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Reaction of Group 17 elements with iron

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Halogens react with hot iron to form iron(III) halides such as 2Fe(s) + 3Cl2(g) → 2FeCl3(s); the reaction becomes less vigorous down the group, showing halogen reactivity decreases from chlorine to iodine.

The Group 17 elements, the halogens, react with hot iron to form an iron halide. This reaction is the standard demonstration that reactivity decreases as you move down the halogen group, matching the way reactivity increases down the metal groups. It is a favourite Periodic Table question in 4541 because the same apparatus gives three clearly different results.

The balanced equation

The general pattern is: halogen + iron → iron halide. Chlorine and bromine are reactive enough to form the iron(III) halide, with correct formulae and state symbols:

2Fe(s) + 3Cl2(g) → 2FeCl3(s)

2Fe(s) + 3Br2(g) → 2FeBr3(s)

Iodine is the least reactive halogen and its reaction is slow; it forms iron(II) iodide rather than an iron(III) salt:

Fe(s) + I2(g) → FeI2(s)

The trend in these equations is the point of the experiment. Each halogen atom gains one electron to become a halide ion, and the more strongly a halogen attracts that electron, the more vigorously and completely it reacts. Chlorine, the smallest atom shown here, reacts most readily; iodine, the largest, reacts least.

Conditions required

Iron wool or fine iron is heated strongly until it is red hot, then the halogen is passed over it: chlorine as a gas, bromine as a brown vapour, and iodine as a violet vapour produced by warming iodine crystals. Heat is needed to start each reaction, and the apparatus is set up in a fume cupboard because the halogens are toxic and corrosive.

What you observe

The three results form a clear sequence. With chlorine the hot iron glows brightly and continues to burn vigorously without further heating, and a brown solid, iron(III) chloride, is deposited. With bromine the iron glows less brightly and the reaction is slower, giving a reddish-brown solid, iron(III) bromide. With iodine the iron only glows dimly and the reaction is slow and needs continued heating, forming iron(II) iodide. The decreasing brightness and speed from chlorine to iodine is the evidence that reactivity falls down the group.

Where it appears in the SPM exam

In 4541/1 you may be asked to order the halogens by reactivity or predict which reacts most vigorously with iron. In 4541/2 you write the balanced equations, describe and compare the three observations, and explain the trend in terms of atomic size and the ease of gaining an electron. The experiment is also a set-piece for the reactivity of Group 17, and it pairs naturally with the displacement of halogens you meet in Form 5 redox.

How we teach it

Our teachers make sure your explanation is the mirror image of the Group 1 argument: down Group 17 the atom gets larger, the nucleus attracts an incoming electron less strongly, and shielding increases, so the halogen accepts an electron less easily and is less reactive. Students most often lose marks by describing the colours without explaining the trend, or by writing the same product for every halogen. Tying each observation to “how strongly the atom attracts an electron” keeps the equations, the colours and the explanation consistent.

Quick summary

Chlorine, bromine and iodine react with hot iron with steadily decreasing vigour, and chlorine and bromine give iron(III) halides while iodine gives iron(II) iodide. Learn the equations with state symbols, the ordered observations, and the size-and-attraction explanation of falling reactivity, and this dependable topic is secured.

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Frequently asked questions

Why does reactivity decrease down Group 17?

Going down the group the atom gets larger, so the nucleus attracts an incoming electron less strongly and the halogen accepts an electron less easily. Gaining an electron drives the reaction, so the halogens become less reactive down the group.

Source: DSKP KSSM Chemistry Form 4 and 5 (English version)

Written by the spmchemistry.com.my editorial teamUpdated: 4 September 2026
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