ in g cm⁻³; $m$ = mass in g; $V$ = volume in cm³): $$\rho = \frac{m}{V}$$

Energy to break bonds ($E$ = bond energy in kJ mol⁻¹, for gases, from the Data Booklet): $$\text{energy} = \text{number of bonds} \times E \qquad \text{e.g. both O-H bonds in H}_2\text{O: } 2 \times 460 = 920\ \text{kJ mol}^{-1}$$

Common Mistakes to Avoid

• Mistake: answering "van der Waals' forces" when asked which intermolecular forces act. Fix: name every type present: between PCl₃ molecules, id-id and pd-pd forces; between NH₃ molecules, id-id forces and hydrogen bonds.
• Mistake: saying covalent bonds break when a simple molecular substance boils. Fix: boiling overcomes intermolecular forces only. Boiling water takes about 41 kJ mol⁻¹, far less than the 920 kJ mol⁻¹ that breaking both O-H bonds would need.
• Mistake: drawing a hydrogen bond to an H atom bonded to carbon. Fix: the H must be bonded to N or O. In ethanol, CH₃CH₂OH, only the H of the O-H group can; the five H atoms on carbon cannot.
• Mistake: a hydrogen-bond diagram with parts missing, drawn as a dative arrow, or showing two hydrogen bonds between one pair of ammonia molecules. Fix: show δ+ on H, δ− on N, the lone pair, and one dashed, labelled bond from the H to that lone pair, with N-H···N in a straight line.
• Mistake: explaining a rise in boiling point down a group with electronegativity, nuclear charge or stronger permanent dipoles. Fix: the rise comes from stronger id-id forces as the number of electrons increases: Cl₂ has 34, Br₂ 70 and I₂ 106. The same is true for HCl, HBr and HI, even though their bonds become less polar.
• Mistake: deciding that polar bonds always make a polar molecule. Fix: check the shape. CO₂ has two polar C=O bonds but is linear, so the dipoles cancel and it is non-polar, like BCl₃ in Worked example 1.
• Mistake: giving two as the most hydrogen bonds one water molecule can form. Fix: one water molecule can take part in up to four; two is the average per molecule, because each bond is shared by two molecules.
• Mistake: calling the attractions in a giant structure "intermolecular forces". Fix: sodium chloride is held by ionic bonding between ions. Use "intermolecular" only for forces between separate molecules.

Exam Tips

• Explain a difference in boiling point in four parts: name the intermolecular forces in each substance; say which are stronger; give the reason (more electrons, a permanent dipole or hydrogen bonds); say that more energy is needed to overcome the stronger forces.
• Draw a hydrogen bond with δ+ and δ−, the lone pair, and a dashed line labelled "hydrogen bond", with O-H···O (or N-H···N) in a straight line. Draw only as many molecules and bonds as the question asks for.
• Describe id-id forces as a chain: uneven electrons at one instant, an instantaneous dipole, an induced dipole in a neighbour, attraction.
• Explain polarity with numbers: give the electronegativity difference, mark δ+ and δ−, then use the shape to say whether the bond dipoles cancel.
• State the comparison exactly: in general, ionic, covalent and metallic bonding are stronger than intermolecular forces.
• Explain each anomalous property of water separately, always through hydrogen bonds: energy to overcome them (melting and boiling points), net inward pull (surface tension), open tetrahedral lattice (density of ice).