Chemguide: Core Chemistry 14 - 16

The Periodic Table - the Halogens

This page introduces the Halogens in Group 7 of the Periodic Table. Group 7 is also known by its more modern name of Group 17.

This is going to be quite a long page. Take your time over it.


Astatine is very radioactive and short-lived. We can, however, predict what its properties might be by exploring the trends in the group.

Physical properties of the Halogens

Fvery pale yellow gas
Clyellow/green gas
Brdark red liquid with a red/brown vapour
Idark grey solid with a purple vapour

This next short video shows chlorine, bromine and iodine. It starts by producing yellow/green chlorine gas and then condensing this to a yellow/green liquid. This is something you are very unlikely to see in the lab.

Reactions of the Halogens with metals

Before we go any further, I want to get the fluorine problem out of the way. Fluorine is extremely reactive and I have never even seen it, let alone used it.

There is an interesting bit of video from the University of Nottingham Periodic Table series which shows that I am not alone. You are unlikely to need this for exam purposes. I include it simply because it is interesting.

You have just seen how vigorously fluorine reacts with almost all elements. The next most reactive element in the group is chlorine and we will start with that.

Chlorine and metals: lithium

The video (and several others below) uses the phrase "highly exothermic reaction". This is a reaction in which a lot of heat is given off.

It is unlikely you would ever be asked about this reaction. I include it just because it is visually interesting.

2Li(s) + Cl2(g)     2LiCl(s)

The video skips over the fact that lithium chloride is a white solid. This is masked by the reaction between lithium and the glass producing black solids.

Chlorine and metals: sodium

2Na(s) + Cl2(g)     2NaCl(s)

Chlorine and metals: aluminium

2Al(s) + 3Cl2(g)     2AlCl3(s)

Chlorine and metals: iron

There are two iron chlorides, iron(II) chloride and iron(III) chloride. The red-brown compound formed when iron reacts with chlorine is iron(III) chloride, FeCl3.

2Fe(s) + 3Cl2(g)     2FeCl3(s)

Reacting the other halogens with iron

The next video repeats the reaction of iron with chlorine, but extends it to include the reactions with bromine and iodine as well.

The reactions tend to get less vigorous as you go from chlorine to bromine to iodine.

You have just seen the equation for the chlorine and iron reaction. The bromine one is similar.

2Fe(s) + 3Br2(g)     2FeBr3(s)

Iron(III) bromide is formed.

For reasons we will look at later on this page, iodine only converts the iron to iron(II) iodide, FeI2.

Fe(s) + I2(g)     FeI2(s)

At this level, you are unlikely to need to know this. You are most likely to come across the reaction with chlorine, but might also be expected to know that the reactions become less vigorous as you go down the group.

Note:  This last video doesn't actually show this change in vigour all that well because it isn't doing all the reactions under the same conditions. It certainly shows that the iodine doesn't react as vigorously as the bromine because it scarcely glows.

In an earlier video you have seen that iron wool burns in fluorine without the need to heat it, and you have also seen a more convincing video of iron burning in a gas jar of chlorine. If you pass chlorine over heated iron wool in a tube, it also burns - but this time you need to heat it to make the reaction start.

Reaction of chlorine with non-metals: hydrogen

Chlorine also reacts with several non-metals, but the only one you are likely to meet at this level is the reaction with hydrogen.

Hydrogen chloride gas is produced explosively when the mixture is exposed to light. In the lab, you have to be very careful if you mix gas jars of hydrogen and chlorine to keep them well away from sunlight.

H2(g) + Cl2(g)     2HCl(g)

But you can also get them to react more gently if you lower a jet of burning hydrogen gas into a gas jar of chlorine.

Hydrogen chloride is a colourless poisonous gas which looks steamy in moist air. It reacts with water vapour in the air to form a fog of hydrochloric acid droplets.

The test used for hydrogen chloride is to expose it to ammonia gas from concentrated ammonia solution. The two gases react together to make white fumes of solid ammonium chloride.

Reactions of bromine and iodine with hydrogen

A bromine and hydrogen mixture will give a mild explosion if you put a lighted taper in it. Hydrogen bromide is formed.

You have to heat a mixture of hydrogen and iodine constantly to get it to react, but the reaction is reversible - so all you get is a mixture of hydrogen, iodine and hydrogen iodide formed.

Hydrogen bromide and hydrogen iodide look and behave like hydrogen chloride.

Again, reactivity gets less as you go from chlorine to bromine to iodine.

Note:  Hydrogen bromide and hydrogen iodide will also produce white fumes with ammonia gas. You are much more likely to come across hydrogen chloride in the lab, and so it is usually safe to assume that you have hydrogen chloride rather than one of the others.

Displacement reactions of the halogens

The halogens get less reactive as you go down the group, and you can get reactivity series type reactions involving them.

The more reactive halogen displaces the less reactive one from one of its salts.

If you react, say, potassium bromide solution with chlorine (usually by using a solution of chlorine - chlorine water), the bromide ions get displaced as bromine molecules. The colourless solution becomes orange coloured due to the free bromine.

2KBr(aq) + Cl2(aq)     2KCl(aq) + Br2(aq)

A similar thing happens if you react potassium iodide solution with chlorine water. This time iodine is formed either as a red solution or, if you use an excess of chlorine, as a grey solid.

2KI(aq) + Cl2(aq)     2KCl(aq) + I2(aq or s)

Typically, you normally use sodium or potassium bromide or iodide in the lab, but as you will see below, all you need is a source of bromide or iodide ions in solution.

Note:  You may not have expected a solution of iodine to be red. The red is because iodine molecules, I2, react with an excess of iodide ions, I-, in the potassium iodide to give a red ion, I3-. Apart from its colour, this ion behaves just like free iodine. You probably don't need to remember that, but you do need to remember the colour of iodine in potassium iodide solution.

The photo shows the bromine formed in these reactions in the left-hand flask and the iodine formed in the right-hand one.

You can also get a displacement reaction if you add bromine solution to potassium iodide solution.

2KI(aq) + Br2(aq)     2KBr(aq) + I2(aq or s)

Again you could get either a red solution of iodine or a grey precipitate of solid iodine if you use enough bromine water.

Sometimes, if you have a very dilute solution of potassium iodide, the iodine solution formed can look orange rather than red, and it looks similar to a solution of bromine.

You can tell the difference by adding a small amount of an organic solvent like hexane. Hexane sits on top of the aqueous layer. Bromine and iodine are soluble in this.

Bromine would dissolve to give an orange solution. Iodine dissolves to give a rather purply solution. This video shows the iodine case.

It would have been more convincing, if he had used less hexane and shaken it for longer!

Halogens as oxidising agents

It is important that you can understand redox reactions in terms of electron transfer - in particular that oxidation is loss of electrons. If you have any doubts about it, you must read the page an introduction to redox reactions before you go on. You probably won't need to read the page suggested at the beginning of that.

Chlorine reactions with metals

Think about the reaction between sodium and chlorine.

2Na(s) + Cl2(g)     2NaCl(s)

The sodium starts of as sodium atoms and ends up as sodium ions. The sodium has lost an electron.

Na    Na+ + e-

Loss of electrons is oxidation (OIL RIG), so the sodium has been oxidised to sodium ions.

The oxidising agent (the substance that is doing the oxidising) is the chlorine. In the process the chlorine molecules are reduced to chloride ions by gaining electrons (OIL RIG again!).

Cl2 + 2e-     2Cl-

The same thing is true for all the other cases where chlorine is reacting with a metal.

Reactions of the other halogens with iron

You will remember that as you go from chlorine to bromine to iodine, the reactions with iron get less vigorous. Chlorine and bromine produce iron(III) chloride or bromide, but iodine produces iron(II) iodide.

The fall in reactivity is because the halogens become less good oxidising agents as you go down the group.

In the cases of chlorine and bromine, the iron atoms are oxidised to iron(III) ions. Electrons are removed by the chlorine or bromine.

Fe    Fe3+ + 3e-

But with iodine you only get iron(II) ions.

Fe    Fe2+ + 2e-

The iodine isn't a strong enough oxidising agent to remove the third electron from the iron.

Chlorine reaction with hydrogen

It is a bit more complicated when chlorine reacts with hydrogen because the hydrogen chloride formed is a gas, and so is covalent. There isn't an obvious transfer of electrons.

H2(g) + Cl2(g)     2HCl(g)

However, hydrogen chloride is a polar molecule - the bonding pair of electrons is pulled towards the chlorine end of the bond.

So although the hydrogen hasn't completely lost its electron, it has lost some share of it. We still count this as oxidation.

Note:  If you aren't sure about bond polarity, you should follow this link.

To understand this bit about the hydrogen-chlorine reaction being a redox reaction properly, you really need to know about oxidation states. That's a problem for a higher level.

Displacement reactions

Let's look in more detail at the reaction between chlorine and potassium bromide solution.

2KBr(aq) + Cl2(aq)     2KCl(aq) + Br2(aq)

Potassium bromide and potassium chloride are ionic compounds, containing potassium ions and chloride or bromide ions.

The potassium ions aren't changed in this reaction - they are just spectator ions in this. The ionic equation (missing out the state symbols to keep it simpler) is

2Br- + Cl2     2Cl- + Br2

The chlorine has removed electrons from the bromide ions, and gained them itself.

2Br-  Br2 + 2e-

Cl2 + 2e-  2Cl-

Chlorine is oxidising the bromide ions to bromine molecules because chlorine is better at forming negative ions than bromine is.

It is much better at forming negative ions than iodine is, and so chlorine will oxidise iodide ions in potassium iodide to iodine in a similar way.

The reaction between bromine and potassium iodide can be explained in the same way.

2KI(aq) + Br2(aq)     2KBr(aq) + I2(aq or s)

Bromine is better at forming negative ions than iodine is. So bromine takes electrons from the iodide ions and leaves iodine molecules. This is yet another redox reaction.

Why does the oxidising ability of the halogens decrease down the group?

The usual explanation of this which is given to students at this level is over-simplistic and faulty. In truth, you can only explain this properly once you know a lot more chemistry.

But this is the faulty explanation anyway!

Suppose you have got an atom of chlorine and an atom of bromine. They have the following electronic structures:

Cl: 2,8,7

Br: 2,8,18,7

Now suppose you add another electron to the outer level to make chloride and bromide ions:

Cl-: 2,8,8

Br-: 2,8,18,8

In a chloride ion, the outer electrons feel a pull of 17 protons in the nucleus, partially shielded by 10 electrons - a net pull of 7+.

In a bromide ion, the outer electrons feel a pull of 35 protons in the nucleus, partially shielded by 28 electrons - a net pull of 7+.

In all the halogens the net pull from the nucleus on the outer electrons is always 7+.

But in the bromine case, the incoming electron is further away from the nucleus and so is less strongly attracted.

This stronger attraction from the nucleus for the outer electrons if the atom is smaller has two consequences.

Remember that an oxidising agent oxidises something else by removing electrons from it.

  • Chlorine will accept another electron to make an ion more readily than bromine will. That makes chlorine a better oxidising agent than bromine.

  • It will be easier to remove an electron from a bromide ion than a chloride ion. That means that bromide ions are more easily oxidised by electron loss than chloride ions are.

If you move the argument on to iodine, the outer electrons are even further from the nucleus but feel the same net pull of 7+.

The atom won't accept an electron as easily as chlorine or bromine, and it will be much easier to remove an electron from an iodide ion.

So iodine is a weaker oxidising agent than the other two, and iodide ions are very easily oxidised by electron loss.

Note:  If you are a teacher or an A level standard student reading this, I have discussed this problem at length on this page in the main part of Chemguide.


For chlorine, bromine and iodine, make sure you know:

  • That they form diatomic molecules, X2.

  • Their colours and physical states (gas, liquid or solid).

  • Their reactions with metals to form salts (chlorides, bromides or iodides).

  • Using the reaction with iron as an example, that the reactivity falls from chlorine to bromine to iodine.

  • Their reaction with hydrogen to form hydrogen halides (HCl, HBr and HI), and that reactivity falls as you go down the group.

  • That a halogen higher in the group can displace one lower down from one of its salts. Be able to describe what you would see in the examples higher up this page.

  • That these reactions are examples of redox reactions, and that oxidising ability of the halogen falls as you go down the group.

What might you expect about astatine?

Astatine is at the very bottom of the group.

From the trends you have seen, it is likely to be a dark coloured solid with a melting point higher than iodine and a molecule At2.

It will be the least reactive element in the group, and not a very good oxidising agent. If it reacts with, say, sodium it will form an ionic salt NaAt.

All the other halogens would displace astatine from salts like NaAt.

And finally . . .

There is a lot of serious chemistry on this page. You may not remember it all at your first read-through, but to be successful it is important that you understand what you have read. Don't leave the page until you are reasonably confident.

Where would you like to go now?

To the Periodic Table menu . . .

To the Chemistry 14-16 menu . . .

To Chemguide Main Menu . . .

© Jim Clark 2020