Properties of the Halogens — AQA A-level Chemistry

SPEC 3.2.3.1 • GROUP 7(17), THE HALOGENS

Halogen Trends, Oxidising & Reducing Ability

Physical trends, displacement reactions, halide reactions with H₂SO₄, and the silver nitrate test

The Halogens - Group 7 Elements

Definition

The halogens are the Group 7 (or Group 17) elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). They are highly reactive non-metals that exist as diatomic molecules (X₂).

The halogens share a common electronic configuration: they all have seven electrons in their outer shell (ns²np⁵). This means they need to gain just one electron to achieve a stable noble gas configuration, which explains their high reactivity and tendency to form X⁻ ions.

Practical Uses

The halogens have important everyday uses: chlorine is used in water purification and as a bleaching agent; bromine is used in flame-retardants and fire extinguishers; iodine is used as an antiseptic and disinfectant agent.

Understanding the trends in Group 7 is essential because it explains why chlorine can displace bromide ions from solution, why iodide ions are more easily oxidised than chloride ions, and why different halides produce different products when reacted with concentrated sulfuric acid.

Key Concept

At room temperature: F₂ and Cl₂ are gases, Br₂ is a liquid, and I₂ is a solid. This reflects the increasing strength of van der Waals forces as molecular size increases.

Physical Trends Down Group 7

As you descend Group 7, several physical properties show clear trends. Understanding the underlying reasons for these trends is crucial for explaining observations and predicting behaviour.

PropertyTrend Down GroupExplanation

Atomic radiusIncreasesEach element has an additional electron shell

ElectronegativityDecreasesAlthough nuclear charge increases, atomic radius and shielding effects are greater

First ionisation energyDecreasesOuter electron further from nucleus, more shielding

Boiling pointIncreasesLarger molecules have stronger van der Waals forces

The boiling point trend deserves particular attention. Halogens exist as simple covalent molecules (X₂), held together by van der Waals forces (specifically, induced dipole-dipole interactions). As you go down the group, the molecules become larger in size and relative atomic mass. This allows for stronger induced dipole-dipole forces to develop between molecules, requiring more energy to overcome.

Exam Tip

When explaining boiling point trends, always mention: (1) the type of intermolecular force (van der Waals / London dispersion forces / instantaneous dipole-induced dipole forces - these are all equivalent terms), (2) why it increases (more electrons = more polarisable = stronger temporary dipoles), and (3) the consequence (more energy needed to separate molecules).

Common Mistake

Don't confuse the boiling point trend with reactivity! Boiling point increases down the group, but reactivity decreases. These are independent properties with different underlying causes.

Bond Strength & Volatility

Halogens exist as diatomic molecules (X₂) in which covalent bonds are formed by overlapping orbitals. The bonding pair of electrons is attracted to the nuclei on either side, holding the molecule together.

Going down the group, atomic size increases, so the bonding pair gets further from the nuclei and the bond becomes weaker. This is reflected in the bond enthalpies:

The Fluorine Anomaly

Notice that F-F has a lower bond enthalpy than Cl-Cl. This is because fluorine atoms are so small that when two bond together, their lone pairs get very close and repel each other. This repulsion weakens the bond, making it easier to break than expected.

Volatility

Volatility refers to how easily a substance evaporates. A volatile substance has a low boiling point. Going down Group 7, volatility decreases because van der Waals forces increase, making it harder for molecules to escape into the gas phase.

Reactivity of Halogens

Key Trend

Reactivity decreases down Group 7. Fluorine is the most reactive halogen; iodine is the least reactive (of the commonly studied halogens).

To react, a halogen atom must gain an electron to form a halide ion (X⁻). This requires attracting an electron from another species. The ability to do this depends on how strongly the nucleus can attract an incoming electron.

Consider what happens as you go down the group:

Atomic radius increases - the incoming electron would be further from the nucleus

Shielding increases - more inner electron shells repel the incoming electron

Nuclear charge increases - but this effect is outweighed by the increased radius and shielding

Electrostatic attraction decreases - harder to attract the electron needed to form a negative ion

The increase in atomic radius and shielding outweigh the increase in nuclear charge, so it becomes harder for larger halogens to attract an incoming electron. Fluorine, with its small atomic radius and minimal shielding, can attract electrons most effectively and is the most reactive.

Key Concept

Reactivity is linked to oxidising ability. The more reactive a halogen, the better it is at gaining electrons (being reduced) and therefore the stronger oxidising agent it is.

Halogens as Oxidising Agents

Definition

An oxidising agent is a substance that oxidises another species by accepting electrons from it. In the process, the oxidising agent itself is reduced.

X_{2} + 2 e^{-} \to 2 X^{-}

This electron gain results in the halogen being reduced, as its oxidation number decreases from 0 to -1.

When a halogen acts as an oxidising agent, each halogen atom gains one electron and is reduced to a halide ion. The species losing electrons is oxidised. Think of it this way: the halogen "steals" electrons from its victim, oxidising them in the process.

The strength of halogens as oxidising agents is directly linked to their electronegativity - how strongly they attract electrons in a covalent bond. Electronegativity decreases down the group for three reasons:

Atomic radius increases - more electron shells are added, so bonding electrons are further from the nucleus

Shielding increases - inner electron shells repel the bonding pair, reducing nuclear attraction

Nuclear charge increases - but this effect is outweighed by the increased distance and shielding

With increasing atomic size going down the group, outer shells are further from the nucleus and experience more shielding from inner electrons. The net effect is that bonding electrons (or incoming electrons) feel less electrostatic attraction to the nucleus, so the halogen's ability to attract electrons decreases.

Key Concept

Oxidising ability decreases down Group 7 because larger atoms with more shielding are less able to attract and gain electrons. Fluorine is the strongest oxidising agent of all elements.

Memory Aid

Remember: OIL RIG - Oxidation Is Loss, Reduction Is Gain. When halogens oxidise something, they cause it to lose electrons. The halogen itself gains those electrons (is reduced).

Halogen Displacement Reactions

Key Principle

A more reactive halogen will displace a less reactive halide from solution. This is because the more reactive halogen is a stronger oxidising agent.

In a displacement reaction, the more reactive halogen oxidises the halide ion, taking its electrons. The displaced halogen is released as the element. This provides clear evidence for the trend in oxidising ability.

Halogen AddedTo KCl(aq)To KBr(aq)To KI(aq)

Cl₂No reactionYellow (Br₂)Orange/brown (I₂)

Br₂No reactionNo reactionOrange/brown (I₂)

I₂No reactionNo reactionNo reaction

Notice the pattern: chlorine can displace both bromide and iodide; bromine can only displace iodide; iodine cannot displace anything. This directly reflects the order of oxidising ability: Cl₂ > Br₂ > I₂.

Key Rule

A halogen will displace any halide ion below it in Group 7. This is because the higher halogen is a stronger oxidising agent and can remove electrons from the halide ion below it.

Writing Equations

For chlorine displacing bromide ions:

Cl_{2} (aq) + 2 Br^{-} (aq) \to 2 Cl^{-} (aq) + Br_{2} (aq)

The chlorine is reduced (gains electrons), the bromide is oxidised (loses electrons).

Worked Example01

Chlorine water is added to potassium iodide solution. Write the ionic equation for this reaction and identify the oxidising and reducing agents.

1

Identify the reactants: Cl₂ (chlorine) and I⁻ (iodide ions from KI)

2

Chlorine is more reactive than iodine, so it will displace iodide ions.

3

Write the ionic equation:

Cl_{2} (aq) + 2 I^{-} (aq) \to 2 Cl^{-} (aq) + I_{2} (aq)

4

Identify oxidation state changes:

Chlorine: 0 → -1 (gain of electrons = reduction)

Iodine: -1 → 0 (loss of electrons = oxidation)

5

Oxidising agent: Cl₂ (accepts electrons)

Reducing agent: I⁻ (donates electrons)

Observation: The colourless solution turns brown as iodine is produced.

Halide Ions as Reducing Agents

Definition

A reducing agent is a substance that reduces another species by donating electrons to it. In the process, the reducing agent itself is oxidised.

X^{-} \to \frac{1}{2} X_{2} + e^{-}

This electron loss results in the halide being oxidised, as its oxidation number increases from -1 to 0.

Halide ions (F⁻, Cl⁻, Br⁻, I⁻) can act as reducing agents by losing their extra electron and being oxidised back to the halogen. The ease with which they do this depends on how tightly the electron is held.

The reducing power of halide ions increases down Group 7 because:

Ionic radius increases - more electron shells are added

Outer electrons further from nucleus - the increasing ionic radius leads to weaker nuclear attraction

More shielding - inner electron shells shield the outer electrons from nuclear charge

Weaker electrostatic attraction - the outer electrons are held less tightly

Easier to lose electrons - the increase in radius and shielding outweigh the increase in nuclear charge

Key Trend

Reducing ability increases down Group 7: I⁻ > Br⁻ > Cl⁻ > F⁻

Iodide ions are the strongest reducing agents because the outer electron is furthest from the nucleus and most easily lost.

As ionic radius increases down the group, the outermost electrons get further from the nucleus and experience more shielding from inner electron shells. This means they are held less tightly and can be lost (donated) more easily, making the ion a better reducing agent.

Etymology

The word "halogen" comes from the Greek meaning salt-maker, a reminder of their tendency to react with metals to form salts (e.g. NaCl, KBr).

Important

Notice this is the opposite trend to oxidising ability! The best oxidising agents (halogens) form the worst reducing agents (halide ions), and vice versa.

Reactions of Halides with Conc. H₂SO₄

When solid sodium halides (NaCl, NaBr, NaI) are reacted with concentrated sulfuric acid, the products depend on the reducing ability of the halide ion. This provides excellent evidence for the trend in reducing ability.

Safety Warning

These reactions produce toxic gases (HCl, HBr, HI, SO₂, H₂S) and should be carried out in a fume cupboard with appropriate safety precautions.

Initial Reaction

All halides first undergo an acid-base reaction to produce the hydrogen halide:

NaX(s) + H_{2} SO_{4} (l) \to NaHSO_{4} (s) + HX(g)

What happens next depends on whether the halide ion is a strong enough reducing agent to reduce the sulfuric acid. The S in H₂SO₄ has an oxidation state of +6, which can potentially be reduced to lower oxidation states.

HalideReducing AbilityCan it reduce H₂SO₄?

Cl⁻WeakNo - only acid-base reaction

Br⁻ModeratePartially - S: +6 → +4

I⁻StrongYes - S: +6 → +4, 0, -2

Key Principle

The stronger the reducing agent, the more products are formed and the further the sulfur is reduced.

Sodium Chloride + Conc. H₂SO₄

Chloride ions are too weak as reducing agents to reduce sulfuric acid. Only the acid-base reaction occurs, producing hydrogen chloride gas.

Reaction

NaCl(s) + H_{2} SO_{4} (l) \to NaHSO_{4} (s) + HCl(g)

This is not a redox reaction. No oxidation states change - it is simply a proton transfer (acid-base) reaction.

ObservationExplanation

White/steamy fumesHCl gas fumes in moist air

Choking smellHCl is an irritant gas

Test for HCl

Hold ammonia-soaked glass rod near the gas. Dense white fumes of NH₄Cl form:

HCl(g) + NH_{3} (g) \to NH_{4} Cl(s)

Key Point

Chloride ions are not strong enough to reduce sulfur in H₂SO₄. The only product is HCl.

Sodium Bromide + Conc. H₂SO₄

Bromide ions are moderate reducing agents. The acid-base reaction occurs first, then HBr reduces the sulfuric acid.

Reaction 1: Acid-Base

NaBr(s) + H_{2} SO_{4} (l) \to NaHSO_{4} (s) + HBr(g)

Reaction 2: Redox

2 HBr + H_{2} SO_{4} \to Br_{2} + SO_{2} + 2 H_{2} O

In the redox reaction: Br goes from -1 to 0 (oxidised), S goes from +6 to +4 (reduced).

ObservationProduct

Orange/brown fumesBr₂ (bromine vapour)

Colourless choking gasSO₂ (sulfur dioxide)

Some white fumesHBr (escapes before oxidised)

Key Point

Bromide reduces sulfur from +6 to +4 only. It cannot reduce it further.

Sodium Iodide + Conc. H₂SO₄

Iodide ions are strong reducing agents. Unlike HCl and HBr, hydrogen iodide decomposes readily because I⁻ is so easily oxidised. The HI immediately reduces the sulfuric acid through a series of reactions, producing multiple products:

Acid-Base

NaI(s) + H_{2} SO_{4} (l) \to NaHSO_{4} (s) + HI(g)

Redox Reactions

To SO₂ (S: +6 → +4):

2 HI + H_{2} SO_{4} \to I_{2} + SO_{2} + 2 H_{2} O

To S (S: +6 → 0):

6 HI + H_{2} SO_{4} \to 3 I_{2} + S + 4 H_{2} O

To H₂S (S: +6 → -2):

8 HI + H_{2} SO_{4} \to 4 I_{2} + H_{2} S + 4 H_{2} O

ObservationProduct

Purple/violet vapourI₂ (iodine)

Yellow solid depositsS (sulfur)

Rotten egg smellH₂S

Choking gasSO₂

Summary: Halides + Conc. H₂SO₄

HalideProductsS Oxidation StatesObservations

Cl⁻HCl only+6 (no change)White steamy fumes

Br⁻HBr, Br₂, SO₂+6 → +4Orange fumes, choking gas

I⁻HI, I₂, SO₂, S, H₂S+6 → +4, 0, -2Purple vapour, yellow solid, rotten eggs

Pattern

The number of products and extent of sulfur reduction increases with reducing ability: Cl⁻ < Br⁻ < I⁻

Exam Tip

You must know observations AND equations. Include state symbols and balance correctly.

Memory Aid

For iodide products: "I₂ SO₂ S H₂S" - ordered by how far sulfur is reduced.

Examiner Insight

It gets easier to oxidise the hydrogen halides going down Group 7: the halides become stronger reducing agents. This is why HI produces more products than HBr, which produces more than HCl.

Worked Example02

Explain why sodium iodide produces more products than sodium bromide when reacted with concentrated sulfuric acid, and calculate the total change in oxidation state of sulfur when H₂S is produced from H₂SO₄.

1

Comparison: I⁻ is a stronger reducing agent than Br⁻ because iodide has a larger ionic radius and more shielding, so its outer electron is held less tightly and more easily lost.

2

Consequence: Br⁻ can only reduce S from +6 to +4 (producing SO₂), but I⁻ can reduce S further to 0 (S) and -2 (H₂S).

3

Oxidation state change: In H₂SO₄, S = +6. In H₂S, S = -2.

4

Total change: +6 → -2 = 8 units reduction

Testing for Halide Ions

The Silver Nitrate Test

Aqueous silver nitrate solution (AgNO₃) is used to test for the presence of halide ions (Cl⁻, Br⁻, I⁻) in a sample.

Method:

1

Add dilute nitric acid to the sample to remove interfering ions that would also form precipitates with silver nitrate (such as carbonate CO₃²⁻, hydroxide OH⁻, sulfite SO₃²⁻, and sulfate SO₄²⁻), which would give false positive results

2

Add silver nitrate solution ; silver ions react with halide ions to form silver halide precipitates

3

Observe the precipitate colour to identify which halide ion is present

Ag^{+} (aq) + X^{-} (aq) \to AgX(s)

Where X = Cl, Br or I

Halide IonSilver HalidePrecipitate Colour

Cl⁻AgClWhite

Br⁻AgBrCream

I⁻AgIYellow

Memory Aid

Colours get "warmer" down the group: White → Cream → Yellow.

Confirmatory Test with Ammonia

Precipitate colours can be hard to distinguish. Ammonia provides a confirmatory test based on silver halide solubility.

Principle

Silver halides dissolve in ammonia by forming the soluble diamminesilver(I) complex, [Ag(NH₃)₂]⁺.

Precipitate+ Dilute NH₃+ Conc. NH₃

AgCl (white)DissolvesDissolves

AgBr (cream)Remains insolubleDissolves

AgI (yellow)Remains insolubleRemains insoluble

Dissolution equation:

AgCl(s) + 2 NH_{3} (aq) \to [Ag(NH_{3})_{2}]^{+} (aq) + Cl^{-} (aq)

Why the Difference?

AgI has the highest lattice enthalpy due to the high polarisability of the large iodide ion, making it insoluble even in concentrated ammonia.

Worked Example03

A cream precipitate forms with acidified AgNO₃. It doesn't dissolve in dilute NH₃ but does in concentrated NH₃. Identify the halide and write ionic equations.

1

Identity: Cream precipitate = AgBr, so the halide is bromide (Br⁻)

2

Confirm: AgBr doesn't dissolve in dilute NH₃ but does in conc. NH₃ ✓

3

Precipitation:

Ag^{+} (aq) + Br^{-} (aq) \to AgBr(s)

4

Dissolution:

AgBr(s) + 2 NH_{3} (aq) \to [Ag(NH_{3})_{2}]^{+} (aq) + Br^{-} (aq)

Practice

Question 1: Bromine water is added to KCl(aq) and separately to KI(aq). State observations and explain in terms of oxidising ability.

1

Br₂ + KCl: No reaction. Bromine is weaker oxidising agent than chlorine.

2

Br₂ + KI: Brown colour (I₂ formed). Bromine is stronger oxidising agent than iodine.

3

Br_{2} (aq) + 2 I^{-} (aq) \to 2 Br^{-} (aq) + I_{2} (aq)

Practice

Question 2: Write the equation for NaI + conc. H₂SO₄ producing H₂S. Calculate the S oxidation state change and explain why this occurs with I⁻ but not Cl⁻.

1

8 HI + H_{2} SO_{4} \to 4 I_{2} + H_{2} S + 4 H_{2} O

2

S: +6 → -2 (reduction of 8 units)

3

I⁻ is a much stronger reducing agent than Cl⁻ (larger radius, more shielding, electron more easily lost). Cl⁻ is too weak to reduce S at all.

Practice

Question 3: Acidified AgNO₃ gives a white precipitate that dissolves in dilute NH₃. Write ionic equations and identify the halide.

1

Halide: Cl⁻ (chloride) - white precipitate dissolving in dilute NH₃ is unique to AgCl

2

Precipitation:

Ag^{+} (aq) + Cl^{-} (aq) \to AgCl(s)

3

Dissolution:

AgCl(s) + 2 NH_{3} (aq) \to [Ag(NH_{3})_{2}]^{+} (aq) + Cl^{-} (aq)

Summary

Key Trends in Group 7

PropertyTrend Down Group

Atomic radiusIncreases (more electron shells)

Boiling pointIncreases (stronger van der Waals)

ReactivityDecreases (harder to gain electrons)

Oxidising ability (X₂)Decreases (Cl₂ > Br₂ > I₂)

Reducing ability (X⁻)Increases (I⁻ > Br⁻ > Cl⁻)

Key Exam Skills

Write ionic equations for displacement and precipitation reactions

Identify oxidation state changes in halide + H₂SO₄ reactions

Describe observations and products for each halide with conc. H₂SO₄

Apply silver nitrate and ammonia tests to identify halide ions