Explain the trend in the boiling temperatures of the elements on descending group 7, from fluorine to iodine.

Understanding the question:

The process of boiling is heating the mixture until the liquid turns into gas. This process requires breaking intermolecular forces (interactions between molecules (eg: hydrogen bonds, london forces/Van der Waal and permanent dipole dipole). NOTE: not breaking the bonds within molecules - intramolecular forces

Group 7 elements tend to exist as diatomic molecules (X2). Therefore, when you are boiling the molecules you are breaking the intermolecular reactions between these Xmolecules.

Group 7 elements cannot hydrogen bond (as there is no hydrogens involved). They cannot form permanent dipole-dipole interactions either as there is no permanent dipoles in the X2 molecules. Hence, their only interaction is london forces.

The strength of london forces increases when the number of electrons in the molecules increase.

You can obtain the number of electrons in a molecule by multiplying the atomic number by the number of atoms of that element present in the molecule.

Answering the question:

As you progress down group 7 boiling temperature increases(the first mark):

Flourine - F= 18 electrons

Chlorine - Cl2 = 34 electrons

Bromine - Br2 = 70 electrons 

Iodine - I2 = 106 electrons

This is because there are more electrons per X2 molecule going from fluorine to iodine (second mark).

Therefore, the strength of london forces increases from fluorine to iodine (third mark).

Therefore, more energy is required to break the intermolecular forces (fourth mark)

BY
Answered by Berna Y. Chemistry tutor

14125 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

A 20cm³ sample of lithium hydroxide solution of unknown concentration is neutralised by 12.25cm³ of 0.15mol/dm³ of sulfuric acid. Calculate the concentration of the lithium hydroxide solution.


Why are teachers now saying electrons are in orbitals? I thought they moved around shells?


The standard enthalpy of formation of glucose is -1273.3kJ/mol, and for carbon dioxide it is -393.5kJ/mol, and for water -285.8 kJ/mol. What is the standard enthalpy of combustion of glucose, C6H12O6?


Calculate the pH of a 0.025 mol dm-​3​ solution of methanoic acid. For HCOOH, Ka = 1.58 x 10-​4​ mol dm-​3


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning