Topic 7 · page 10 of 14

The hydrogen halides

From HCl to HI the boiling temperature rises because London forces grow with the number of electrons. HF breaks the pattern and boils highest, because it forms hydrogen bonds. A three-step method ranks any set of molecules.

13 min read7.57.5(iv)

Read first: Alkane chains and branching, Hydrogen bonds

HCl to HI, and then HF 7.57.5(iv)

The hydrogen halides are HF, HCl, HBr and HI. All four are polar, because the halogen is more electronegative than hydrogen.

Hydrogen halide Electrons Boiling temperature / °C Forces between molecules
HF 10 20 hydrogen bonds, London
HCl 18 −85 permanent dipole–dipole, London
HBr 36 −67 permanent dipole–dipole, London
HI 54 −35 permanent dipole–dipole, London
Bar chart of boiling temperatures of the hydrogen halides. HF, 10 electrons, 20 degrees Celsius, drawn in orange for hydrogen bonds. HCl, 18 electrons, minus 85. HBr, 36 electrons, minus 67. HI, 54 electrons, minus 35. From HCl to HI the boiling temperature rises as the number of electrons rises, but HF, with the fewest electrons, boils highest.
Boiling temperatures of HF, HCl, HBr and HI. HF is out of line with the other three.

HCl to HBr to HI. The number of electrons increases down the group (HCl 1 + 17 = 18, HBr 1 + 35 = 36, HI 1 + 53 = 54), so the London forces get stronger. The H–X bond also gets less polar down the group, because the halogen becomes less electronegative, so the permanent dipole–dipole interactions get weaker. The two effects pull in opposite ways. The increase in London forces is the larger one, so the boiling temperature rises: −85, −67, −35 °C.

HF. HF has the fewest electrons, so it has the weakest London forces. But it boils highest, at 20 °C, because its molecules form hydrogen bonds: H is bonded to F, and F has lone pairs. More energy is needed to overcome the hydrogen bonds than the London forces and dipole–dipole forces in the others.

The order of boiling temperatures is therefore:

HF>HI>HBr>HCl \text{HF} > \text{HI} > \text{HBr} > \text{HCl}

A method for any boiling-temperature question 7.5

You now have three forces and two ways of making London forces stronger (more electrons and more surface contact). Use them in this order.

MethodComparing boiling temperatures

  1. List the forces in each substance: London forces always; permanent dipole–dipole if it has a net dipole; hydrogen bonds if an H is bonded to N, O or F.
  2. If the substances have similar numbers of electrons, the one with the strongest type of force boils higher: hydrogen bonds, then permanent dipole–dipole, then London forces only.
  3. If they have the same type of force, the one with more electrons boils higher. If the electron counts are also equal (isomers), the one with less surface contact (more branched) boils lower.
  4. Finish with: more energy is needed to overcome the stronger forces between molecules, so the boiling temperature is higher.

Worked exampleRank butane, propanone and propan-1-ol

Their boiling temperatures are −1 °C, 56 °C and 97 °C, in some order.

  1. Electrons: butane C₄H₁₀ = 34; propanone C₃H₆O = 18 + 6 + 8 = 32; propan-1-ol C₃H₈O = 18 + 8 + 8 = 34. All are similar, so compare the type of force.
  2. Butane is non-polar: London forces only. Propanone is polar (C=O) but has no H on O: permanent dipole–dipole and London. Propan-1-ol has an O–H bond: hydrogen bonds and London.
  3. Order of strength: London < dipole–dipole < hydrogen bonds. So butane (−1 °C) < propanone (56 °C) < propan-1-ol (97 °C).

Check yourself

  1. Work out the number of electrons in HBr and in HI.
  2. Explain the trend in boiling temperature from HCl (−85 °C) to HI (−35 °C).
  3. HF has only 10 electrons but boils at 20 °C, higher than HI (54 electrons, −35 °C). Explain.
  4. Put HF, HCl, HBr and HI in order of increasing boiling temperature.
  5. A student writes: "HF has the highest boiling temperature because the H–F bond is the strongest." Give the mistake and the correct reason.
  6. Use the method to put these in order of increasing boiling temperature, with a reason: methanol (CH₃OH), ethane (C₂H₆) and fluoromethane (CH₃F). Each has about the same number of electrons (18).

Answers cover these until you have tried the questions

  1. HBr: 1 + 35 = 36. HI: 1 + 53 = 54.
  2. More electrons, so stronger London forces between molecules, and these increase by more than the permanent dipole–dipole interactions weaken. More energy is needed to separate the molecules, so the boiling temperature rises.
  3. HF forms hydrogen bonds (H bonded to F, with lone pairs on F), as well as London forces. Hydrogen bonds are stronger than the London forces and dipole–dipole forces in HI, even though HI has more electrons, so more energy is needed to separate HF molecules.
  4. HCl < HBr < HI < HF (−85, −67, −35, 20 °C).
  5. The mistake is to refer to the covalent H–F bond, which is not broken on boiling. The correct reason is the hydrogen bonds between HF molecules, which need more energy to overcome.
  6. Ethane < fluoromethane < methanol. Electrons are about equal (C₂H₆ 18; CH₃F 6 + 3 + 9 = 18; CH₃OH 18), so compare the type of force. Ethane is non-polar: London forces only. CH₃F is polar but has no H on F, so it has permanent dipole–dipole interactions and London forces. Methanol has an O–H bond: hydrogen bonds and London forces, the strongest.