• Colour gets darker down the group: F₂ pale yellow → Cl₂ green/yellow → Br₂ orange/brown → I₂ grey/black solid (purple vapour).
• Volatility decreases down the group (melting/boiling points increase) because London dispersion forces get stronger between bigger molecules.
• Electronegativity and oxidising power decrease down the group — bigger atoms = more shielding = weaker pull on incoming electrons.
• Reactivity with hydrogen gets less vigorous down the group (F₂ explosive in the dark → I₂ needs heat and reaches equilibrium).
• Displacement reactions: a more reactive (higher) halogen displaces a less reactive (lower) halide ion from solution. Chlorine displaces both bromide and iodide; bromine displaces iodide.
• Redox with metals: halogens oxidise Group 1/2 metals to form ionic metal halides.
• Redox with iron: Cl₂ and Br₂ oxidise Fe²⁺ → Fe³⁺, but I₂ can't — instead Fe³⁺ oxidises I⁻ → I₂.
• Disproportionation: chlorine reacts with cold/hot alkali and with water, getting oxidised AND reduced simultaneously in the same reaction.
• Concentrated H₂SO₄ + halide ions produces different products depending on halide size — this is the classic "identify the mystery gas" topic, and iodide (the best reducing agent) produces the widest range of products.
• Reducing power of halide ions increases down the group (I⁻ > Br⁻ > Cl⁻) — the opposite trend to oxidising power of the halogens themselves.
• Silver nitrate + ammonia test identifies halide ions by precipitate colour and solubility in ammonia.

Every halogen has its own signature colour, and they get progressively darker as you move down the group. This isn't just a fun fact to memorise — it's actually how you identify which halogen has formed in a displacement reaction (more on that in Section 2), so get this table locked into memory.

Why the darkening trend? As atoms get bigger, they have more electrons that are more loosely held, which changes how they absorb and reflect visible light — the bonding electrons need less energy to get excited, shifting absorption toward longer (redder) wavelengths, giving darker colours.

Volatility just means "how easily something evaporates" — a highly volatile substance has a low boiling point (it turns to gas easily). Fluorine is the most volatile halogen; iodine is the least.

Here's the reasoning chain, step by step, because this is a classic "explain the trend" exam question:

• Halogens exist as diatomic molecules (X₂) — simple molecular structures.
• Between these molecules, the only forces holding them together are weak London (van der Waals) dispersion forces, caused by temporary/instantaneous dipoles.
• An instantaneous dipole happens when electrons momentarily bunch up on one side of a molecule. This can then induce a dipole in a neighbouring molecule — and the two attract.
• The more electrons a molecule has, the stronger these instantaneous dipole-induced dipole forces become.
• Going down the group, molecules get bigger (more electrons) → stronger London forces → more energy needed to separate them → higher melting and boiling points → lower volatility.

Electronegativity is how strongly an atom attracts a bonding pair of electrons toward itself. It decreases down Group 7: F ≈ 4.0 → Cl ≈ 3.0 → Br ≈ 2.8 → I ≈ 2.5.

Why? Going down the group, atomic radius increases, so the outer shell (where the "incoming" electron would go) sits further from the nucleus. There's also more shielding from inner-shell electrons blocking the pull of the positive nucleus. Net result: the nucleus's grip on any electron trying to join weakens.

This directly explains why reactivity of the halogens themselves decreases down the group — bigger atoms find it harder to grab the extra electron needed to form the 1⁻ ion.

A brilliant, concrete way to see the reactivity trend in action:

Notice the iodine reaction uses a reversible arrow (⇌), not a one-way arrow — iodine is so unreactive that the reaction doesn't even finish; it settles into an equilibrium.

This is the practical, "watch the colour change" side of reactivity. The rule is simple:

Chlorine + bromide:

Ionic equation (removing spectator Na⁺):

Chlorine + iodide:

Bromine + iodide:

Checking oxidation numbers proves these are redox reactions: the halide ion goes from −1 → 0 (oxidised), and the free halogen goes from 0 → −1 (reduced). Spectator ions like Na⁺ or K⁺ never change oxidation number.

Colour observations are commonly examined — know these cold:

Halogens react with reactive metals to form ionic metal halide salts. In every case, the metal is oxidised (loses electrons) and the halogen is reduced (gains electrons) — the halogen is acting as an oxidising agent.

This is a really elegant illustration of the oxidising power trend.

Chlorine and bromine are both strong enough oxidising agents to pull an electron off Fe²⁺, turning it into Fe³⁺.

But iodine can't do this — it's too weak an oxidising agent. In fact, the reaction runs the opposite way: Fe³⁺ is a strong enough oxidising agent to oxidise iodide ions into iodine.

A disproportionation reaction is one where the same species is simultaneously oxidised and reduced. Chlorine reacting with alkali is the textbook example, and the products depend entirely on the temperature.

(a) Cold dilute alkali (≈15 °C):

Ionic form: Cl₂(aq) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(aq)

• Chlorine oxidised: 0 → +1 (in ClO⁻, chlorate(I) ion)
• Chlorine reduced: 0 → −1 (in Cl⁻)

Half-equations:

(b) Hot concentrated alkali (≈70 °C):

Ionic form: Cl₂(aq) + 6OH⁻(aq) → 5Cl⁻(aq) + ClO₃⁻(aq) + H₂O(l)

• Chlorine oxidised: 0 → +5 (in ClO₃⁻, chlorate(V) ion — the bleach-active species is different at higher temp)
• Chlorine reduced: 0 → −1 (in Cl⁻)

Water treatment relies on this same disproportionation idea:

Chlorine is reduced (0 → −1, in HCl) and oxidised (0 → +1, in HClO) at the same time. HClO (chloric(I) acid) is the star of the show — it sterilises water by killing bacteria. It can further dissociate:

The ClO⁻ ion produced also has sterilising power, so both species contribute to making water safe to drink.

This is one of the most heavily examined practicals in the whole topic. Solid sodium halides react with concentrated H₂SO₄ (added dropwise), and the products get progressively more dramatic as you go from chloride → bromide → iodide.

Chloride — reaction stops here.

Observation: misty white fumes of HCl gas. Chloride ions are too weak a reducing agent to reduce the sulfuric acid any further — the reaction stops at this simple acid–base step.

Bromide — HBr forms, then gets oxidised further.

Observations: misty fumes of HBr, PLUS a choking gas (SO₂) and reddish-brown Br₂ gas. Bromide is a stronger reducing agent than chloride, strong enough to reduce sulfuric acid down to sulfur (oxidation state +4, i.e. SO₂).

Iodide — the full show. Three extents of reduction.

Iodide is the strongest reducing agent of the three, so sulfuric acid gets reduced all the way through several oxidation states of sulfur: +6 (H₂SO₄) → +4 (SO₂) → 0 (S) → −2 (H₂S).

This confuses a lot of students, so let's be crystal clear about the direction:

Why? As you go down Group 7, halide ions get bigger (Cl⁻ → Br⁻ → I⁻). Their outer electrons are further from the nucleus and more shielded, so the nucleus holds onto them less tightly. This makes it easier for the larger ion to lose an electron — and losing an electron to reduce something else is exactly what a reducing agent does.

This is precisely why iodide can reduce sulfuric acid so much further than chloride can — it's simply a much stronger electron donor.

A classic qualitative analysis test: dissolve the unknown solution in dilute nitric acid (to remove interference from carbonate/hydroxide ions), then add silver nitrate solution.

A precipitate forms whose colour gives a first clue, and whose solubility in ammonia confirms the identity:

Logic in plain English: AgCl is the most soluble/weakest precipitate, so even weak (dilute) ammonia can dissolve it. AgBr needs the stronger, more concentrated ammonia to break it apart. AgI is so insoluble that even concentrated ammonia can't touch it.

Fluorine and astatine aren't studied in as much detail, but exam questions often ask you to predict their behaviour by extrapolating the trend. Follow the pattern — but watch for the flagged exception!