Group 17 — The Halogens

— Chemistry · AS Level (7405) · Ruhab Iqbal, Crash Course —

★ Exam Tip: Sir's words: Group 17 is a very important chapter and the halide ions are its main topic. Expect: colours/states, displacement, silver-nitrate + ammonia test, halides + conc. H₂SO₄ (observations!), and chlorine disproportionation. Learn the observations word-for-word.

1. The Halogens — Introduction & Appearance

Group 17 = the halogens (called Group 7 ins). Sir: "funky name, my favourite group".
All exist as diatomic molecules: F₂, Cl₂, Br₂, I₂, At₂ — two atoms joined by a covalent bond.
Focus on Cl₂, Br₂, I₂ — then predict F₂ and At₂ from the trend.

Appearance at room temperature & pressure

Halogen	State	Colour	Marking note (sir)
Cl₂	gas	yellow-green	"pale green" ✓, "green" ✓ — but not just "yellow" ✗
Br₂	liquid	red-brown	"orange" also accepted, but red-brown is better
I₂	solid	black	(grey-black solid)

going down: colour gets darker, state gas → liquid → solid

Sir's reminder: these are simple molecules — melting/boiling breaks intermolecular forces only, NOT the Cl–Cl bond.

Trend in m.p. / b.p. (why?)

Down the group: colour gets darker and m.p./b.p. increase.

Reason: number of electrons ↑ ⇒ temporary dipole forces (instantaneous dipole – induced dipole) get stronger ⇒ more energy to melt/boil.

Why ONLY temporary dipoles? Homonuclear diatomic (same atom both sides) ⇒ no electronegativity difference ⇒ non-polar ⇒ no permanent dipoles; no H attached to N/O/F ⇒ no hydrogen bonding.

Exam answer pattern (m.p. trend): "Cl₂, Br₂, I₂ are simple molecular. Only temporary dipole forces. Number of electrons increases down the group ⇒ temporary dipole forces stronger ⇒ more energy needed ⇒ m.p./b.p. increase." Never say the Cl–Cl covalent bond breaks — only intermolecular forces do.

Predictions (uses the trend)

Fluorine (above Cl)lighter than yellow-green ⇒ yellow gas. Fewer electrons ⇒ weaker forces ⇒ still gas (no other state to go to).

Astatine (below I)black solid — can't get blacker than black (sir writes it as "black solid" since we can't tell I₂ from At₂). Solid: m.p. even higher.

2. Reactivity of the Halogens

Key idea: Halogens are non-metals — they react by gaining (accepting) electrons (or sharing them). So "reactivity" = how easily the atom pulls in an extra electron.

Reactivity of halogens DECREASES down the group: F₂ > Cl₂ > Br₂ > I₂ > At₂

Why? (learn these as a set)

Nuclear charge increases down the group.
Shielding also increases (more inner-shell electrons).
These two cancel each other out (net effect ≈ nothing) — so the deciding factor is:
Atomic radius increases (more shells) ⇒ nucleus is far from the valence shell ⇒ cannot easily attract a new electron in ⇒ less reactive.

small atom: nucleus grabs the incoming electron easily • big atom: nucleus too far away ⇒ less reactive

Trick: "reactivity of halogens ↓, reactivity of halides ↑" — opposite trends! (see §3)

Halogens = oxidising agents

Halogens accept electrons ⇒ they are reduced themselves ⇒ they oxidise other species ⇒ oxidising agents.

Sir's slip (corrected in class): halogens are OXIDISING agents — not reducing agents. Examiner may phrase it as "oxidising ability decreases down the group".

Example (Na + Cl₂):

2Na_(s) + Cl₂_(g) → 2NaCl_(s)

Oxidation numbers: Na 0 → +1 (oxidised) | Cl 0 → −1 (reduced).

Cl₂ was reduced ⇒ Cl₂ = oxidising agent. Na was oxidised. (Sir halved it: ½Cl₂ or 2Na + Cl₂ — both fine.)

Remember: element on its own (Na, Cl₂, I₂…) always has oxidation number 0. Rule of thumb: OIL RIG — Oxidation Is Loss (of e⁻), Reduction Is Gain.

3. Halide Ions & Their Reducing Ability

Halide = a halogen ion (X⁻): Cl⁻ chloride, Br⁻ bromide, I⁻ iodide. (Group 17 is special: we study the ions too — they are the main topic.)

A halide X⁻ has an extra (8th) electron that it gained from another species when it bonded. It can give this electron away to something that needs it ⇒ it is oxidised and reduces the other species ⇒ reducing agent.

Reducing ability of halides INCREASES down the group: F⁻ < Cl⁻ < Br⁻ < I⁻ < At⁻

Why? (same story, different view)

Nuclear charge ↑ and shielding ↑ — cancel out again.
Atomic radius ↑ (more shells) ⇒ nucleus has lost control of the outer electron.
So the extra electron is easy to lose and pass to another species ⇒ better reducing agent.

the extra electron (red ring): F⁻ nucleus grips it • I⁻ nucleus is far away ⇒ electron easily donated

Sir's summary: halogens ⇒ "how reactive / how good an OXIDISING agent" ↓ down the group. halides ⇒ "how good a REDUCING agent" (= "reactivity of halides") ↑ down the group. I⁻ = a very, very good reducing agent.

Proof that halides reduce

2HI + H₂SO₄ → SO₂ + I₂ + 2H₂O

H₂SO₄: H +1, O −2x + 2(+1) + 4(−2) = 0S = +6

SO₂: O −2 eachx + 2(−2) = 0S = +4

I: −1 → 0 = oxidised (oxidation number ↑; −1 is smaller than 0) | S: +6 → +4 = reduced. ⇒ I⁻ acted as a reducing agent. (Full study of this in §7.)

4. Displacement Reactions

RuleA more reactive halogen (higher up the group) displaces a less reactive halogen from its salt (kicks it out of solution).

Link to Group 2: same idea as displacing Mg from its salt with Na₂CO₃/Na₂SO₄. Here Cl₂ > Br₂ > I₂ in reactivity.

① Cl₂ + NaBr

Cl₂_(aq) + 2NaBr_(aq) → 2NaCl_(aq) + Br₂_(aq)

Observation: colourless solution turns brown (Br₂ formed). (Cl₂ in water is only faintly greenish.)

Cl₂ + 2Na⁺ + 2Br⁻ → 2Na⁺ + 2Cl⁻ + Br₂ Cl₂ & Br₂ do not dissociate; NaBr, NaCl do. Cancel spectator Na⁺. Cl₂ + 2Br⁻ → Br₂ + 2Cl⁻

Why (aq)? Cl₂ and Br₂ dissolve in water, so they are written aq not l/g. "Soluble liquids/gases become aqueous."

② Cl₂ + KI

Cl₂_(aq) + 2KI_(aq) → 2KCl_(aq) + I₂_(aq)

Observation: colourless solution turns brown (I₂ dissolved in water is brown). Ionic: Cl₂ + 2I⁻ → 2Cl⁻ + I₂ (K⁺ spectator).

Bubbling Cl₂ for several minutes: lots of I₂ made ⇒ brown darkens ⇒ once no more can dissolve, black solid I₂ forms (Paper-4-style Q, see §10).

③ Br₂ + KI

Br₂_(aq) + 2KI_(aq) → 2KBr_(aq) + I₂_(aq) ionic: Br₂ + 2I⁻ → 2Br⁻ + I₂

Observation: brown → brown ⇒ only "there might be a change in the shade of brown". Not a convincing observation!

④ Br₂ + KCl — no reaction

Br₂ is below Cl₂ ⇒ less reactive ⇒ cannot displace Cl⁻. Observation: "no reaction / no observable change" (or "brown solution remains brown").

Problem: both I₂(aq) and Br₂(aq) look brown — how to tell them apart? both look brown in water!

Species	Solid	Liquid	Vapour	In water	In organic solvent
Iodine I₂	black	—	purple	brown	purple
Bromine Br₂	—	red-brown	red-brown	brown	red-brown

I₂: purple layer on top ⇒ iodine present

Br₂: stays brown ⇒ bromine present, no I₂

Test (rote learning!): add a little organic solvent (e.g. cyclohexane). Purple layer ⇒ I₂. No purple layer ⇒ Br₂ (reaction did not happen / only Br₂). Learn the colour table — it is examined in MCQs too (see Q15 below).

Why is I₂ (e.g. "Pyodine") an antiseptic, not Cl₂ or Br₂? Iodine is fairly unreactive (reactivity ↓ down the group) — it won't attack skin cells or blood but still keeps germs out. Cl₂/Br₂/F₂ in blood ⇒ deadly; At is too hard to obtain.

5. Reactions with Hydrogen & Thermal Stability of HX

A) Halogens + H₂ — proof that reactivity decreases

Halogen + H₂	Conditions needed	What it shows
F₂	explosive under all conditions (even dark, cold)	conditions get harsher ⇒ reactivity ↓ down the group
Cl₂	burns in the dark; explosive in UV light
Br₂	200 °C + platinum catalyst
I₂	500–600 °C, and even then only reversibly

H₂ + I₂ ⇌ 2HI (reversible — 2HI breaks back into H₂ + I₂)

Sir: HF forms explosively (complete). With iodine the reaction hardly starts, and even the HI formed decomposes ⇒ "a very weak reaction".

B) Thermal stability of the hydrogen halides

Lab test: insert a hot glass rod into a gas jar of each HX and watch:

bond enthalpy (kJ mol⁻¹) ↓ ⇒ thermal stability ↓ ⇒ more decomposition into H₂ + X₂ on heating

HX	Bond enthalpy	Hot rod
H–F	562 kJ mol⁻¹	nothing happens (very stable)
H–Cl	431 kJ mol⁻¹	nothing happens
H–Br	366 kJ mol⁻¹	some brown vapour seen (Br₂)
H–I	299 kJ mol⁻¹	lots of purple vapour seen (I₂)

Down the group: bond enthalpy ↓ ⇒ thermal stability ↓

Careful: when a student answered "reactivity decreases" for this, sir said NO — the question asks why the bond weakens (bond enthalpy), not about reactivity. Sir's two reasons below (write both).

Why does the H–X bond get weaker? (2 reasons)

Size ↑ / electronegativity ↓: larger halogen atom ⇒ cannot pull the shared pair as strongly ⇒ bond becomes less polar. (H–F is very polar; I and H have almost the same electronegativity so H–I is ≈ non-polar.)
Orbital overlap becomes less efficient: the p orbital of I is in a big, far-out shell — large and diffuse — overlapping poorly with the tiny H 1s orbital. F 2p is similar in size to H 1s ⇒ efficient overlap ⇒ strong, stable bond.

H 1s with F 2p (similar size) vs H 1s with a huge I 5p

Also (sir): HI is so weak it decomposes spontaneously: 2HI ⇌ H₂ + I₂ — the same reversibility seen with I₂ + H₂ above. HI decomposition is why it is "not a strong compound".

6. Testing for Halide Ions (AgNO₃ then NH₃)

Procedure — do in ORDER

Dissolve the halide salt in water (must be a soluble salt; sir used tap water, so purity matters).
Add dilute HNO₃ — removes other ions (e.g. CO₃²⁻) that would give a confusing precipitate.
Add AgNO₃(aq) — a precipitate of the silver halide forms.
Add aqueous NH₃ (dilute first, then concentrated) to confirm.

Why HNO₃? (student question, answered)

e.g. carbonate impurity: MgCO₃ + 2HNO₃ → Mg(NO₃)₂ + CO₂ + H₂O — CO₂ escapes, and nitrate salts are always soluble, so they cannot form a false precipitate with Ag⁺. HNO₃ just displaces the unwanted anion (CO₂ leaves as gas) and leaves only a nitrate. Same reason AgNO₃ is used — silver nitrate is soluble.

Halide ion	+ AgNO₃(aq)	Precipitate	+ dilute NH₃	+ conc. NH₃
Cl⁻	white ppt	AgCl	dissolves (colourless soln)	(dissolves)
Br⁻	cream ppt	AgBr	✗ stays	dissolves (colourless soln)
I⁻	yellow ppt	AgI	✗ stays	✗ does NOT dissolve in any conc.

Sir's memory tip: just learn the ORDER — white, cream, yellow | dilute, concentrated, none. "No short-cut, students pick it up easily."

Problem: white / cream / yellow are almost the same shade — cream vs yellow especially. Not a reliable test on its own ⇒ add ammonia.

Equation (learn the pattern)NaCl + AgNO₃ → NaNO₃ + AgCl_(s)
Ag⁺ + Cl⁻ → AgCl_(s) (same for Br⁻ → AgBr, I⁻ → AgI)

Na⁺ + Cl⁻ + Ag⁺ + NO₃⁻ → Na⁺ + NO₃⁻ + AgCl_(s) AgCl is an insoluble salt — does NOT dissociate; Na⁺ and NO₃⁻ are spectators. Ag⁺_(aq) + Cl⁻_(aq) → AgCl_(s)

whole test on one page

Exam logic (sir): if the ppt dissolves in dilute NH₃ ⇒ chloride (and it'd dissolve in conc. too). Dilute fails, conc. works ⇒ bromide. Neither works ⇒ iodide. Use "dissolves to give a colourless solution".

Why do they dissolve? (for completeness — not in syllabus!)

NH₃ = ligand (donates its lone pair in a dative bond to a metal ion)

Two NH₃ molecules form dative bonds to Ag⁺ ⇒ soluble complex [Ag(NH₃)₂]⁺ (+ Cl⁻ / Br⁻ / I⁻). Whether the ppt dissolves depends on how insoluble the halide is. Ligand = species that uses its lone pair to form a dative bond to a metal atom/ion. You will meet this properly in A2 transition metals — don't memorise now.

7. Halide Ions + Concentrated H₂SO₄

★ Very, very important: these reactions are the proof that reducing ability increases down the group (Cl⁻ < Br⁻ < I⁻). Learn the equations and the observations.

Method: react the sodium halide (NaCl / NaBr / NaI) with concentrated H₂SO₄. Two kinds of reaction can happen:

Type	When?
① Acid–base	ALWAYS happens for all three halides (independent of reducing ability). Product is NaHSO₄ (not Na₂SO₄!) + HX gas.
② Redox	Depends on the halide's reducing ability: may not happen (Cl⁻), or happen a little (Br⁻), or a lot (I⁻). The HX gas from ① is what reduces the H₂SO₄.

① Acid–base (same pattern for all — just swap Cl → Br → I)NaX_(aq) + H₂SO₄_(aq) → NaHSO₄_(aq) + HX_(g)

Why "steamy fumes"? HX gas fumes in moist air — looks like white smoky/steamy fumes. Always quote it: "steamy fumes of HCl / HBr / HI". HX is a gas (mixture warms up so it escapes).

Not neutralisation reactions — sir: acid-base here, but the further steps are not neutralisation. No need to memorise balancing of these.

A) NaCl + conc. H₂SO₄

NaCl_(aq) + H₂SO₄_(aq) → NaHSO₄_(aq) + HCl_(g)

Observation: steamy fumes of HCl.

Redox: NONE — Cl⁻ is a weak reducing agent (top of the group), it cannot reduce H₂SO₄. Reaction stops at the acid–base step.

B) NaBr + conc. H₂SO₄

Acid–base: NaBr_(aq) + H₂SO₄_(aq) → NaHSO₄_(aq) + HBr_(g)

Obs (①): steamy fumes of HBr.

Redox: 2HBr_(g) + H₂SO₄_(aq) → SO₂_(g) + Br₂_(g) + 2H₂O_(l)

BromineHBr → Br₂−1 → 0oxidised

SulfurH₂SO₄ → SO₂+6 → +4reduced

Obs (redox): red-brown vapour (Br₂) + pungent smell of SO₂.

Crude way to remember: H₂SO₄ had 4 O on S; SO₂ has 2 ⇒ removal of 2 oxygens = reduction. HBr is strong enough to do exactly this. Must learn this equation.

C) NaI + conc. H₂SO₄ — the "ladder"

Acid–base: NaI_(aq) + H₂SO₄_(aq) → NaHSO₄_(aq) + HI_(g) Obs: few steamy fumes of HI

Only few steamy fumes seen: HI is used up in the redox steps, and the mixture gets contaminated by so many products (sir: "such a mess because of sulfur").

I⁻ is the best reducing agent, so HI reduces the H₂SO₄ three different ways:

	Equation (balanced)	I	S: change	Observation
① to SO₂	2HI + H₂SO₄ → SO₂ + I₂ + 2H₂O	−1→0	+6 → +4	purple vapour of I₂ + pungent smell of SO₂
② to S	6HI + H₂SO₄ → S + 3I₂ + 4H₂O	−1→0	+6 → 0	yellow solid seen (S_(s))
③ to H₂S	8HI + H₂SO₄ → H₂S + 4I₂ + 4H₂O	−1→0	+6 → −2	rotten-egg smell of H₂S

In every case the iodine (I⁻ → I₂) is oxidised and sulfur is reduced. S goes down by 2, 6, then 8 units ⇒ I⁻ gives away up to 8 electrons per S — huge reducing power.

Sir's "ladder of reduction": how far each halide takes H₂SO₄ down

Sir's memory trick — the ladder: Step 1: acid–base (all three). Step 2: take 2 O out ⇒ SO₂ (Br⁻, I⁻). Step 3: take 2 more O out ⇒ S (I⁻ only). Step 4: no O left, so add 2 H ⇒ H₂S (I⁻ only). Cl⁻ can't climb any step, Br⁻ climbs one, I⁻ climbs all. This is only a learning aid — the real equations are HI + H₂SO₄ each time (not SO₂ + HI).

Extension: astatide (below I) is an even stronger reducing agent ⇒ would go at least as far as H₂S (Q4 below).

Exam summary: Cl⁻ — steamy fumes only. Br⁻ — steamy fumes + red-brown Br₂ + pungent SO₂. I⁻ — few fumes + purple I₂ vapour + pungent SO₂ + yellow solid S + rotten-egg H₂S.

8. Disproportionation of Chlorine

DefinitionDisproportionation = simultaneous oxidation and reduction of the same element in a reaction.

(Not covered at — new at AS.) Know 3 reactions: Cl₂ + cold NaOH, Cl₂ + hot NaOH, Cl₂ + water.

oxidation number of Cl: from 0 to −1 (reduced) and to +1 or +5 (oxidised)

① Cl₂ + cold NaOH → bleach

Cl₂_(g) + 2NaOH_(aq) → NaCl_(aq) + NaClO_(aq) + H₂O_(l)

Bleach! NaCl + NaClO mixture = household bleach — removes stains/colour from clothes. NaClO = sodium chlorate(I).

Cl in Cl₂ = 0 → in NaCl = −1 ⇒ reduced Cl in Cl₂ = 0 → in NaClO: (+1) + x + (−2) = 0 ⇒ x = +1 ⇒ oxidised

Cl₂ + 2Na⁺ + 2OH⁻ → Na⁺ + Cl⁻ + Na⁺ + ClO⁻ + H₂O Cl₂ doesn't dissociate; NaOH, NaCl, NaClO do (ClO⁻ = hypochlorite ion). Cancel Na⁺. Cl₂_(g) + 2OH⁻_(aq) → Cl⁻_(aq) + ClO⁻_(aq) + H₂O_(l)

Balancing is easy — just put 2 in front of NaOH; sir: no need to memorise.

② Cl₂ + hot NaOH

3Cl₂_(g) + 6NaOH_(aq) → 5NaCl_(aq) + NaClO₃_(aq) + 3H₂O_(l)

Cl: 0 → −1 (reduced, in NaCl) and 0 → +5 (oxidised, in NaClO₃). NaClO₃ = sodium chlorate(V): (+1) + x + 3(−2) = 0 ⇒ x = +5.

Learn this ratio: 6 : 5 : 1 — 6 NaOH, 5 NaCl, 1 NaClO₃ (sodium ratio). This reaction "comes up a lot" — hard to balance on the spot, so memorise 6-5-1, then balance Cl (3Cl₂) and H/O (3H₂O).

3Cl₂ + 6Na⁺ + 6OH⁻ → 5Na⁺ + 5Cl⁻ + Na⁺ + ClO₃⁻ + 3H₂O cancel 6Na⁺ on both sides 3Cl₂_(g) + 6OH⁻_(aq) → 5Cl⁻_(aq) + ClO₃⁻_(aq) + 3H₂O_(l)

Balancing ② using oxidation numbers (if you forget 6-5-1)

Half-eq 1 (0 → −1)Cl₂ + 2e⁻ → 2NaClgains e⁻; balance Cl only

Half-eq 2 (0 → +5)Cl₂ → 2NaClO₃ + 10e⁻2 × (+5) = 10 e⁻ lost

equalise e⁻ (LCM of 2 and 10 = 10): half-eq 1 × 5 5Cl₂ + 10e⁻ → 10NaCl add half-eq 2 & cancel 10e⁻: 6Cl₂ → 10NaCl + 2NaClO₃ ÷ 2 : 3Cl₂ → 5NaCl + NaClO₃ Na: 5 + 1 = 6 ⇒ 6NaOH ; H: 6 ⇒ 3H₂O

③ Cl₂ + water — water purification

Cl₂_(g) + H₂O_(l) → HCl_(aq) + HClO_(aq)

Cl: 0 → −1 (HCl, reduced) and 0 → +1 (HClO: (+1)+x+(−2)=0, oxidised). HClO = chloric(I) acid (hypochlorous acid) — a weak acid.

HCllowers pH ⇒ kills some bacteria, but is a strong acid — can't leave it in drinking water (harmful!). So use very little Cl₂ and neutralise/remove the HCl.

HClO ← the real disinfectantstays behind; does not lower pH much; enters bacterial cells and attacks them ⇒ purifies water (tanks, swimming pools).

Note: "HCl reduces pH of water and kills bacteria" is what the board says — sir adds that HClO is the main bacteria-killer.

9. Past-Paper Questions — MCQs

★ Technique (sir): can't spot the right answer? Cancel the options you KNOW are wrong. One cut ⇒ 25% → 33% chance; two cuts ⇒ 50%. Always do this if it turns into a guessing game.

MCQ Q1MCQ · Cl vs I properties

Property Q is smaller for chlorine than for iodine. Which is Q?

A ✗ oxidising ability of the element — Cl₂ is the better oxidising agent (reactivity ↓ down) ⇒ Cl's value is larger B ✗ solubility of silver halide in NH₃(aq) — AgCl dissolves in dilute NH₃, AgI in none ⇒ Cl's is larger C ✓ strength of van der Waals forces between molecules — I₂ has far more electrons ⇒ stronger temporary dipoles; Cl₂ weaker D ✗ thermal stability of the hydrogen halide — HCl is very stable, HI decomposes ⇒ Cl's is larger

Answer: C

MCQ Q2MCQ · Cl₂ bubbled through KI(aq)

Cl₂ + 2KI → 2KCl + I₂Cl: 0 → −1 (reduced) ; I: −1 → 0 (oxidised)

A ✗ "chlorine is oxidised to chloride ions" — no! 0 → −1 is reduction B ✗ "hydrochloric acid is formed" — HCl forms nowhere C ✓ iodide ions are oxidised to iodine (−1 → 0) D ✗ "potassium iodide is reduced to iodine" — I⁻ is oxidised; never say the whole compound KI is reduced

Answer: C (displacement — Cl₂ more reactive kicks out I₂)

MCQ Q4MCQ · Astatine (below iodine) — educated guess

Use trends: At is less reactive halogen but astatide is an even stronger reducing agent than iodide.

A ✗ At₂ + KCl → KAt + Cl₂: At₂ is below Cl₂, can't displace it (it's the reverse: Cl₂ displaces At⁻) B ✗ KAt + hot dilute H₂SO₄ → only white fumes of HAt: that's what Cl⁻ does. At⁻ is more reducing than I⁻ ⇒ goes to at least H₂S C ✗ AgAt dissolves in excess dilute NH₃: AgI doesn't dissolve at all, so AgAt won't D ✓ NaAt + hot conc. H₂SO₄ → At₂ (like I₂ formed at every step)

Answer: D

MCQ Q12MCQ · Chile saltpetre (NaNO₃ with NaI impurity)

AgNO₃(aq) added, then concentrated NH₃(aq).

NaI + AgNO₃ ⇒ yellow ppt of AgI (so a ppt must form: A and B wrong). AgI does not dissolve in any concentration of NH₃ ⇒ precipitate remains: C (dissolves) wrong.

	with acidified AgNO₃	with conc. NH₃
A	no ppt ✗	no further reaction
B	no ppt ✗	ppt forms
C	ppt forms	ppt dissolves ✗
D	ppt forms	ppt remains ✓

Answer: D

MCQ Q7MCQ · which graph shows a halogen trend? (X = halogen atom)

Eliminate: C ✗, then A ✗, then D ✗ ⇒ B ✓.

A ✗ bond length in X₂ must increase (bigger atoms: Cl smallest radius → I biggest) — graph falls B ✓ strength of van der Waals forces increases (more electrons) C ✗ boiling point of X₂ must increase: Cl₂ gas, Br₂ liquid, I₂ solid — graph shows it falling (upside down) D ✗ bond energy of HX must decrease: HCl strong (efficient overlap, high electronegativity), HI weak (decomposes) — graph rises

Answer: B

MCQ Q15MCQ · newspaper report on bromine

"Drums of bromine broke open… purple gaseous bromine drifted… denser than air… firemen sprayed water, dissolving the bromine." What's wrong?

Bromine is red-brown, not purple. Purple vapour is iodine (reporter didn't study AS chemistry!). Bromine does dissolve in water, does vapourise readily (volatile liquid) and is denser than air, so A, B, C are wrong statements about the report.

Answer: D — bromine is not purple.

MCQ Q19MCQ · photographic film & sodium thiosulphate

AgCl + 2Na₂S₂O₃ → 4Na⁺ + Cl⁻ + [Ag(S₂O₃)₂]³⁻. Function of thiosulphate?

Check oxidation numbers on both sides — Ag (+1), Cl (−1), Na (+1), O (−2), S = ?

Na₂S₂O₃ (neutral): 2(+1) + 2x + 3(−2) = 0 2x = 4 ⇒ x = +2 [Ag(S₂O₃)₂]³⁻: (+1) + 4x + 6(−2) = −3 4x = 8 ⇒ x = +2 S = +2 both sides, Ag = +1 both sides ⇒ nothing oxidised or reduced

A ✓ to make the silver ions soluble — forms a soluble complex ion (AgCl is insoluble, so the unexposed AgCl washes out) B ✗ oxidise the silver ions — no change in Ag oxidation number C ✗ oxidise silver metal — no D ✗ reduce silver ions — no change

Answer: A

MCQ Q20MCQ · X, Y, Z halogens — nine experiments

	X⁻	Y⁻	Z⁻
X₂	no reaction	no reaction	no reaction
Y₂	X₂ formed	no reaction	Z₂ formed
Z₂	X₂ formed	no reaction	no reaction

X₂ displaces nothing ⇒ least reactive ⇒ X = I Y₂ displaces both X⁻ and Z⁻ ⇒ most reactive ⇒ Y = Cl Z₂ displaces X⁻ but not Y⁻ ⇒ in between ⇒ Z = Br reducing strength: I⁻ > Br⁻ > Cl⁻ ⇒ X⁻ > Z⁻ > Y⁻ ⇒ B

Careful: the question asks for reducing strength of the ions — that runs I⁻ > Br⁻ > Cl⁻, the opposite of reactivity of the halogens.

Answer: B (X⁻, Z⁻, Y⁻)

MCQ Q24MCQ · NaCl / NaI with conc. H₂SO₄ vs conc. H₃PO₄

	NaCl	NaI
conc. H₂SO₄	colourless acidic gas	purple vapour formed
conc. H₃PO₄	colourless acidic gas	colourless acidic gas

H₂SO₄ + I⁻ ⇒ purple I₂ ⇒ I⁻ was oxidised ⇒ H₂SO₄ acted as an oxidising agent. H₃PO₄ only gives HX (acid–base) — it could not oxidise I⁻ ⇒ weak oxidising agent. H₂SO₄ can't oxidise Cl⁻ either.

A ✗ H₃PO₄ stronger than H₂SO₄ — would have oxidised I⁻ too; it didn't B ✗ H₃PO₄ stronger oxidising agent than iodine — I⁻ wasn't oxidised, so no C ✗ H₂SO₄ stronger than chlorine — Cl⁻ wasn't oxidised, so no D ✓ H₂SO₄ is a stronger oxidising agent than iodine (it oxidised I⁻ to I₂)

Answer: D

MCQ Q49MCQ · students P and Q draw bar charts

P: solubility of AgX in NH₃(aq) — bars drawn Cl < Br < I. Q: ΔH° for H₂(g) + X₂(g) → 2HX(g) — Cl most negative, I smallest.

P: solubility of AgX in NH₃

Q: ΔH° of H₂ + X₂ → 2HX

P ✗ AgCl most soluble (dilute NH₃), AgBr next, AgI least ⇒ bars should fall. Q ✓ HCl has the strongest bond (highest bond energy) ⇒ forming it releases most energy (bond formation is exothermic): highly exo; HBr mid; HI mild. Q is correct.

Answer: C (Q only)

10. Past-Paper Questions — Structured

Paper 2 · Q2HX made from NaX + anhydrous H₃PO₄

NaX + H₃PO₄ → NaH₂PO₄ + HX. A hot glass rod placed in the HX sample immediately gave a red/orange colour.

(a) identity of NaX [1]: NaBr — hot rod decomposes HBr ⇒ red-brown/orange Br₂ vapour (HCl: nothing; HI would give purple).

(b) gas other than HX if conc. H₂SO₄ used [1]: SO₂ — (Br₂ is also correct; both are formed: 2HBr + H₂SO₄ → SO₂ + Br₂ + 2H₂O).

H₃PO₄ is a weaker oxidising agent than conc. H₂SO₄ ⇒ only the acid–base reaction, producing pure HX. H₂SO₄ gives redox reactions with HX (SO₂, Br₂, I₂, S, H₂S…) which contaminate the mixture and use up HX — board: "conc. H₂SO₄ is a stronger oxidising agent than conc. H₃PO₄ and so produces many more products using up HX".

Paper 2 · Q6(a)K, L, M = halogen atom / halide ion

(i) Conc. H₂SO₄ added to solid NaK: white fumes that turn damp blue litmus red; no other colour change [3]

K = Cl (only Cl⁻ stops at acid–base; Br⁻ gives Br₂/SO₂, I⁻ gives I₂/SO₂/S/H₂S).

Equation: NaCl_(aq) + H₂SO₄_(aq) → NaHSO₄_(aq) + HCl_(g) (NaHSO₄, not Na₂SO₄).

Explanation: the fumes are HCl gas (steamy fumes); it dissolves in the damp litmus paper's water and acts as an acid ⇒ blue → red (shows an acidic gas).

(ii) AgNO₃ added to aqueous NaL: precipitate remains after conc. NH₃ [3]

L = I (only AgI never dissolves). Colour of ppt = yellow (AgI — not the colour of iodine!). Equation: NaI_(aq) + AgNO₃_(aq) → AgI_(s) + NaNO₃_(aq) (already balanced).

(iii) M₂ is a liquid at r.t.; b.p. higher than Cl₂, lower than I₂ [2]

M = Br. Br₂ has more electrons than Cl₂ but fewer than I₂ ⇒ its temporary dipole forces are stronger than Cl₂'s but weaker than I₂'s (so more energy than Cl₂ but less than I₂ to overcome).

Sir: b.p./m.p. of a simple molecule ⇒ talk about intermolecular forces (temporary dipoles), not "dative" or any bonds. Dative bonding is a chemical bond, not an intermolecular force!

Paper 2 · Q5(a)–(b)Group 7 variation in properties

(a)(i) colours at r.t.: chlorine yellow-green (m.p. −101 °C) · bromine reddish-brown (−7 °C) · iodine black (114 °C).

(ii) m.p. increases Cl → I [4]: "Down the group the number of electrons increases, hence the strength of the temporary dipole forces gets stronger ⇒ more energy required to melt."

(b)(i) complete the electronic configurations:

Cl: 1s² 2s² 2p⁶ 3s² 3p⁵ Br: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵

d-block fills one shell "behind" (3d before 4p, after 4s). The only anomalous ones to memorise at AS are Cr and Cu.

Paper 2 · Q5(b)(ii)dot-and-cross diagram of BrCl

One single covalent bond (each atom needs 1 e⁻ for a full shell). Show outer electrons only, all 8 around each atom.

Sir's rules: no charges (±) on atoms in a covalent bond (δ+/δ− only if asked). Draw the bigger halogen bigger (Br larger than Cl) — you may lose marks otherwise, even if examiners usually don't deduct.

Paper 2 · Q5(c)BrCl — interhalogen compound

Interhalogen = two different halogens covalently bonded (e.g. BrCl, ICl, IF₅). Behave like the halogens — no need to study any in detail.

(i) physical state: gas. Electrons: Cl = 17, Br = 35 ⇒ BrCl = 52; Cl₂ = 34, Br₂ = 70.

Explanation: number of electrons in BrCl is less than in Br₂ ⇒ weaker temporary dipole forces ⇒ lower b.p. ⇒ still a gas (liquid only from ≈ Br₂'s electron count).

(ii) colour: orange — yellow-green + red-brown blend ⇒ lighter shade of brown ≈ orange.

Paper 2 · Q5(d)Cl₂ and BrCl with aqueous KI

(i) Cl₂ bubbled through KI(aq) for several minutes: initially → brown solution formed (I₂ dissolved; no purple vapour); after several minutes → black solid forms (so much I₂ made that it no longer dissolves and comes out of solution; the brown darkens first).

(ii) Cl₂_(g) + 2KI_(aq) → 2KCl_(aq) + I₂_(aq)board line left it unbalanced as Cl₂ + KI → KCl + I₂ — needs 2KI and 2KCl

(iii) BrCl + 2KI → KCl + KBr + I₂both Cl and Br displace I; iodine forced out by both halogens

(iv) how do Cl₂ and BrCl behave? oxidising agents — they oxidise I⁻ (−1) to I₂ (0).

Paper 2 · Q8(a)successive ionisation energies F → Br

every element: IE rises with each electron; at every step F > Cl > Br

(i) Why first IE decreases down the group [3]:

Nuclear charge ↑ and shielding ↑ (more inner-shell electrons) — they cancel each other out.
Atomic radius ↑ (more shells).
So the outer electron is less attracted by the nucleus ⇒ easier to lose ⇒ less energy needed.

(ii) Why successive IE of F increase [2]:

As electrons are removed, protons outnumber electrons (protons per electron ↑) ⇒ remaining electrons are attracted more strongly (a positive ion holds electrons harder).
Repulsion between the remaining electrons decreases ⇒ they move closer ⇒ atomic radius decreases ⇒ even harder to remove the next one.

Paper 2 · Q8(b)b.p. trend in Group 7 [4]

Board: "Already done" (covered in §1). Sir's mark scheme for 4 marks:

State the trend: b.p. increases down the group.
Evidence: Cl₂ gas, Br₂ liquid, I₂ solid (at r.t.) — "1 mark".
They are simple molecular; only temporary dipole forces between molecules.
Number of electrons ↑ down the group ⇒ temporary dipole forces stronger ⇒ more energy needed ⇒ higher b.p.

"Don't make it short — the more you write (correctly), the better." Temporary dipole = at any random moment electrons may sit unevenly, creating a dipole → attraction.

Paper 2 · Q8(c)(i)interhalogen molecular formula (M_r = 222; F 42.8 %, I 57.2 %)

Method 1 — empirical formula moles F = 42.8 ÷ 19 = 2.2526 moles I = 57.2 ÷ 126.9 = 0.45075 ÷ smallest ⇒ F : I = 5 : 1 empirical formula = IF₅ (M = 126.9 + 5×19 ≈ 221.9) n = 222 ÷ 221.9 ≈ 1 ⇒ IF₅

Method 2 — Rumesa's shortcut mass of F = 42.8 % × 222 = 95.0 number of F = 95.0 ÷ 19 = 5 mass of I = 57.2 % × 222 = 127.0 number of I = 127.0 ÷ 126.9 ≈ 1 molecular formula = IF₅

Molecular formula = empirical formula × n; n = M_r ÷ empirical-formula mass (n isn't a mole count!). Don't treat % as moles: assume 100 g ⇒ % becomes grams.

Paper 2 · Q8(c)(ii)dot-and-cross ICl — polar?

Polar — Cl is much more electronegative than I ⇒ shared pair pulled towards Cl ⇒ δ⁻ on Cl, δ⁺ on I; the molecule is asymmetric (two different atoms).

Oxidation numbers ≠ full transfer: "I +1, Cl −1" only means Cl controls the shared pair a bit more — no electron has moved fully. Covalent compounds can still be assigned oxidation numbers.

Paper 2 · Q8(d)reaction scheme of chlorine and compounds

W = NaClO · X = NaClO₃ · Y = HCl · Z = AgCl

(i) formulae [4]: W = NaClO (cold NaOH) | X = NaClO₃ (hot NaOH) | Y = HCl (H₂ + Cl₂ in UV; observation: steamy fumes of HCl) | Z = AgCl (white ppt).

Both NaOH reactions = disproportionation (NOT acid–base): NaCl (−1) + NaClO (+1) or NaClO₃ (+5). Confirm Z = AgCl: dissolves in dilute NH₃ (AgBr: only conc.; AgI: never).

(ii) Cl₂ + hot NaOH [2]3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O6 : 5 : 1 — the 5 goes with NaCl and 1 with NaClO₃ (the answer line on the board looks as if these are swapped; this is the correct balanced equation)

(iii) oxidation numbers of Cl [2]: start 0 (Cl₂); end +5 (in NaClO₃) and −1 (in NaCl).

(iv) ionic equation Y + AgNO₃ [1]HCl + AgNO₃ → AgCl + HNO₃H⁺ + Cl⁻ + Ag⁺ + NO₃⁻ → AgCl_(s) + H⁺ + NO₃⁻Ag⁺_(aq) + Cl⁻_(aq) → AgCl_(s)HCl and AgNO₃ dissociate; AgCl doesn't; HNO₃ dissociates back; H⁺ and NO₃⁻ cancel

11. Quick Recap — Group 17 Cheat-Sheet

✓ Trends down Group 17 (Cl → Br → I)

Colour darker: yellow-green gas → red-brown liquid → black solid (F₂ yellow gas, At₂ black solid)
m.p./b.p. ↑: more electrons ⇒ stronger temporary dipole forces (only these — non-polar, no H-bond)
Halogen reactivity / oxidising ability ↓: nuclear charge & shielding cancel; atomic radius ↑ ⇒ harder to attract a new electron
Halide reducing ability ↑: same reasoning — nucleus loses control of the extra electron, so it's easy to give away
Thermal stability of HX ↓: bond enthalpy 562 (HF) > 431 (HCl) > 366 (HBr) > 299 kJ mol⁻¹ (HI) — size ↑, less polar, poorer orbital overlap

✓ Reactions & observations to know

Displacement: Cl₂ + 2Br⁻ → Br₂ + 2Cl⁻ (colourless → brown) | Cl₂ + 2I⁻ → I₂ + 2Cl⁻ (→ brown) | Br₂ + 2I⁻ → I₂ + 2Br⁻ (shade of brown) | Br₂ + Cl⁻ → no reaction
Organic solvent test: I₂ ⇒ purple layer; Br₂ stays red-brown. I₂: black solid / purple vapour & organic solvent / brown in water. Br₂: red-brown everywhere.
AgNO₃ test (after dil. HNO₃): white AgCl · cream AgBr · yellow AgI. NH₃: AgCl dissolves in dilute; AgBr only in conc.; AgI never.
NaX + conc. H₂SO₄ always ⇒ NaHSO₄ + HX (steamy fumes). Cl⁻: stops. Br⁻: 2HBr + H₂SO₄ → SO₂ + Br₂ + 2H₂O (red-brown vapour, pungent SO₂). I⁻: → SO₂ (+ purple I₂), → S (yellow solid), → H₂S (rotten eggs).
Ladder for H₂SO₄: acid–base → −2 O → −2 O → +2 H. S: +6 → +4 → 0 → −2.
NaX + H₃PO₄ ⇒ only acid–base ⇒ clean HX (H₃PO₄ = weak oxidising agent).
H₂: F₂ explodes (dark) · Cl₂ UV · Br₂ 200 °C + Pt · I₂ 500–600 °C reversible.

✓ Chlorine disproportionation (0 → −1 and up)

cold NaOH: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (bleach) ; ionic Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
hot NaOH: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (learn 6-5-1) ; ionic 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O
water: Cl₂ + H₂O → HCl + HClO (HClO = the disinfectant; weak acid; pools/water treatment)
Oxidation numbers of Cl: 0 → −1 (reduced), 0 → +1 or +5 (oxidised)

Exam habits (sir): 1) Learn every observation with the correct colour word — "steamy fumes", "pungent smell", "purple vapour". 2) Melting/boiling of simple molecules ⇒ intermolecular forces + electron count. 3) Ionic equations: only dissociate soluble salts; cancel spectators; keep (s), (l), (g), (aq). 4) Draw dot-and-cross with proper sizes. 5) Attempt every MCQ — eliminate wrong options first.

Board note: the "Group 17 crash course" board and this transcript cover the AS syllabus only; ligand/complex-ion ideas ([Ag(NH₃)₂]⁺) are just "for completeness" — A2 will revisit them.