Group 7(17), the halogens

In a nutshell

The halogens in Group 7(17) are reactive non-metals that exist as diatomic molecules (F₂, Cl₂, Br₂, I₂).

Two opposite trends run the whole topic: going down the group the halogens become weaker oxidising agents, while the halide ions become stronger reducing agents. You also need the test for halide ions, and the reactions and uses of chlorine.

Assumed knowledge: Periodicity, Oxidation, reduction and redox equations, Bonding.

Core content

Trend in electronegativity

Electronegativity is the power of an atom to attract the pair of electrons in a covalent bond.

Electronegativity decreases down Group 7.

The reason, in the order that earns credit:

• Going down the group the atomic radius increases (each element has an extra electron shell, so there is more shielding).
• The bonding pair of electrons is therefore further from the nucleus and more shielded from it.
• So the nucleus attracts the shared (bonding) pair of electrons less strongly, which is a lower electronegativity.

Trend in boiling point

The boiling point of the elements increases down Group 7.

This is about the forces between the molecules, not the covalent bonds inside them:

• Down the group the molecules are larger and have more electrons.
• So there are stronger van der Waals forces between the molecules.
• More energy is needed to overcome the forces between the molecules, so the boiling point rises.

At room temperature this trend is why F₂ and Cl₂ are gases, Br₂ is a liquid and I₂ is a solid.

Oxidising ability of the halogens: displacement reactions

A halogen gains electrons when it reacts, so a halogen is an oxidising agent (an electron acceptor).

Oxidising ability decreases down the group (Cl₂ > Br₂ > I₂), for the same reason as electronegativity: the larger atom attracts an incoming electron less strongly.

A more reactive halogen will displace a less reactive halide ion from solution:

• Chlorine displaces both bromide and iodide ions.
• Bromine displaces iodide ions only.
• Iodine displaces none of them.

The ionic equations (each halogen is reduced; each displaced halide is oxidised):

• Cl₂(aq) + 2Br^-(aq) → 2Cl^-(aq) + Br₂(aq)
• Cl₂(aq) + 2I^-(aq) → 2Cl^-(aq) + I₂(aq)
• Br₂(aq) + 2I^-(aq) → 2Br^-(aq) + I₂(aq)

The colour of the free halogen produced tells you which reaction has happened. Note that iodine looks brown in water, and only looks violet or purple when an organic solvent (such as cyclohexane) is added and shaken:

Reducing ability of the halide ions

A halide ion can lose (donate) an electron, so a halide ion can act as a reducing agent (an electron donor).

Reducing ability increases down the group (I^- > Br^- > Cl^-). Leading with the point that earns the mark:

• Down the group the halide ion is larger, with more electron shells so the outer electron is further from the nucleus and there is more shielding.
• So the outer electron is less strongly held by the nucleus.
• A less strongly held electron is more easily lost (donated), so the ion is a better reducing agent.

Still don't get it? · why reducing power increases down the group

Think of holding a balloon on a string. If the balloon is close to your hand and nothing is in the way, you grip it firmly. If the string is long and there is a crowd of people standing between your hand and the balloon, your grip feels much weaker and it is easy for someone to take it.

Now the chemistry. To act as a reducing agent, a halide ion has to give away one of its outer electrons. Going down the group the ion gets bigger: the outer electron sits in a shell that is further from the nucleus, and there are more inner shells shielding it. Both of these mean the nucleus holds that outer electron less tightly.

An electron that is held less tightly is handed over more easily. So in exam wording: the iodide ion is larger, its outer electron is further from the nucleus and more shielded, so the electron is less strongly held and more easily lost. Make the comparison explicit (iodide versus chloride) and always talk about electrons, because that is what the examiner is looking for.

Reactions of solid sodium halides with concentrated sulfuric acid

Reacting the solid sodium halide with concentrated sulfuric acid shows the reducing-power trend directly. The further down the group, the more strongly the halide reduces the sulfur in H₂SO₄ (which starts at oxidation state +6).

Every halide first does the same acid-base step (H₂SO₄ acting as an acid, a proton donor):

$\text{NaX(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HX(g)}$

What happens next depends on how good a reducing agent the halide is:

Redox equations (H₂SO₄ now acting as the oxidising agent):

Bromide reduces sulfur only as far as SO₂:

• 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

Iodide is the strongest halide reducing agent and reduces sulfur three ways:

• 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
• 6HI + H₂SO₄ → 3I₂ + S + 4H₂O
• 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O

So concentrated sulfuric acid plays two different roles: an acid in the first step, and an oxidising agent in the redox steps.

Testing for halide ions: acidified silver nitrate

To identify a halide ion in solution:

1. Add dilute nitric acid.
2. Then add silver nitrate solution, AgNO₃(aq).

A coloured precipitate of the silver halide forms:

• Chloride: white precipitate. Ag⁺(aq) + Cl^-(aq) → AgCl(s)
• Bromide: cream precipitate. Ag⁺(aq) + Br^-(aq) → AgBr(s)
• Iodide: yellow precipitate. Ag⁺(aq) + I^-(aq) → AgI(s)
• Fluoride: no precipitate (silver fluoride is soluble).

Why each step is done (these are the explain marks):

• Why silver nitrate is used: the silver ion forms insoluble silver halides of different colours, so the colour of the precipitate identifies the ion.
• Why it is acidified (with nitric acid): the acid reacts with and removes carbonate and hydroxide ions, which would otherwise form their own precipitates with silver (such as Ag₂CO₃) and give a false positive. Nitric acid is used because it adds no interfering ion: hydrochloric acid would add Cl^-, and sulfuric acid would form a silver sulfate precipitate.
• Why ammonia is added: the cream and yellow (and white) precipitates can look alike, so ammonia is used to tell them apart by their different solubilities.

The solubility of the silver halides in ammonia decreases down the group:

The silver halide dissolves because it forms a soluble complex ion, for example:

• AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl^-(aq)

Required practical 4: test-tube tests for ions

Required practical 4 is to carry out simple test-tube reactions to identify some cations and anions. The halide test above is the Group 7 part of it.

When several ions might be present, test the anions in a set order (carbonate, then sulfate, then halide), because an ion found earlier would interfere with the later tests:

The carbonate must be removed first because it would also give a precipitate with both barium ions and silver ions. The barium chloride is acidified with hydrochloric acid, and the silver nitrate with nitric acid, for the same reason. Group 2 cations are covered in Group 2, the alkaline earth metals.

Chlorine and water: disproportionation

When chlorine dissolves in water a little of it reacts to form chloride ions and chlorate(I) ions:

Cl₂(aq) + H₂O(l) ⇌ HClO(aq) + HCl(aq)

or, in ionic form:

Cl₂(aq) + H₂O(l) ⇌ 2H⁺(aq) + Cl^-(aq) + ClO^-(aq)

This is disproportionation: the same element, chlorine, is simultaneously oxidised and reduced. Chlorine starts at oxidation state 0 and ends up at both +1 and -1:

• In HClO (chloric(I) acid) and ClO^- (chlorate(I)): chlorine is +1 (oxidised).
• In HCl and Cl^- (chloride): chlorine is -1 (reduced).

Still don't get it? · disproportionation

Picture a class where every pupil starts on exactly the same grade. After one test, half of them are moved up a grade and the other half are moved down a grade, even though they all began at the same place. One single event has pushed the same group in two opposite directions at once.

Oxidation state is the "grade". In a Cl₂ molecule every chlorine atom is at 0. When it reacts with water, some chlorine atoms end up at +1 (in HClO or ClO^-, which is oxidation) and the rest end up at -1 (in HCl or Cl^-, which is reduction).

Because the same element has been both oxidised and reduced in the same reaction, we call it disproportionation. In the exam, prove it by writing the oxidation states on the chlorine: 0 going to +1 and 0 going to -1.

In bright sunlight a different reaction happens, forming chloride ions and oxygen:

2Cl₂(aq) + 2H₂O(l) → 4H⁺(aq) + 4Cl^-(aq) + O₂(g)

Here chlorine is only reduced (0 to -1) and it is the oxygen in water that is oxidised (-2 to 0). The pale green colour of chlorine water fades as the Cl₂ is used up.

Chlorine in water treatment

Chlorine is added to drinking water and swimming pools because it kills bacteria (it acts as a disinfectant).

Chlorine is itself toxic, and it can react with organic compounds in the water to form small amounts of harmful (chlorinated) products. Society weighs these disadvantages against the benefits, and the accepted judgement is that the benefits to health of treating water with chlorine outweigh its toxic effects.

Chlorine with cold, dilute sodium hydroxide

Chlorine reacts with cold, dilute sodium hydroxide solution. This is also a disproportionation (chlorine goes to both -1 and +1):

Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

The product NaClO is sodium chlorate(I) (sometimes written NaOCl). The solution formed is bleach, used as a disinfectant and for whitening.

Naming reminder: oxygen-containing chlorine compounds are named as chlorates with the oxidation state of chlorine in roman numerals, for example NaClO is sodium chlorate(I) and NaClO₃ is sodium chlorate(V).

Worked examples

Building and balancing the redox equations for sodium iodide with concentrated sulfuric acid

This is the equation-writing skill examiners most often see go wrong, so work it in stages.

Step 1, the acid-base step that every halide does (no change in oxidation state):

$\text{NaI(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HI(g)}$

Step 2, write the oxidation half-equation (iodide loses electrons):

$2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$

Step 3, write the reduction half-equation for the sulfur product you want. For sulfur going from +6 to +4 (SO₂), two electrons are gained:

$\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}$

Step 4, the electrons already match (2 each), so add the two half-equations and cancel the electrons:

$2\text{I}^- + \text{H}_2\text{SO}_4 + 2\text{H}^+ \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$

Check: atoms balance (I 2 = 2, S 1 = 1, O 4 = 4, H 4 = 4) and charge balances (left 2- plus 2+ = 0, right 0).

For the deeper reductions, scale the iodide half-equation to match the electrons gained:

$6\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 3\text{I}_2 + \text{S} + 4\text{H}_2\text{O}$

$8\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}$

Where the marks are lost: forgetting the H⁺ ions or the water (which leaves the equation unbalanced), not balancing the charge, and writing atomic I instead of I₂.

Model answer: explain the trend in boiling point down Group 7 (3 marks)

A full-mark answer needs three linked points, in this order:

1. Going down the group the molecules are larger and have more electrons.
2. So the van der Waals forces between the molecules get stronger.
3. More energy is needed to overcome the forces between the molecules, so the boiling point increases.

The second and third points only score if you say the forces are between the molecules. An answer that talks about breaking covalent bonds, or about ionisation energy, scores nothing.

Common exam mistakes

• Confusing the halogen molecule, the halogen atom and the halide ion: Cl₂, Cl and Cl^- are different species, and the name and formula must match. A very common slip is calling the oxidising agent in a displacement reaction "chloride" when it is chlorine (Cl₂).
• Describing iodine in water as a "purple solution" or "purple fumes": in water an iodine solution is brown, and it only looks violet or purple in a cyclohexane layer.
• In the boiling-point explanation, not stating that the van der Waals forces are between the molecules, or wrongly explaining the trend by ionisation energy or by breaking covalent bonds.
• Writing unbalanced redox equations, especially leaving the charges unbalanced, forgetting the H⁺ ions or the water, or using atomic Br/I and only one halide ion instead of two.
• Saying that bromide ions are reduced by concentrated sulfuric acid: the halide ions are oxidised; it is the sulfur that is reduced.
• Forgetting that concentrated (not dilute) sulfuric acid is needed to show the reducing ability of the halides, and giving the wrong oxidation state of sulfur (it is +4 in SO₂, 0 in S and -2 in H₂S).
• For the silver nitrate test, writing a "white colour" or "white solution" instead of a white precipitate: both the colour and the word precipitate (or solid) are needed, and the precipitate colours (white, cream, yellow) are often muddled.
• Not stating whether the ammonia is dilute or concentrated when distinguishing the silver halides, and using hydrochloric acid to acidify the silver nitrate (it adds Cl^- and ruins the test).
• For chlorine with sodium hydroxide, writing the acids (HCl, HClO) instead of the sodium salts (NaCl, NaClO), and not knowing that chlorine is +1 in NaClO, which is named sodium chlorate(I).
• Thinking that HClO is responsible for the green colour of chlorine water: the green colour is dissolved Cl₂.

Key definitions

• Electronegativity is the power of an atom to attract the pair of electrons in a covalent bond.
• Oxidising agent: a species that accepts electrons (an electron acceptor).
• Reducing agent: a species that donates electrons (an electron donor). It is not enough to say "electron pair donor".
• Disproportionation: a reaction in which the same element is simultaneously oxidised and reduced.

Specification

• I can explain the trend in electronegativity down Group 7.
• I can explain the trend in the boiling point of the elements in terms of their structure and bonding.
• I can describe and explain the trend in oxidising ability of the halogens, including the displacement reactions of halide ions in aqueous solution, with equations and observations.
• I can describe and explain the trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid.
• I can describe the use of acidified silver nitrate to identify and distinguish halide ions, and explain why silver nitrate is used, why it is acidified, and why ammonia is added.
• I can state the trend in solubility of the silver halides in ammonia.
• I can write the equation for the reaction of chlorine with water to form chloride ions and chlorate(I) ions, and recognise it as disproportionation.
• I can write the equation for the reaction of chlorine with water in sunlight to form chloride ions and oxygen.
• I can describe the use of chlorine in water treatment and explain why its benefits are judged to outweigh its toxic effects.
• I can write the equation for the reaction of chlorine with cold, dilute NaOH and state the uses of the solution formed.
• I can carry out simple test-tube reactions to identify the anions (halides, OH^-, CO₃^2-, SO₄^2-) and the cations (Group 2, NH₄⁺) in Required practical 4.

Related notes

• Periodicity
• Group 2, the alkaline earth metals
• Oxidation, reduction and redox equations
• Bonding