12.1
Physical Properties of Group 17 Elements
Overview of the Halogens
Group 17 (halogens — Greek: "salt-formers"): Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At). All have outer electron configuration ns²np⁵ — 7 valence electrons, one electron short of a noble gas. They are the most electronegative group (F most electronegative element, EN = 4.0).
All halogens exist as diatomic molecules (X₂) in their elemental form. They react vigorously with metals to form ionic halide salts and with non-metals to form covalent halides.
| Element | Z | Config | State (25°C) | Colour | M.p. (°C) | B.p. (°C) | EN |
|---|---|---|---|---|---|---|---|
| Fluorine (F₂) | 9 | [He]2s²2p⁵ | Gas | Pale yellow | −220 | −188 | 4.0 |
| Chlorine (Cl₂) | 17 | [Ne]3s²3p⁵ | Gas | Yellow-green | −101 | −34 | 3.0 |
| Bromine (Br₂) | 35 | [Ar]3d¹⁰4s²4p⁵ | Liquid | Red-brown | −7 | +59 | 2.8 |
| Iodine (I₂) | 53 | [Kr]4d¹⁰5s²5p⁵ | Solid | Dark grey/purple vapour | +114 | +184 | 2.5 |
| Astatine (At₂) | 85 | [Xe]4f¹⁴5d¹⁰6s²6p⁵ | Solid (radioactive) | Black (predicted) | ~302 | ~337 | 2.2 |
Trends in Physical Properties
- State at room temperature — F₂ and Cl₂ are gases; Br₂ is the only liquid non-metallic element; I₂ is a solid. This reflects increasing London dispersion forces (more electrons → stronger vdW).
- Colour — deepens down the group: pale yellow → yellow-green → red-brown → dark grey/violet
- Melting and boiling points — increase down the group (stronger London forces with more electrons)
- Atomic radius — increases (more shells)
- Electronegativity — decreases: F (4.0) > Cl (3.0) > Br (2.8) > I (2.5)
- Oxidising power — decreases: F₂ > Cl₂ > Br₂ > I₂ (key trend!)
- Bond dissociation enthalpy of X₂ — generally decreases (F–F is anomalously low at 158 kJ/mol due to lone pair–lone pair repulsion in the small F₂ molecule)
- X–H bond strength — decreases: H–F (565) > H–Cl (432) > H–Br (366) > H–I (298) kJ/mol
Why is Bromine the only liquid non-metal at room temperature?
Br₂ has 70 electrons — strong enough London dispersion forces to keep it liquid at 25°C (b.p. = 59°C). F₂ and Cl₂ have too few electrons (18 and 34) for strong enough London forces → gases. I₂ has 106 electrons but forms a close-packed solid crystal → solid.
12.2
Oxidising Power and Displacement Reactions
Trend in Oxidising Power
The halogens are excellent oxidising agents — they gain electrons readily (X₂ + 2e⁻ → 2X⁻). Oxidising power decreases down the group: F₂ > Cl₂ > Br₂ > I₂.
Explanation: Going down the group, atomic radius increases. The incoming electron is further from the nucleus and is more shielded → less attracted → less easily gained → weaker oxidising agent.
| Half-equation | E° (V) | Oxidising power |
|---|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 | Strongest oxidising agent |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 | Strong |
| Br₂ + 2e⁻ → 2Br⁻ | +1.07 | Moderate |
| I₂ + 2e⁻ → 2I⁻ | +0.54 | Weakest oxidising agent |
Halogen Displacement Reactions
A more reactive (stronger oxidising) halogen can displace a less reactive one from its salt solution:
Cl₂ + 2KBr(aq) → 2KCl(aq) + Br₂ (Cl₂ displaces Br⁻ → orange/brown solution)
Cl₂ + 2KI(aq) → 2KCl(aq) + I₂ (Cl₂ displaces I⁻ → brown solution; blue-black with starch)
Br₂ + 2KI(aq) → 2KBr(aq) + I₂ (Br₂ displaces I⁻ → brown → blue-black with starch)
Br₂ + 2KCl(aq) → NO REACTION (Br₂ cannot displace Cl⁻ — Cl₂ stronger oxidising agent)
I₂ + 2KCl(aq) → NO REACTION (I₂ weakest — cannot displace Br⁻ or Cl⁻)
Memory tip — displacement reactions
A halogen can only displace halides BELOW it in Group 17. F displaces all; Cl displaces Br and I; Br displaces only I; I displaces none. "Stronger halogens kick out weaker ones."
Confirming displacement — solvent extraction test: Add hexane (cyclohexane) to the reaction mixture and shake. The organic layer separates and colours indicate:
- Cl₂ dissolved in hexane → very pale yellow (almost colourless)
- Br₂ dissolved in hexane → orange-red
- I₂ dissolved in hexane → violet/purple
Worked Example 12.1 — Displacement + Organic Layer
Question: Aqueous KI is treated with Cl₂(aq), then hexane is added and shaken. Describe and explain what is observed.
1
Cl₂ is a stronger oxidising agent than I₂: Cl₂ + 2I⁻ → 2Cl⁻ + I₂. Iodine is displaced — the aqueous layer turns brown (I₂ in water).
2
When hexane is added and shaken, I₂ preferentially dissolves in the non-polar hexane layer (like dissolves like — I₂ is non-polar).
3
The hexane layer (top) turns violet/purple — I₂ dissolved in hexane. The aqueous layer (bottom) becomes paler.
4
Adding starch solution to the aqueous layer gives a blue-black colour — confirms the presence of I₂ (iodine-starch complex).
12.3
Reactions of the Halogens
Reactions with Metals
All halogens react with metals to form ionic metal halides (except some halides of less reactive metals which have covalent character):
2Na + Cl₂ → 2NaCl (burns in Cl₂ — bright white flame)
2Fe + 3Cl₂ → 2FeCl₃ (iron burns in Cl₂ → Fe³⁺; in I₂ → FeI₂ (Fe²⁺ only — I₂ weaker oxidiser))
2Al + 3Cl₂ → 2AlCl₃ (AlCl₃ — covalent solid, hydrolyses in water)
Cu + Cl₂ → CuCl₂ (copper(II) — Cl₂ strong enough to oxidise to +2)
Cu + I₂ → NO CuI₂; instead: 2Cu + I₂ → 2CuI (I₂ too weak to oxidise Cu²⁺ → CuI forms)
Iron + Halogens: Fe²⁺ vs Fe³⁺
Cl₂ is a strong enough oxidising agent to oxidise Fe to Fe³⁺ (FeCl₃). Br₂ also gives FeBr₃. I₂ is a weaker oxidising agent and can only oxidise Fe to Fe²⁺ (FeI₂) — it cannot achieve Fe³⁺ because I₂ is not strong enough (and Fe³⁺ would actually oxidise I⁻ back to I₂).
Reactions with Non-metals and Hydrogen
With hydrogen (vigour decreases F > Cl > Br > I):
H₂ + F₂ → 2HF (explosive, even in the dark — room temperature)
H₂ + Cl₂ → 2HCl (explosive in UV light; slow in dark; 300°C in diffuse daylight)
H₂ + Br₂ → 2HBr (reversible; 300°C, Pt catalyst)
H₂ + I₂ ⇌ 2HI (reversible equilibrium; high T — never complete)
With water:
F₂ + H₂O → 2HF + ½O₂ (or: 2F₂ + 2H₂O → 4HF + O₂)
Cl₂ + H₂O ⇌ HCl + HOCl (chlorine water — reversible; HOCl = hypochlorous acid)
Br₂ + H₂O ⇌ HBr + HOBr (less extent than Cl₂)
I₂ + H₂O → negligible reaction (I₂ barely reacts with water)
With sodium hydroxide (cold, dilute):
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O (bleach — sodium chlorate(I))
With NaOH (hot, concentrated):
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (sodium chlorate(V))
Reactions of Fluorine — Special Cases
Fluorine is so reactive that it shows unique behaviour:
- Fluorine reacts with water (unlike other halogens which only partially dissolve): 2F₂ + 2H₂O → 4HF + O₂
- Fluorine can react with noble gases: Xe + F₂ → XeF₂ (xenon difluoride — impossible for other halogens)
- Fluorine cannot be made by chemical oxidation — it is prepared only by electrolysis of molten KHF₂ (Moissan process, 1886)
- F₂ has anomalously low bond energy (158 kJ/mol) compared to Cl₂ (243), Br₂ (193), I₂ (151 kJ/mol) — due to lone pair–lone pair repulsion in the very small F₂ molecule
- Fluorine shows only −1 oxidation state (most electronegative — cannot be oxidised to positive OS)
12.4
Hydrogen Halides (HX)
Properties of Hydrogen Halides
| Property | HF | HCl | HBr | HI |
|---|---|---|---|---|
| Boiling point (°C) | +19.5 | −85 | −67 | −35 |
| Bond energy (kJ/mol) | 565 | 432 | 366 | 298 |
| Acid strength in water | Weak (Ka = 6.3×10⁻⁴) | Strong | Strong | Strongest |
| Reducing power of X⁻ | None | None (HCl with MnO₂ → Cl₂ only under forcing) | Moderate — HBr reduced by H₂SO₄ | Strong — HI reduces H₂SO₄ to H₂S |
| Thermal stability | Very stable (strong H–F) | Stable | Less stable | Decomposes above ~300°C |
Anomalous boiling point of HF
HF has an anomalously HIGH boiling point (+19.5°C) compared to HCl (−85°C), HBr (−67°C), HI (−35°C). This is because HF molecules form strong F–H···F hydrogen bonds — F is the most electronegative element and has a small atomic radius. HCl, HBr, HI cannot form proper H-bonds → only van der Waals forces → lower b.p. The trend from HCl → HI is a regular increase because London forces increase with electron count.
Preparation of Hydrogen Halides
HF and HCl — from concentrated H₂SO₄
CaF₂ + H₂SO₄(conc.) → CaSO₄ + 2HF↑ (gently warm — HF is a gas)
NaCl + H₂SO₄(conc.) → NaHSO₄ + HCl↑ (H₂SO₄ displaces more volatile HCl)
Why can't HBr or HI be made this way?
Hot concentrated H₂SO₄ is an oxidising agent. It would oxidise HBr → Br₂ and HI → I₂ (and further to H₂S). Therefore, a non-oxidising acid (H₃PO₄) must be used instead for HBr and HI.
HBr and HI — from H₃PO₄ (non-oxidising acid)
NaBr + H₃PO₄ → NaH₂PO₄ + HBr↑ (H₃PO₄ non-oxidising — won't oxidise Br⁻)
NaI + H₃PO₄ → NaH₂PO₄ + HI↑ (H₃PO₄ non-oxidising — won't oxidise I⁻)
Alternative — direct synthesis
H₂ + X₂ → 2HX (F₂: room temp; Cl₂: UV; Br₂: 300°C + Pt; I₂: 300°C, equilibrium)
Reducing Power of Halide Ions — Reaction with Concentrated H₂SO₄
This is a key comparison test. The reducing power of X⁻ ions increases: F⁻ < Cl⁻ < Br⁻ < I⁻ (opposite to oxidising power of X₂).
NaF + H₂SO₄(conc.) → NaHSO₄ + HF↑ (no redox — F⁻ cannot reduce H₂SO₄)
NaCl + H₂SO₄(conc.) → NaHSO₄ + HCl↑ (no redox — Cl⁻ cannot reduce H₂SO₄; only acid-base)
NaBr + H₂SO₄(conc.) → NaHSO₄ + HBr↑ (some HBr initially)
then: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O (Br⁻ reduces S from +6→+4; Br₂ produced — orange fumes)
NaI + H₂SO₄(conc.) → NaHSO₄ + HI↑ (HI initially)
then: 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O (I⁻ strong reducing agent — reduces S all the way to −2 in H₂S; I₂ — black solid/purple vapour)
Summary of observations with conc. H₂SO₄:
NaF → steamy HF fumes only. NaCl → steamy HCl fumes only. NaBr → steamy fumes then orange/brown Br₂ fumes + SO₂ (choking). NaI → steamy fumes then violet I₂ vapour + rotten-egg H₂S smell (very distinctive). The further down the group, the greater the reducing power and the more severe the oxidation of H₂SO₄.
12.5
Metal Halides — Properties and Trends
Ionic vs Covalent Halides
The bonding in metal halides depends on the metal's charge density and the halide's polarisability:
- Alkali metal halides (NaX, KX) — ionic; NaCl structure or CsCl structure; high m.p.; conduct when molten/dissolved
- Alkaline earth metal halides (MgX₂, CaX₂) — ionic; high m.p.; conduct
- Higher oxidation state metal halides (AlCl₃, FeCl₃, SnCl₄) — covalent; lower m.p.; hydrolyse in water
- Fluorides are more ionic than the corresponding chloride — F⁻ is small, less polarisable. Iodides are more covalent than fluorides (I⁻ large and very polarisable)
Fajans' Rules — Covalent Character in Halides
Covalent character in a "ionic" compound increases when: (1) the cation has a high charge and small radius (high charge density) → AlCl₃ more covalent than NaCl; (2) the anion is large and easily polarised → iodides more covalent than fluorides. A small, highly charged cation distorts the electron cloud of the anion → charge density on cation approaches that needed for covalent bond.
Silver Halides — Solubility and Stability
| Salt | Colour | Solubility in water | Solubility in NH₃(aq) | Light sensitivity |
|---|---|---|---|---|
| AgF | White | Soluble | Soluble | Stable |
| AgCl | White | Insoluble | Dissolves in dilute NH₃: AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻ | Turns grey in UV |
| AgBr | Cream/pale yellow | Insoluble | Dissolves in conc. NH₃ only | Sensitive — used in photography |
| AgI | Yellow | Insoluble | Insoluble in NH₃ | Very sensitive |
The trend in solubility (AgF soluble → AgI insoluble) reflects increasing covalent character: As the anion gets larger (F⁻ → I⁻), it is more polarisable → more covalent character → less lattice energy released on dissolution → less soluble.
12.6
Oxoacids of the Halogens and Halates
Oxoacids of Chlorine
| Name | Formula | OS of Cl | Acid strength | Notes |
|---|---|---|---|---|
| Hypochlorous acid | HOCl (HClO) | +1 | Weak (Ka = 4.7×10⁻⁸) | Formed when Cl₂ dissolves in water; oxidising/bleaching agent |
| Chlorous acid | HClO₂ | +3 | Weak | Unstable, rarely encountered |
| Chloric acid | HClO₃ | +5 | Strong | Formed in hot NaOH + Cl₂; basis of KClO₃ |
| Perchloric acid | HClO₄ | +7 | Strongest common acid | Stable; thermally sensitive at high conc.; made from NaClO₄ |
Acid strength increases with oxidation state of Cl
As more O atoms are bonded to Cl (higher OS), the O–H bond is weaker → easier to ionise → stronger acid. HClO₄ (+7) is a much stronger acid than HOCl (+1). This is a general trend for oxoacids: more O atoms on the central atom = stronger acid.
Chlorates and Perchlorates
Chlorate(I) — hypochlorite: NaOCl (bleach), OCl⁻, OS = +1
Chlorate(V) — chlorate: KClO₃, ClO₃⁻, OS = +5
Perchlorate: — perchlorate: KClO₄, ClO₄⁻, OS = +7
Thermal decomposition of KClO₃:
2KClO₃ → 2KCl + 3O₂ (MnO₂ catalyst, 200–300°C) — lab source of O₂
4KClO₃ → 3KClO₄ + KCl (slow, above 400°C — disproportionation)
KClO₃ is used in matches, flares, and fireworks (strong oxidising agent)
KClO₄ (perchlorate) — stable oxidising agent; rocket propellant
12.7
Disproportionation Reactions
Definition — Disproportionation
A reaction in which the same element is simultaneously oxidised and reduced. The element is in an intermediate oxidation state and reacts with itself — some atoms are oxidised to a higher state and some are reduced to a lower state.
Disproportionation of Halogens in Alkali
Cl₂ + 2NaOH(cold, dilute) → NaCl + NaOCl + H₂O
Cl₂: OS = 0 → NaCl (Cl = −1, reduced) + NaOCl (Cl = +1, oxidised)
This is disproportionation: 0 → −1 and +1 simultaneously
3Cl₂ + 6NaOH(hot, conc.) → 5NaCl + NaClO₃ + 3H₂O
0 → −1 (in 5NaCl) and +5 (in NaClO₃) — more extensive disproportionation
Similarly for bromine:
Br₂ + 2NaOH(cold) → NaBr + NaOBr + H₂O
3Br₂ + 6NaOH(hot) → 5NaBr + NaBrO₃ + 3H₂O
Iodine in alkali
I₂ + 2NaOH → NaI + NaIO + H₂O (cold — NaIO, iodate(I), unstable, rapidly converts to iodate(V)). 3I₂ + 6NaOH → 5NaI + NaIO₃ + 3H₂O even at room temperature (iodine prefers the +5 state in basic solution). This is why the iodine/alkali reaction gives iodate(V) rather than iodate(I) as the main product.
Other Disproportionation Examples
Comproportionation (reverse of disproportionation — two different OS → one):
Cl₂ + SO₂ + 2H₂O → HCl + H₂SO₄ (not a disproportionation of Cl)
True disproportionation examples:
2H₂O₂ → 2H₂O + O₂ (H₂O₂: O = −1 → H₂O (O = −2) + O₂ (O = 0))
3MnO₄²⁻ + 2H₂O → 2MnO₄⁻ + MnO₂ + 4OH⁻ (manganate disproportionates in water)
Cu⁺ + Cu⁺ → Cu + Cu²⁺ (Cu⁺ disproportionates in water — this is why CuI is stable but CuCl not)
12.8
Uses of Halogens and Tests for Halide Ions
Industrial and Everyday Uses of Halogens
| Halogen / Compound | Use | Reason |
|---|---|---|
| Cl₂ (chlorine gas) | Water treatment / sterilisation | Cl₂ + H₂O → HOCl — kills bacteria and viruses |
| NaOCl (bleach) | Household bleach, disinfectant | HOCl releases [O] — oxidises colour molecules and bacteria |
| PVC (polyvinylchloride) | Pipes, flooring, clothing, insulation | Addition polymer from chloroethene (vinyl chloride) |
| HCl | Metal pickling, food processing (HCl in stomach), PVC production | Strong acid; reaction with metal oxides |
| CFCs (Freons) | Now banned — formerly refrigerants, aerosol propellants | Inert, non-toxic, low b.p. — but destroy ozone layer |
| HFCs | Modern refrigerants (replacing CFCs) | No Cl → no ozone depletion; still greenhouse gases |
| PTFE (Teflon) | Non-stick cookware, cables | Very inert (strong C–F bonds), low friction |
| AgBr | Photographic film (now mostly replaced by digital) | Light-sensitive: 2AgBr + light → 2Ag + Br₂ (darkens) |
| NaI (iodide) | Iodised salt — prevents goitre (iodine deficiency) | I⁻ absorbed by thyroid gland — needed for thyroxine hormone |
| I₂ | Antiseptic (tincture of iodine); starch test | Kills bacteria; I₂ + starch → blue-black complex |
| F₂ / HF | Production of uranium hexafluoride (UF₆) for nuclear fuel enrichment; etching glass | HF attacks SiO₂: SiO₂ + 4HF → SiF₄ + 2H₂O |
Tests for Halide Ions (X⁻)
Add dilute nitric acid (to remove CO₃²⁻, SO₃²⁻ and other interferences) then add silver nitrate solution (AgNO₃):
Ag⁺ + F⁻ → NO precipitate (AgF is soluble)
Ag⁺ + Cl⁻ → AgCl↓ (white precipitate)
Ag⁺ + Br⁻ → AgBr↓ (cream/pale yellow precipitate)
Ag⁺ + I⁻ → AgI↓ (yellow precipitate)
Distinguishing the precipitates — add ammonia solution:
AgCl + 2NH₃(dil.) → [Ag(NH₃)₂]⁺ + Cl⁻ (dissolves in DILUTE NH₃)
AgBr: dissolves only in CONCENTRATED NH₃
AgI: does NOT dissolve in NH₃ (even concentrated)
| Test | F⁻ | Cl⁻ | Br⁻ | I⁻ |
|---|---|---|---|---|
| AgNO₃/HNO₃ | No ppt | White ppt | Cream ppt | Yellow ppt |
| + dilute NH₃ | — | Dissolves | Remains | Remains |
| + conc. NH₃ | — | Dissolves | Dissolves | Remains |
Memory: AgF, AgCl, AgBr, AgI colours
"WCCY" — White, Cream, Cream-yellow, Yellow. Or: "Further down, deeper colour." AgF = soluble (no ppt). AgCl = white. AgBr = cream. AgI = yellow.
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Exercises
- Explain the trend in oxidising power across Group 17 from F₂ to I₂. Give one displacement reaction equation to illustrate this trend.
Oxidising power decreases: F₂ > Cl₂ > Br₂ > I₂. Reason: going down group, atomic radius increases and the outer shell is more shielded. The incoming electron (to form X⁻) is added to a larger shell, further from the nucleus, with more shielding → less attracted by the nucleus → less readily gained → weaker oxidising agent. Example: Cl₂ + 2KBr(aq) → 2KCl(aq) + Br₂ (Cl₂ displaces Br⁻ because Cl₂ is the stronger oxidising agent). Ionic equation: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
- Describe what you would observe when concentrated H₂SO₄ is added to (a) NaCl (b) NaBr (c) NaI. Write equations and explain the differences in terms of reducing power of the halide ions.
(a) NaCl + H₂SO₄ → NaHSO₄ + HCl↑. Observation: steamy white/colourless HCl fumes. No redox — Cl⁻ too weak a reducing agent to reduce H₂SO₄. (b) NaBr + H₂SO₄ → NaHSO₄ + HBr↑; then 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. Observation: steamy fumes initially, then orange-brown Br₂ fumes + choking SO₂ smell. Br⁻ reduces S from +6 → +4. (c) NaI + H₂SO₄ → NaHSO₄ + HI↑; then 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O. Observation: steamy fumes, then black solid I₂ / violet vapour + rotten-egg H₂S gas. I⁻ is strongest reducing agent — reduces S all the way from +6 to −2. Reducing power: I⁻ > Br⁻ > Cl⁻ (increasing atomic radius, weaker bond to H, more easily oxidised).
- Explain what is meant by disproportionation. Write the equation for the reaction of Cl₂ with cold dilute NaOH. Identify the oxidation state changes.
Disproportionation: a reaction in which the SAME element is simultaneously oxidised (OS increases) AND reduced (OS decreases) in the same reaction. Cl₂ + 2NaOH(cold, dilute) → NaCl + NaOCl + H₂O. OS changes: Cl₂ has Cl in OS = 0. In NaCl: Cl = −1 (REDUCED). In NaOCl: Cl = +1 (OXIDISED). So Cl goes from 0 → −1 (reduction) and 0 → +1 (oxidation) simultaneously. This is a disproportionation. The product NaOCl (sodium hypochlorite) is household bleach.
- Explain why HF has a much higher boiling point (19.5°C) than HCl (−85°C) even though HF has lower molar mass. Why does the boiling point then increase from HCl → HBr → HI?
HF high b.p.: fluorine is very small (Period 2) and highly electronegative (4.0). The H–F bond is very polar → HF molecules form strong F–H···F hydrogen bonds. Many hydrogen bonds must be broken to vaporise HF → high b.p. (19.5°C). HCl lower b.p.: Cl is larger (Period 3, EN=3.0), H–Cl bond is less polar → no significant H-bonds → only weak London dispersion forces → easy to vaporise → b.p. = −85°C. HCl → HBr → HI increase: none of these can form H-bonds (Cl, Br, I not small/electronegative enough). They only have van der Waals (London) forces. Going from HCl → HBr → HI: more electrons, larger molecules → stronger London forces → higher b.p. (−85 → −67 → −35°C).
- Describe how you would distinguish between solutions of NaCl, NaBr, and NaI using silver nitrate solution and ammonia. State observations and equations for each.
Step 1: Add dilute HNO₃ (to remove interferents), then AgNO₃(aq). NaCl: white precipitate (AgCl). NaBr: cream precipitate (AgBr). NaI: yellow precipitate (AgI). Equations: Ag⁺ + Cl⁻ → AgCl↓ (white); Ag⁺ + Br⁻ → AgBr↓ (cream); Ag⁺ + I⁻ → AgI↓ (yellow). Step 2: Add dilute NH₃: AgCl dissolves (AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻); AgBr and AgI remain. Step 3: Add concentrated NH₃: AgBr dissolves; AgI remains. Summary: Cl⁻ = dissolves in dilute NH₃; Br⁻ = requires conc. NH₃; I⁻ = insoluble in NH₃. Also colour of precipitate alone distinguishes them.
- Why can fluorine NOT be prepared by chemical oxidation of F⁻ ions, whereas Cl₂, Br₂, and I₂ can? How is F₂ manufactured industrially?
F₂ is the strongest oxidising agent known (E° = +2.87V). No chemical oxidising agent is strong enough to oxidise F⁻ to F₂ — F⁻ simply cannot be chemically oxidised by any reagent that exists. In contrast, Cl⁻ can be oxidised to Cl₂ by MnO₂ + HCl, Br⁻ by Cl₂, I⁻ by Cl₂ or Br₂. Industrial production of F₂: electrolysis of molten potassium hydrogen difluoride (KHF₂) dissolved in liquid HF at ~70°C. This is the Moissan process (Henri Moissan, Nobel Prize 1906). Anode: 2F⁻ → F₂ + 2e⁻. Steel or copper containers are used (they form a protective fluoride coating — passivation).
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Multiple Choice Quiz — 25 Questions
Unit 12: Group 17 — The Halogens
25 Questions · Select one answer eachQ1
The outer electron configuration of all Group 17 elements is:
Group 17 halogens: ns²np⁵ — 7 valence electrons. One electron short of a noble gas configuration. This drives them to gain one electron (forming X⁻, OS = −1) or share electrons in covalent bonds.
Q2
Which halogen is a red-brown liquid at room temperature?
Bromine (Br₂) is a red-brown liquid — it is the only non-metallic element that is a liquid at room temperature. F₂ and Cl₂ are gases (pale yellow and yellow-green respectively). I₂ is a dark grey/black solid.
Q3
The oxidising power of halogens decreases in the order:
F₂ > Cl₂ > Br₂ > I₂. F₂ is the strongest oxidising agent in all of chemistry (E° = +2.87V). Down the group: atomic radius increases, shielding increases → incoming electron less attracted → weaker oxidising agent.
Q4
Chlorine water (Cl₂ dissolved in water) contains which species?
Cl₂ + H₂O ⇌ H⁺ + Cl⁻ + HOCl (reversible equilibrium). Chlorine water contains unreacted Cl₂, plus the products of the disproportionation: H⁺, Cl⁻, and HOCl (hypochlorous acid). HOCl is a weak acid but a strong oxidising/bleaching agent — it is responsible for bleaching action.
Q5
When chlorine water is added to potassium bromide solution, the solution turns:
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. Br₂ is displaced — it produces an orange-brown colour in aqueous solution. If hexane is added, the organic layer turns orange-red (Br₂ more soluble in non-polar hexane). Blue-black would indicate I₂ with starch.
Q6
Iodine solution turns starch:
I₂ + starch → intense blue-black complex (I₃⁻ ions fit into the helical structure of amylose starch). This is a very sensitive and specific test for I₂ (or I₃⁻). Used as the endpoint indicator in iodometric titrations. The blue-black colour disappears on heating (I₂ evaporates) and returns on cooling.
Q7
Why does HF have a higher boiling point than HCl despite having a lower molar mass?
F is small (Period 2) and highly electronegative (4.0) → very polar H–F bond → strong hydrogen bonds F–H···F. Many H-bonds must be broken to boil HF → b.p. = +19.5°C. HCl: Cl too large/not electronegative enough to form H-bonds → only van der Waals → b.p. = −85°C. Same anomaly as H₂O vs H₂S and NH₃ vs PH₃.
Q8
The reaction of Cl₂ with cold dilute NaOH is described as disproportionation because:
Cl₂ + 2NaOH → NaCl + NaOCl + H₂O. Cl₂ (Cl = 0) → NaCl (Cl = −1, reduced) and NaOCl (Cl = +1, oxidised). The SAME element (Cl) undergoes both oxidation (+) and reduction (−) simultaneously. Definition of disproportionation: one element in one oxidation state converts to two different oxidation states.
Q9
Silver halide precipitates with AgNO₃/HNO₃: which is yellow?
AgI is yellow. AgCl = white; AgBr = cream (pale yellow). AgF is soluble (no precipitate). Memory: deeper colour going down the group because increasing covalent character causes greater light absorption. AgI is also the most insoluble, insoluble even in concentrated NH₃.
Q10
HBr cannot be made by adding concentrated H₂SO₄ to NaBr because:
Hot conc. H₂SO₄ is an oxidising agent: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. The HBr produced is immediately oxidised to Br₂ — you get Br₂ vapour instead of HBr. Therefore H₃PO₄ (non-oxidising) must be used instead. Same problem for HI (8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O).
Q11
The oxidation state of Cl in household bleach (NaOCl) is:
NaOCl: Na = +1, O = −2, Cl = ? Na(+1) + O(−2) + Cl = 0 → Cl = +1. Also written as sodium chlorate(I). This is the +1 state that results from disproportionation of Cl₂ in cold NaOH. The bleaching/disinfecting action is due to Cl(+1) being a powerful oxidising agent.
Q12
When I₂ is shaken with hexane and aqueous KI solution, I₂ preferentially moves into the hexane layer because:
Like dissolves like: I₂ is a non-polar diatomic molecule (only London forces between I₂ molecules). Hexane is non-polar — London forces between I₂ and hexane molecules → I₂ dissolves in hexane (violet layer). Water is polar and can only form weak interactions with I₂ → I₂ prefers non-polar hexane. This extraction test identifies which halogen is present from the colour of the organic layer.
Q13
The reaction between iron and iodine produces:
I₂ is the weakest halogen oxidising agent — it can only oxidise Fe to Fe²⁺ (FeI₂). It cannot achieve Fe³⁺ because I₂'s oxidising power (E° = +0.54V) is insufficient. Also, Fe³⁺ would oxidise I⁻ back to I₂ — so FeI₃ is thermodynamically unstable: 2FeI₃ → 2FeI₂ + I₂. Contrast: Cl₂ → FeCl₃; Br₂ → FeBr₃.
Q14
What is the observation when AgCl is treated with dilute ammonia solution?
AgCl + 2NH₃(dil.) → [Ag(NH₃)₂]⁺ + Cl⁻. The white AgCl precipitate dissolves completely in dilute ammonia, forming the soluble diamminesilver(I) complex ion. This is used to distinguish AgCl (dissolves in dilute NH₃) from AgBr (dissolves only in concentrated NH₃) and AgI (insoluble in any concentration of NH₃).
Q15
Which halogen reacts with water to produce oxygen gas?
F₂ is so powerfully oxidising that it oxidises water: 2F₂ + 2H₂O → 4HF + O₂. F₂ oxidises O from −2 (in H₂O) to 0 (in O₂). Cl₂, Br₂, I₂ react much more gently with water via disproportionation (Cl₂ + H₂O ⇌ HCl + HOCl) but do not produce O₂. This is another indication of F₂'s extraordinary reactivity.
Q16
The acid strength of the oxoacids of chlorine increases in the order:
Acid strength increases with the number of oxygen atoms (higher OS on Cl): HOCl (+1) weakest → HClO₂ (+3) → HClO₃ (+5) → HClO₄ (+7) strongest. Each additional O atom withdraws more electron density from the O–H bond → H⁺ more easily released → stronger acid. HClO₄ (perchloric acid) is one of the strongest known acids.
Q17
Fluorine shows only the −1 oxidation state in its compounds. This is because:
Fluorine: most electronegative element (4.0) → always attracts electrons → always gains electrons → always −1. Cannot be oxidised to positive OS because: (1) No element is more electronegative than F to force it to lose electrons; (2) F is in Period 2 — no available d orbitals to expand its octet and accommodate extra bonds (unlike Cl which can form ClO₃⁻ (+5), ClO₄⁻ (+7) using 3d orbitals). F only shows 0 (in F₂) and −1.
Q18
The trend in HX bond strength across the hydrogen halides is:
H–X bond energy decreases: H–F (565) > H–Cl (432) > H–Br (366) > H–I (298) kJ/mol. As X gets larger going down Group 17, the bonding orbitals overlap less effectively (smaller H 1s orbital vs larger X np orbital) → weaker bond. This explains why HI is the strongest acid (weakest H–I bond → easiest to donate H⁺) and why the reducing power of X⁻ increases I⁻ > Br⁻ > Cl⁻.
Q19
Which of the following reactions is NOT a displacement reaction?
A displacement reaction involves a more reactive halogen pushing out a less reactive halide ion. Option C (Cl₂ + NaOH → NaCl + NaOCl + H₂O) is a disproportionation reaction — Cl₂ reacts with alkali, not with a halide ion, and the same element (Cl) is both oxidised and reduced. Options A, B, D are all displacement reactions.
Q20
Bleaching powder (CaOCl₂) releases its bleaching action because:
Bleaching powder (Ca(OCl)Cl): CaOCl₂ + CO₂ + H₂O → CaCO₃ + 2HOCl. HOCl (hypochlorous acid, Cl = +1) is the actual bleaching agent. HOCl decomposes: HOCl → HCl + [O] (nascent oxygen) — the reactive oxygen atom oxidises coloured organic dye molecules, destroying the chromophore → colour lost → bleached.
Q21
The reducing power of the halide ions increases in the order:
F⁻ < Cl⁻ < Br⁻ < I⁻ (reducing power increases going down). This is the REVERSE of oxidising power of X₂. I⁻ has the largest radius → H–I bond weakest → most easily oxidised (gives up electron most easily) → strongest reducing agent. F⁻ cannot be oxidised at all under normal conditions. Demonstrated by reactions with conc. H₂SO₄: only Br⁻ and I⁻ reduce it.
Q22
AgCl turns grey in sunlight because:
2AgCl → 2Ag + Cl₂ (photodecomposition in UV light). Silver metal (Ag⁰) is grey/black. This light sensitivity forms the basis of traditional photographic film (AgBr is even more sensitive). In photography: light hits AgBr crystals → Ag⁰ forms (latent image) → developing amplifies this → image forms.
Q23
The reaction 3Cl₂ + 6NaOH(hot, concentrated) → 5NaCl + NaClO₃ + 3H₂O gives which product containing Cl in an oxidation state of +5?
NaClO₃: Na(+1) + Cl + 3O(−2) = 0 → Cl = +5. Sodium chlorate(V). Compare with NaOCl (Cl = +1) from cold dilute NaOH. Hot concentrated NaOH gives more extensive disproportionation: Cl goes from 0 → −1 (in 5NaCl) and 0 → +5 (in NaClO₃). KClO₃ is used in matches, fireworks, and thermal decomposition to make O₂.
Q24
Iodine solution in organic solvents such as hexane appears:
I₂ dissolved in non-polar solvents (hexane, CCl₄, cyclohexane) gives a violet/purple colour. In polar solvents (water, ethanol), I₂ appears brown (complex formed with solvent molecules). The colour difference helps identify the halogen present in solvent extraction tests. Cl₂ in hexane = very pale yellow; Br₂ in hexane = orange-red; I₂ in hexane = violet.
Q25
The anomalously low bond dissociation energy of F₂ (158 kJ/mol) compared to Cl₂ (243 kJ/mol) is explained by:
F₂ has an anomalously weak F–F bond (158 kJ/mol). F is very small — the two F atoms are close together in F₂. The three lone pairs on each F interact strongly with the lone pairs on the other F atom (lone pair–lone pair repulsion) → this weakens the F–F σ bond. Cl₂ (243 kJ/mol), Br₂ (193), I₂ (151) follow the expected decreasing trend with increasing size. F₂'s weakness explains its extreme reactivity despite the strong H–F bond in products.
📝
Unit Test — 50 Marks
Section A — Short Answer
30 marksQ1 [5 marks]
Explain the trend in oxidising power of the halogens from F₂ to I₂. Use standard electrode potentials to illustrate your answer. Write one displacement reaction equation and state what you would observe. [5]
Trend: F₂ > Cl₂ > Br₂ > I₂ (decreasing oxidising power). [1] Explanation: going down group, atomic radius increases and nuclear shielding increases. The X₂ molecule is less able to attract incoming electrons to form X⁻ because the electron would be added to a larger, more shielded shell → electron affinity decreases → oxidising power decreases. [2] E° values: F₂ = +2.87V, Cl₂ = +1.36V, Br₂ = +1.07V, I₂ = +0.54V — confirms F₂ strongest, I₂ weakest. [1] Example: Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂. Observation: colourless solution → brown (I₂ in water); blue-black with starch; organic layer turns violet with hexane. [1]
Q2 [5 marks]
Describe what happens when concentrated H₂SO₄ is added separately to (a) NaCl, (b) NaBr, (c) NaI. Write equations and explain the differences in terms of reducing power of X⁻. [5]
(a) NaCl + H₂SO₄ → NaHSO₄ + HCl↑. Steamy colourless fumes. Cl⁻ cannot reduce H₂SO₄ → only acid-base, no redox. [1] (b) NaBr: NaBr + H₂SO₄ → NaHSO₄ + HBr↑; then 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. Steamy fumes, then orange-brown Br₂ fumes + choking SO₂. Br⁻ reduces S(+6) to S(+4). [2] (c) NaI: NaI + H₂SO₄ → NaHSO₄ + HI↑; then 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O. Steamy fumes, then black solid I₂ and violet I₂ vapour + rotten-egg H₂S smell. I⁻ strongest reducer — reduces S all the way to −2. [2] Order of reducing power: I⁻ > Br⁻ > Cl⁻ (increasing radius, weaker X–H bond, easier to lose electron).
Q3 [5 marks]
Describe the test for halide ions using silver nitrate. Include: (a) the reagent added before AgNO₃ and why, (b) observations for Cl⁻, Br⁻, I⁻, (c) how to distinguish between the precipitates using ammonia solution. Write equations for each step. [5]
(a) Add dilute HNO₃ first — acidification removes interfering ions (CO₃²⁻ → CO₂; SO₃²⁻ → SO₂; OH⁻ → H₂O) that would also precipitate with Ag⁺, giving false positives. [1] (b) AgNO₃ + Cl⁻ → AgCl↓ (white); Ag⁺ + Br⁻ → AgBr↓ (cream); Ag⁺ + I⁻ → AgI↓ (yellow). [1.5] (c) Add dilute NH₃: AgCl dissolves (AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻); AgBr and AgI remain. Then add concentrated NH₃: AgBr dissolves; AgI remains. Summary: Cl⁻: dissolves in dilute NH₃; Br⁻: dissolves only in conc. NH₃; I⁻: insoluble in any strength NH₃. [2.5]
Q4 [5 marks]
Explain what disproportionation means and give two examples from Group 17 chemistry. For each example, write the equation and identify the oxidation state changes. [5]
Disproportionation: the same element simultaneously undergoes both oxidation (OS increases) and reduction (OS decreases) in one reaction. [1] Example 1: Cl₂ + 2NaOH(cold) → NaCl + NaOCl + H₂O. Cl (0) → Cl(−1) in NaCl [REDUCED] and Cl(+1) in NaOCl [OXIDISED]. [2] Example 2: 3Cl₂ + 6NaOH(hot) → 5NaCl + NaClO₃ + 3H₂O. Cl(0) → Cl(−1) in NaCl [REDUCED] and Cl(+5) in NaClO₃ [OXIDISED]. [2] (Other valid examples: Cl₂ + H₂O → H⁺ + Cl⁻ + HOCl; Cu⁺ + Cu⁺ → Cu + Cu²⁺).
Q5 [5 marks]
Compare the properties of the hydrogen halides HF, HCl, HBr, HI with respect to: (a) boiling points and the anomaly of HF; (b) acid strength in water; (c) thermal stability; (d) reducing power of the corresponding halide ion. [5]
(a) B.p.: HF (+19.5) >> HCl (−85) < HBr (−67) < HI (−35)°C. HF anomaly: F–H···F hydrogen bonds (F small, highly electronegative) → much more energy to boil. HCl–HI: only van der Waals, increasing with electron count → regular increase. [1.5] (b) Acid strength: HCl, HBr, HI are all strong acids (complete ionisation). HF is a weak acid (Ka = 6.3×10⁻⁴) — anomalously: H–F bond (565 kJ/mol) is very strong, HF dimerises in solution (F–H···F⁻ remains stable). Acid strength increases down: HCl < HBr < HI (H–X bond weakens). [1.5] (c) Thermal stability decreases: HF > HCl > HBr > HI (H–I bond only 298 kJ/mol, decomposes at ~300°C; H–F 565 kJ/mol, stable to very high T). [1] (d) Reducing power of X⁻: F⁻ (none) < Cl⁻ (none normally) < Br⁻ (moderate) < I⁻ (strong). As X–H bond weakens going down, X⁻ is more easily oxidised. [1]
Q6 [5 marks]
Describe a halogen displacement experiment. Include: (a) the method with three test tubes (KCl/KBr/KI with Cl₂ and Br₂ water, then hexane extraction); (b) all observations; (c) full ionic equations for reactions that occur. [5]
(a) Method: set up 6 test tubes with: (1) KCl + Cl₂(aq); (2) KBr + Cl₂(aq); (3) KI + Cl₂(aq); (4) KCl + Br₂(aq); (5) KBr + Br₂(aq); (6) KI + Br₂(aq). Add hexane, shake, observe organic layer colour. [1] (b) Observations: (1) hexane = very pale/colourless (Cl₂ present, no displacement); (2) hexane = orange (Br₂ extracted); (3) hexane = violet (I₂ extracted, aqueous = blue-black with starch); (4) hexane = pale (no displacement — Br₂ cannot displace Cl⁻); (5) hexane = orange (Br₂, no displacement); (6) hexane = violet (I₂ extracted). [2] (c) Equations for reactions that occur: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂; Cl₂ + 2I⁻ → 2Cl⁻ + I₂; Br₂ + 2I⁻ → 2Br⁻ + I₂. No reaction: Br₂ + 2Cl⁻ → no reaction; I₂ + 2Br⁻ → no reaction; I₂ + 2Cl⁻ → no reaction. [2]
Section B — Extended Answer
20 marksQ7 [10 marks]
(a) Describe the industrial uses of chlorine and its compounds. Give five specific applications with explanations. [4]
(b) Explain the chemistry of water chlorination — what reactions occur, what species are formed, and why it is effective for sterilisation. Discuss any disadvantages. [3]
(c) Compare and contrast the silver halides (AgCl, AgBr, AgI) — formation, colour, solubility in water and in ammonia, light sensitivity. Explain the trends in terms of bonding. [3]
(a) Uses of Cl₂ (any 5): (1) Water purification: Cl₂ added to drinking water kills pathogens; Cl₂ + H₂O → HOCl (bactericidal). (2) Bleach (NaOCl): from Cl₂ + NaOH; used as household disinfectant, laundry bleach, hospital sterilisation. (3) PVC manufacture: Cl₂ + C₂H₄ → C₂H₄Cl₂ → (cracking) → CH₂=CHCl → polymerisation → PVC (pipes, insulation, flooring). (4) HCl production: H₂ + Cl₂ → 2HCl; used in metal pickling, food processing, pharmaceutical synthesis. (5) Solvents: CCl₄ (historical), CHCl₃ (chloroform — anaesthetic). (6) Agrochemicals: chlorinated pesticides (DDT — now banned; various herbicides). [4] (b) Cl₂ + H₂O ⇌ HCl + HOCl. HOCl (hypochlorous acid, Cl = +1) is the active sterilising species — it is a small, neutral molecule able to penetrate bacterial cell walls. Once inside, HOCl oxidises vital enzymes and disrupts metabolism → bacteria killed. More effective at slightly acidic pH (HOCl dissociation: HOCl ⇌ H⁺ + OCl⁻; at higher pH more OCl⁻ which is less bactericidal). Disadvantages: chlorine taste/odour in water; forms trihalomethanes (THMs) such as CHCl₃ (by-products from organic matter + HOCl) — potential carcinogens at high concentration; some bacteria develop resistance. [3] (c) Formation: Ag⁺ + X⁻ → AgX↓ (from AgNO₃ + KX). All insoluble except AgF. Colour: AgCl white, AgBr cream, AgI yellow — deepening colour reflects increasing covalent character and decreased band gap → absorbs visible light at longer wavelengths. Solubility in water: AgF soluble; AgCl, AgBr, AgI increasingly insoluble — as I⁻ is more polarisable → more covalent → less ionic → lattice harder to break by water. Ammonia: AgCl dissolves in dilute NH₃ ([Ag(NH₃)₂]⁺ formed); AgBr requires conc. NH₃; AgI insoluble in any NH₃ (very low Ksp). Light: all sensitive (2AgX → 2Ag + X₂ under UV); AgBr most widely used in photography (appropriate sensitivity + solubility balance). Trend in covalent character: F⁻ small/hard/non-polarisable → AgF ionic → soluble. I⁻ large/soft/polarisable → AgI largely covalent → very insoluble, strongly light-sensitive. [3]
Q8 [10 marks]
(a) Explain why fluorine is the most reactive element in the periodic table. Your answer should refer to: electronegativity, bond dissociation energy of F₂, electrode potential, and the anomalously high reactivity compared with the expected trend. [4]
(b) Describe the reactions of Cl₂ with (i) cold dilute NaOH (ii) hot concentrated NaOH (iii) water. For each, write the equation, identify the type of reaction, and state the oxidation states. [3]
(c) Explain the uses of fluorine in the chemical industry, giving three specific examples with chemical details. [3]
(a) Fluorine most reactive element: (1) Highest electronegativity (4.0) — attracts electrons from ANY other element → always acts as oxidising agent → OS always −1. (2) Very large positive electrode potential (E° = +2.87V) → F₂ is the strongest oxidising agent of all elements. (3) F₂ bond dissociation energy is anomalously LOW (158 kJ/mol) — F atoms are so small that lone pair–lone pair repulsion between the two F atoms weakens the F–F bond → F₂ breaks apart easily → reactive. (4) Products (HF, metal fluorides) have very strong bonds → reactions are highly exothermic → thermodynamically strongly favoured. Expected trend would suggest F₂ should have higher bond energy than Cl₂ (as it is smaller, closer bond) — but lone-pair repulsion in the very small molecule reverses this. [4] (b) (i) Cl₂ + 2NaOH(cold, dilute) → NaCl + NaOCl + H₂O. Type: disproportionation. Cl(0) → Cl(−1) in NaCl and Cl(+1) in NaOCl. Product NaOCl is bleach. [1] (ii) 3Cl₂ + 6NaOH(hot, conc.) → 5NaCl + NaClO₃ + 3H₂O. Type: disproportionation (more extensive). Cl(0) → Cl(−1) and Cl(+5). Product NaClO₃ = sodium chlorate(V). [1] (iii) Cl₂ + H₂O ⇌ H⁺ + Cl⁻ + HOCl. Type: disproportionation (equilibrium). Cl(0) → Cl(−1) in HCl and Cl(+1) in HOCl. [1] (c) Industrial uses of fluorine/HF (any 3): (1) Uranium enrichment: UF₆ (uranium hexafluoride) is used for gaseous diffusion or centrifuge enrichment of U-235. F₂ + U → UF₆ (gas). The only volatile uranium compound — used for isotope separation. [1] (2) PTFE (polytetrafluoroethylene/Teflon): monomer CF₂=CF₂ (tetrafluoroethylene) from CHClF₂ by pyrolysis. Polymerisation → PTFE. Non-stick, inert, high-temperature resistant. F₂ essential for making CF₂=CF₂. [1] (3) HF for etching glass/semiconductors: HF + SiO₂ → SiF₄ + 2H₂O. Used to etch silicon wafers in semiconductor manufacture (circuit patterns), frost glass, and in chemical analysis. [1] Other valid: refrigerants (HFCs — replacing CFCs; contain F but no Cl → no ozone depletion); fluoride in toothpaste and water (NaF, SnF₂ → F⁻ ion hardens tooth enamel: Ca₅(PO₄)₃OH + F⁻ → Ca₅(PO₄)₃F). [3 total]