Electronegativity, polarity and intermolecular forces: the chain from a bond's electronegativity difference to a molecule's boiling point

Electronegativity is how strongly an atom pulls shared electrons toward itself, and the difference between two bonded atoms' values tells you the bond: near zero, nonpolar covalent; moderate, polar covalent; large, ionic. A molecule is polar only if its polar bonds do not cancel by symmetry. The kind of polarity then picks the intermolecular force, and the force sets the boiling point. Below, that chain is run on three liquids of almost the same mass whose boiling points are 120 degrees apart, then on three you can rank yourself.

Most courses teach these as three chapters. They are one chain, and a test question that asks you to rank boiling points is asking you to run all of it.

Step 1: the electronegativity difference names the bond

Use these values on Pauling's scale, stated here as givens. They are the one-decimal values Pauling published in 1960 and that many Grade 12 textbooks print.

Element	H	C	N	O	F	Cl	Br	Na
Electronegativity	2.1	2.5	3.0	3.5	4.0	3.0	2.8	0.9

Current reference tables print two decimals (oxygen 3.44, chlorine 3.16, and so on). Your course will name one table; use that one, and the differences below will move by a tenth at most.

The cut-offs most courses use are a convention, not a law: a difference of about 0 to 0.4 is treated as nonpolar covalent, about 0.4 to 1.7 as polar covalent, and above about 1.7 as ionic. Bonds do not switch type at 1.7. Bond character changes gradually, and the numbers are a way to sort the common cases. The ideas this step sits in are laid out on the SCH4U course map.

Four bonds, worked line by line:

• C–H: subtract, 2.5 − 2.1 = 0.4. At the edge of the first band, treated as nonpolar.
• O–H: subtract, 3.5 − 2.1 = 1.4. Polar covalent, with oxygen as the negative end.
• C–O: subtract, 3.5 − 2.5 = 1.0. Polar covalent, oxygen negative.
• Na–Cl: subtract, 3.0 − 0.9 = 2.1. Above 1.7, ionic.

The more electronegative atom always carries the partial negative charge, written δ−. The other end is δ+.

The common wrong line in step 1

"Fluorine is the most electronegative element, so any bond to fluorine is ionic." Run the numbers instead. For C–F, 4.0 − 2.5 = 1.5, which is polar covalent. A large value on one atom does not make a large difference; the difference is what you read.

Step 2: polar bonds do not make a polar molecule

A polar bond has a bond dipole, an arrow from the δ+ end toward the δ− end. A molecule is polar only when its bond dipoles add up to something. Whether they cancel is a question about the shape, and the shape comes from the shape that decides whether the dipoles cancel, which comes from the Lewis structure you draw first. This page does not re-derive VSEPR; it uses the shapes.

• CO₂: two C=O bonds, difference 1.0 each, so both polar. The molecule is linear. The two dipoles point in opposite directions along one line and cancel. Net dipole zero: nonpolar.
• H₂O: two O–H bonds, difference 1.4 each. The molecule is bent at about 104.5°. Both dipoles point toward the oxygen at an angle, so they add. Net dipole toward the oxygen: polar.
• CCl₄: four C–Cl bonds, difference 3.0 − 2.5 = 0.5 each, polar. Tetrahedral, four identical dipoles pointing to the four corners: they cancel. Nonpolar.
• CHCl₃: still tetrahedral, but one of the four bonds is C–H instead of C–Cl. The four dipoles are no longer identical, so they no longer cancel. Polar.

The common wrong line in step 2

"CO₂ has polar bonds, so CO₂ is polar." This is the most common lost mark on this unit. Polar bonds are necessary for a polar molecule but not enough. Check the shape before you answer.

Step 3: polarity picks the intermolecular force

Intermolecular forces are attractions between molecules. They are not the bonds inside a molecule, and they are much weaker. There are three to know, listed weakest to strongest for molecules of similar size.

• London dispersion forces. Every molecule has them, polar or not. Electrons shift for an instant, making a brief dipole that induces one in a neighbour. They grow with the number of electrons and with the contact area, so a long chain has more than a compact molecule of the same formula.
• Dipole–dipole forces. These need a permanent dipole, so only a polar molecule has them. The δ+ end of one molecule attracts the δ− end of the next. They act on top of dispersion.
• Hydrogen bonding. This needs a hydrogen bonded directly to N, O or F, and a lone pair on an N, O or F atom of a neighbouring molecule. The O–H, N–H or F–H bond is so polar that the hydrogen is nearly a bare proton. Hydrogen bonding is the strongest of the three for molecules of similar size.

Two honest limits. First, the ranking holds for molecules of similar size: dispersion in a large molecule can outweigh dipole–dipole in a small one, so compare like with like. Second, these forces are between molecules. The wrong line here is "hydrogen bonds hold the atoms of water together." The O–H bonds inside a water molecule are covalent bonds; hydrogen bonds hold one water molecule to the next, and boiling breaks those, not the covalent ones.

Worked example: hydrogen bonding and three liquids of the same size, 120 degrees apart

Rank propane (C₃H₈), dimethyl ether (CH₃OCH₃) and ethanol (CH₃CH₂OH) by boiling point, then explain the order. Given: atomic masses C 12.01, H 1.008, O 16.00 g/mol; the electronegativity values in step 1.

First, show the molecules are the same size, so that only the forces differ.

• Propane, molar mass: multiply and add, 3 × 12.01 + 8 × 1.008 = 36.03 + 8.064 = 44.09 g/mol.
• Dimethyl ether and ethanol, both C₂H₆O: 2 × 12.01 + 6 × 1.008 + 16.00 = 24.02 + 6.048 + 16.00 = 46.07 g/mol.
• Electrons per molecule: propane 3 × 6 + 8 × 1 = 26; C₂H₆O 2 × 6 + 6 × 1 + 8 = 26. Same electron count, so similar dispersion.

Now run the chain on each.

Propane.

• Bonds: C–H (difference 0.4) and C–C (difference 0.0). Both nonpolar by the convention.
• Shape: no polar bonds to add, so the molecule is nonpolar.
• Force: London dispersion only.
• Prediction: lowest boiling point.

Dimethyl ether.

• Bonds: two C–O bonds, difference 1.0 each, polar; the rest C–H.
• Shape: the oxygen has two bonds and two lone pairs, so the C–O–C unit is bent. The two C–O dipoles point toward the oxygen at an angle and add. Net dipole: polar.
• Force: dipole–dipole on top of dispersion. No hydrogen is bonded to the oxygen, so no hydrogen bonding.
• Prediction: middle.

Ethanol.

• Bonds: one O–H bond, difference 1.4, very polar; one C–O, polar; the rest C–H and C–C.
• Shape: bent at the oxygen, net dipole, polar.
• Force: an H bonded to O, and lone pairs on the O of every neighbour, so hydrogen bonding, on top of dipole–dipole and dispersion.
• Prediction: highest, by a wide margin.

Given the normal boiling points, check the prediction:

Liquid	Molar mass	Strongest force	Boiling point
Propane	44.09 g/mol	London dispersion	about −42 °C
Dimethyl ether	46.07 g/mol	Dipole–dipole	about −24 °C
Ethanol	46.07 g/mol	Hydrogen bonding	about 78 °C

The chain predicted the order. The numbers show the size of each step. Adding a dipole moved the boiling point up by 18 degrees: −24 − (−42) = 18. Adding hydrogen bonding moved it up by 102 degrees: 78 − (−24) = 102. Top to bottom, 78 − (−42) = 120 degrees, between molecules with the same number of electrons.

The common wrong lines in the worked example

• "Ethanol is heavier, so it boils higher." It is not heavier than dimethyl ether: both are 46.07 g/mol. Mass cannot explain a 102-degree gap between two molecules with the same formula.
• "The ether is more polar because it has two polar bonds." Polarity is the net dipole, not the count of polar bonds. And polarity is not the deciding factor here anyway: the hydrogen on the oxygen is.

The whole chain on one screen

Step	What you read	What it tells you
1. Bond	The electronegativity difference	Nonpolar (about 0 to 0.4), polar (about 0.4 to 1.7), ionic (above about 1.7), as a convention
2. Molecule	The shape: do the bond dipoles cancel?	Polar or nonpolar
3. Force	Polarity, and H on N, O or F	Dispersion only, plus dipole–dipole, plus hydrogen bonding
4. Property	The strongest force, among molecules of similar size	The boiling-point rank

The summary in four lines:

• Subtract the electronegativities to name each bond.
• Check the shape before calling a molecule polar; CO₂ and CCl₄ have polar bonds and are nonpolar.
• Pick the force: everyone has dispersion; polar adds dipole–dipole; H on N, O or F adds hydrogen bonding.
• Rank only molecules of similar size by force; otherwise count electrons too.

Reading a worked example once and then doing a fresh one cold is the habit in chapter 3 of the free playbook. Sweller and Cooper (1985) found that studying worked examples before solving made later problem-solving faster than solving from the start, which is why the next section gives you a problem rather than another example.

Try one cold: the hydrogen halides

Rank HF, HCl and HBr from highest to lowest boiling point, and explain the one that breaks the pattern. Given: atomic masses H 1.008, F 19.00, Cl 35.45, Br 79.90 g/mol; electronegativities H 2.1, F 4.0, Cl 3.0, Br 2.8. Write your ranking and the reason before you read on.

Check

• Molar masses: HF 1.008 + 19.00 = 20.01 g/mol; HCl 1.008 + 35.45 = 36.46 g/mol; HBr 1.008 + 79.90 = 80.91 g/mol.
• Differences: H–F 4.0 − 2.1 = 1.9; H–Cl 3.0 − 2.1 = 0.9; H–Br 2.8 − 2.1 = 0.7.
• Bond type: H–Cl and H–Br are polar covalent. H–F reads 1.9, above the 1.7 line, yet HF is a molecule of two nonmetals sharing a pair, a polar covalent bond. This is the cut-off convention failing at its edge, and it is why the line is a guide, not a rule. With two-decimal values the difference is 3.98 − 2.20 = 1.78, still over the line, and the answer is the same.
• All three molecules are linear and polar, so all three have dipole–dipole forces and dispersion.
• HBr above HCl: HBr has 36 electrons to HCl's 18, so stronger dispersion.
• HF above both, though it is the lightest: only HF has an H bonded to F, so only HF hydrogen-bonds.
• Order, highest first: HF, HBr, HCl. Given boiling points: HF about 20 °C, HBr about −67 °C, HCl about −85 °C.

If you ranked by mass alone you wrote HBr, HCl, HF, which is the trend for every hydrogen halide except the one with hydrogen bonding. The marks are for the exception.

Where this sits in the course

In the SCH4U pack these ideas run in a line. Before them: Electron configurations and periodic trends, which is where the electronegativity trend comes from (it rises across a period and falls down a group). Then Ionic and covalent bonding; Lewis structures; electronegativity, the idea step 1 belongs to, then Molecular shape (VSEPR) and polarity, the idea step 2 uses, then Intermolecular forces and physical properties, the idea step 3 and the worked example belong to. After them comes Enthalpy changes and calorimetry.

The chain comes back twice. In Acid–base theories, pH and strong/weak acids, bond polarity helps explain why some acids give up their hydrogen more easily. In the organic strand, Organic functional groups and IUPAC nomenclature is where you meet which functional groups hydrogen-bond: alcohols, carboxylic acids and amines do; ethers, ketones and alkanes do not. SBI4U uses the same idea again in Water, hydrogen bonding and the chemistry of life.

How to practise it

Make a set of ten molecules and take each from formula to boiling-point rank: the bonds, the shape, polar or not, the strongest force. Mix the set so that mass and polarity sometimes disagree, as they did for the hydrogen halides. Rohrer and Taylor (2007) found that shuffled practice beat blocked practice on a delayed test, because a shuffled set makes you choose the method each time; a set where the heaviest molecule always wins teaches you nothing about the exception. Mixed practice has the full case.

Write the reason beside each rank, not only the order. "Ethanol: H on O, hydrogen bonding" earns the marks; "ethanol, highest" does not. Chemistry rewards running the same procedure every time, which is the habit in chemistry as procedures.

What this method cannot do: it ranks molecules of similar size well and molecules of very different size poorly, where you have to weigh dispersion against polarity and the numbers can surprise you. When the sizes differ a lot, say so in your answer and count the electrons.