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State Board · Class 12 · Chemistry

Lesson 7: Elements of Groups 16, 17 and 18

7 concepts · 56 practice questions with answers · Free to read

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Occurrence, electronic configuration and atomic properties

From daily life

Salt (Cl), balloon helium (He) and the oxygen we breathe all belong to these groups.

Simple definition

Group 16: ns²np⁴; group 17: ns²np⁵; group 18: ns²np⁶ (He 1s²).

ConfigurationGroups 16, 17, 18: configuration and electronegativityGroupOuter config.Electronegativity16 (O, S, Se, Te, Po)ns² np⁴O 3.5 · S 2.517 (F, Cl, Br, I, At)ns² np⁵F 4.0 · Cl 3.0 · Br 2.8 · I 2.518 (He, Ne, Ar, Kr, Xe, Rn)ns² np⁶ (He 1s²)no tendencyGroup 16: chalcogens (ore formers); Group 17: halogens (salt formers)Group 17 needs 1 e⁻ ⇒ most negative e⁻ gain enthalpy (Cl highest)Group 18: closed shell ⇒ inert (noble gases)Down the group: size ↑, IE ↓, metallic character ↑

In the figure : Configuration and electronegativity.

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Occurrence, electronic configuration and atomic properties

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Rule 1: Group 16 (chalcogens) — O, S, Se, Te, Po need 2 e⁻; O is 21 % of air; S occurs near volcanoes.

  • Easy : O: [He]2s²2p⁴.
  • Medium : S: [Ne]3s²3p⁴.
  • Exam-type : Po is radioactive.

Rule 2: Group 17 (halogens) — F, Cl, Br, I, At need 1 e⁻; Cl, not F, has the most negative electron gain enthalpy (F is too small).

  • Easy : Cl: [Ne]3s²3p⁵.
  • Medium : F is most electronegative, 4.0.
  • Exam-type : Halides occur in sea water and minerals.

Rule 3: Group 18 (noble gases) — He, Ne, Ar, Kr, Xe, Rn: closed shells, monatomic gases, highest IE; Ar is about 0.93 % of air.

  • Easy : He: 1s².
  • Medium : Ne: [He]2s²2p⁶.
  • Exam-type : Ar: [Ne]3s²3p⁶.

Reasoning: why Cl gains an electron more exothermically than F

  • Step 1: F is small (2p), already crowded with electrons.
  • Step 2: The incoming electron feels strong repulsion.
  • Step 3: Cl is larger (3p) ⇒ less repulsion ⇒ more negative.

CET level (MCQ)

  • CET 1) Halogen outer configuration? 1) ns²np⁵ 2) ns²np⁴ 3) ns²np⁶ 4) ns² — Answer: 1).
  • CET 2) Most negative electron gain enthalpy? 1) Cl 2) F 3) Br 4) I — Answer: 1).

Real-life use: Salt, air, balloons, lamps.

Remember

16: ns²np⁴; 17: ns²np⁵; 18: ns²np⁶; Cl beats F in electron gain enthalpy.

Common mistake

Assuming F has the most negative electron gain enthalpy.

CET trap

Small F ⇒ repulsion; Cl is highest.

Shortcut

Memory clue: "16 → p⁴, 17 → p⁵, 18 → p⁶ (full)".

Spot the difference

GroupOuter config.
16ns²np⁴
17ns²np⁵
18ns²np⁶

Practice questions

Easy
  1. 1Name group 16 elements.
  2. 2Halogen configuration.
  3. 3Why noble gases inert?
Medium
  1. 4Configuration of Cl.
  2. 5Most electronegative element.
  3. 6Cl vs F electron gain.
Challenging
  1. 7Ar % in air.
  2. 8Configuration of He.
Show answers (try them yourself first)
  1. 1O, S, Se, Te, Po.
  2. 2ns²np⁵.
  3. 3Closed shell.
  4. 4[Ne]3s²3p⁵.
  5. 5F.
  6. 6Cl is higher.
  7. 7≈ 0.93 %.
  8. 81s².

Physical properties and anomalous behaviour of oxygen and fluorine

From daily life

Water boils at 100 °C but H₂S at about −60 °C — hydrogen bonding at work.

Simple definition

The first element differs: small size, high electronegativity, no d-orbitals ⇒ O covalency ≤ 2, F covalency 1.

AnomalousX–X bond enthalpy (kJ/mol): F₂ is weak!159F₂243Cl₂193Br₂151I₂Small F: strong lone-pair repulsion ⇒ weak F–F ⇒ F₂ very reactiveO and F: no d-orbitals ⇒ O covalency ≤ 2, F covalency 1H₂O boils high (373 K) due to H-bonding; H₂S only 213 KO forms pπ–pπ double bonds (O₂); S prefers rings (S₈)

In the figure : X–X bond enthalpy.

Rule 1: Physical properties — Down the group size, density, m.p./b.p. increase; halogens: F₂ pale yellow, Cl₂ greenish, Br₂ brown liquid, I₂ violet solid.

  • Easy : I₂ sublimes.
  • Medium : Br₂ is the only liquid non-metal.
  • Exam-type : O₂ is a colourless gas.

Rule 2: X–X bond enthalpy — Cl₂ 243 > Br₂ 193 > F₂ 159 > I₂ 151 kJ/mol; F₂ is weak due to lone-pair repulsion.

  • Easy : F₂ is very reactive.
  • Medium : Order: Cl > Br > F > I.
  • Exam-type : I₂ is weakest.

Rule 3: Anomalies — O: H-bonding (H₂O), pπ–pπ double bonds (O₂), no d-orbitals; F: only –1, H-bonding in HF, a weak acid.

  • Easy : HF is weak, HI is strong.
  • Medium : SF₆ exists, OF₆ does not.
  • Exam-type : S₈ rings, O₂ double bond.

Reasoning: why F₂ is so reactive

  • Step 1: The F–F bond is weak (159 kJ/mol).
  • Step 2: F forms very strong bonds with others (e.g. H–F 565).
  • Step 3: Little energy to break, much released on forming ⇒ easy reaction.

CET level (MCQ)

  • CET 1) Highest X–X bond enthalpy? 1) Cl₂ 2) F₂ 3) Br₂ 4) I₂ — Answer: 1).
  • CET 2) Liquid non-metal? 1) Br₂ 2) I₂ 3) Cl₂ 4) F₂ — Answer: 1).

Real-life use: Properties of water, dyes, disinfectants.

Remember

Cl₂ > Br₂ > F₂ > I₂ (bond); O, F lack d-orbitals; HF is weak.

Common mistake

Thinking F₂ has the highest bond enthalpy.

CET trap

F₂ is weak, Cl₂ highest.

Shortcut

Memory clue: "chlorine's bond is strongest; fluorine's is a dud".

Spot the difference

MoleculekJ/mol
F₂159
Cl₂243
Br₂193
I₂151

Practice questions

Easy
  1. 1Colours of halogens.
  2. 2Bond enthalpy order.
  3. 3Why is F₂ reactive?
Medium
  1. 4Covalency of O.
  2. 5Why is HF weak?
  3. 6Why is H₂O's b.p. high?
Challenging
  1. 7Only liquid non-metal.
  2. 8A property of I₂.
Show answers (try them yourself first)
  1. 1F₂ yellow, Cl₂ green, Br₂ brown, I₂ violet.
  2. 2Cl > Br > F > I.
  3. 3Lone-pair repulsion.
  4. 4≤ 2.
  5. 5H-bonding.
  6. 6H-bonding.
  7. 7Br₂.
  8. 8Sublimation.

Oxidation states and chemical reactivity

From daily life

Iron rusts because oxygen oxidises; bleaching powder removes colour because chlorine oxidises.

Simple definition

Group 17: –1, +1, +3, +5, +7; group 16: –2, +2, +4, +6; oxidising power F₂ > Cl₂ > Br₂ > I₂.

OxidationOxidation states and reactivityGroupStatesNote16–2, +2, +4, +6O: –2 (+2 with F)17–1, +1, +3, +5, +7F: only –1E° (X₂/X⁻): F₂ 2.87 V · Cl₂ 1.36 · Br₂ 1.08 · I₂ 0.54Oxidising power: F₂ > Cl₂ > Br₂ > I₂ (higher E° ⇒ stronger)Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ (upper halogen displaces lower)Acidity: HF < HCl < HBr < HI; H₂O < H₂S < H₂Se < H₂TeReason: the H–X bond weakens as X grows

In the figure : States and E°.

Rule 1: Oxidation states — Halogens show odd states (pairs unpair); F: only –1; O: usually –2 (+2 in OF₂).

  • Easy : Cl: –1 to +7.
  • Medium : S: H₂S ⇒ –2, SO₂ ⇒ +4, SO₃ ⇒ +6.
  • Exam-type : In HClO₄, Cl = +7.

Rule 2: Oxidising power — E°: F₂ 2.87, Cl₂ 1.36, Br₂ 1.08, I₂ 0.54 V; an upper halogen displaces a lower one.

  • Easy : Cl₂ + 2KI → 2KCl + I₂.
  • Medium : Br₂ + 2KCl → no reaction.
  • Exam-type : F₂ even oxidises water.

Rule 3: Acidity of hydrides — Weaker H–X (bigger X) ⇒ more acidic: HF < HCl < HBr < HI; H₂O < H₂S < H₂Se < H₂Te; thermal stability is opposite.

  • Easy : HI is the strongest acid.
  • Medium : H₂O is the most stable.
  • Exam-type : Reducing power is highest for H₂Te.

Calculation: oxidation state of Cl in HClO₄

  • Step 1: +1 (H) + x + 4(–2) = 0.
  • Step 2: x – 7 = 0.
  • Step 3: x = +7.

CET level (MCQ)

  • CET 1) Strongest oxidising agent? 1) F₂ 2) Cl₂ 3) Br₂ 4) I₂ — Answer: 1).
  • CET 2) Most acidic? 1) HI 2) HF 3) HCl 4) HBr — Answer: 1).

Real-life use: Bleaching, disinfection, iodine tincture.

Remember

F₂ strongest; displacement; HI most acidic; Cl = +7 in HClO₄.

Common mistake

Treating HF as the strongest acid.

CET trap

HF is weak (H-bonding, strong H–F).

Shortcut

Memory clue: "F is top oxidant; I is bottom, most acidic".

Spot the difference

ElementE° V
F₂2.87
Cl₂1.36
Br₂1.08
I₂0.54

Practice questions

Easy
  1. 1Group 17 states.
  2. 2Why only –1 for F?
  3. 3E° order.
Medium
  1. 4Cl₂ + KBr?
  2. 5Acidity order.
  3. 6Cl in HClO₄.
Challenging
  1. 7S in H₂S, SO₂, SO₃.
  2. 8Thermal stability order.
Show answers (try them yourself first)
  1. 1–1, +1, +3, +5, +7.
  2. 2Most electronegative, no d.
  3. 3F > Cl > Br > I.
  4. 4Br₂ is formed.
  5. 5HI > HBr > HCl > HF.
  6. 6+7.
  7. 7–2, +4, +6.
  8. 8HF > HCl > HBr > HI.

Allotropy of sulfur and oxoacids

From daily life

Match-head sulfur and acids around us — sulfur and chlorine form series of oxoacids.

Simple definition

Allotropy: different physical forms of one element; sulfur: rhombic (α) and monoclinic (β), both S₈.

SulfurSulfur allotropes and oxoacidsS₈crown ringRhombic (α) ⇄ Monoclinic (β)transition temperature 369 K (95.6 °C)α stable below 369 K, β aboveboth contain S₈; m.p. 385 K / 393 KOxoacidsH₂SO₃sulfurousH₂SO₄sulfuricH₂S₂O₇oleumH₂S₂O₈peroxodisulfuricChlorine oxoacids: HOCl < HClO₂ < HClO₃ < HClO₄ (acidity ↑)Oxidation states +1, +3, +5, +7; more O ⇒ more resonance

In the figure : S₈ and oxoacids.

Rule 1: Sulfur allotropes — α (rhombic) is stable below 369 K, β above; both are crown S₈ rings; plastic sulfur has long chains.

  • Easy : Transition 369 K = 95.6 °C.
  • Medium : S dissolves in CS₂.
  • Exam-type : S is insoluble in water.

Rule 2: Sulfur oxoacids — H₂SO₃ (+4), H₂SO₄ (+6), H₂S₂O₇ oleum (+6), H₂S₂O₈ (peroxo, +6).

  • Easy : H₂SO₄ is dibasic.
  • Medium : Oleum = SO₃ in H₂SO₄.
  • Exam-type : H₂S₂O₈ has an O–O bond.

Rule 3: Chlorine oxoacids — HOCl +1, HClO₂ +3, HClO₃ +5, HClO₄ +7; more O ⇒ more acidic (more stable conjugate base).

  • Easy : HClO₄ is strongest.
  • Medium : HOCl is weak.
  • Exam-type : ClO₄⁻ has four resonance forms.

Reasoning: why HClO₄ is stronger than HOCl

  • Step 1: Losing H⁺ gives ClO₄⁻.
  • Step 2: The charge spreads over 4 O atoms (resonance).
  • Step 3: A stable conjugate base ⇒ stronger acid.

CET level (MCQ)

  • CET 1) Sulfur transition temperature? 1) 369 K 2) 273 K 3) 393 K 4) 300 K — Answer: 1).
  • CET 2) Strongest acid? 1) HClO₄ 2) HClO₃ 3) HClO₂ 4) HOCl — Answer: 1).

Real-life use: Matches, fireworks, fertilisers, bleach.

Remember

S₈ crown; 369 K; HClO₄ strongest; oleum = H₂S₂O₇.

Common mistake

Thinking HOCl is stronger than HClO₄.

CET trap

More O ⇒ more acidic.

Shortcut

Memory clue: "more oxygens, more acidity".

Spot the difference

AcidCl state
HOCl+1
HClO₂+3
HClO₃+5
HClO₄+7

Practice questions

Easy
  1. 1What is allotropy?
  2. 2Allotropes of S.
  3. 3Shape of S₈.
Medium
  1. 4What is oleum?
  2. 5Basicity of H₂SO₄.
  3. 6Oxoacid acidity order.
Challenging
  1. 7Cl in HClO₃.
  2. 8Peroxo acid.
Show answers (try them yourself first)
  1. 1Different physical forms.
  2. 2α and β.
  3. 3Crown ring.
  4. 4H₂S₂O₇.
  5. 52.
  6. 6HClO₄ > HClO₃ > HClO₂ > HOCl.
  7. 7+5.
  8. 8H₂S₂O₈.

Dioxygen, oxides and ozone

From daily life

The fresh smell after a thunderstorm is partly ozone; the ozone layer shields us from UV.

Simple definition

O₂: colourless, odourless, paramagnetic; O₃: bent (117°), strong oxidant, 3O₂ ⇌ 2O₃, ΔH = +142 kJ/mol.

OzoneDioxygen, oxides and ozoneOOOO₃ : 117°O–O 128 pmOzone: 3O₂ ⇌ 2O₃ΔH = +142 kJ/mol (unstable)O₃ + 2I⁻ + H₂O → O₂ + I₂ + 2OH⁻Strong oxidant; absorbs UVOxideTypeNa₂O, MgObasicCO₂, SO₂, P₄O₁₀acidicAl₂O₃, ZnOamphotericCO, NO, N₂Oneutral2KClO₃ → 2KCl + 3O₂ (MnO₂, Δ); 2H₂O₂ → 2H₂O + O₂

In the figure : O₃ and oxides.

Rule 1: Preparing O₂ — 2KClO₃ → 2KCl + 3O₂ (MnO₂, Δ); 2H₂O₂ → 2H₂O + O₂; 2HgO → 2Hg + O₂; industrial: fractional distillation of liquid air.

  • Easy : MnO₂ is a catalyst.
  • Medium : 12.25 g KClO₃ ⇒ 4.8 g O₂.
  • Exam-type : Electrolysis of water.

Rule 2: Types of oxides — Basic (Na₂O), acidic (CO₂, SO₂), amphoteric (Al₂O₃, ZnO), neutral (CO, NO, N₂O).

  • Easy : Metal oxides are generally basic.
  • Medium : Non-metal oxides are acidic.
  • Exam-type : Al₂O₃ reacts with both acid and base.

Rule 3: Ozone — Silent electric discharge on O₂ gives O₃; O₃ + 2I⁻ + H₂O → O₂ + I₂ + 2OH⁻ (starch-iodide paper turns blue); mercury ‘tailing’.

  • Easy : O–O 128 pm.
  • Medium : Water disinfection.
  • Exam-type : Absorbs UV.

Calculation: O₂ from 12.25 g KClO₃ (M = 122.5)

  • Step 1: n(KClO₃) = 12.25/122.5 = 0.1 mol.
  • Step 2: 2 mol ⇒ 3 mol O₂, so 0.15 mol O₂.
  • Step 3: 0.15 × 32 = 4.8 g (3.36 L at STP).

CET level (MCQ)

  • CET 1) Amphoteric oxide? 1) Al₂O₃ 2) CO₂ 3) Na₂O 4) CO — Answer: 1).
  • CET 2) O–O–O angle in O₃? 1) 117° 2) 180° 3) 104.5° 4) 120° — Answer: 1).

Real-life use: Hospital oxygen, water purification, UV protection.

Remember

MnO₂ + KClO₃; 4 oxide types; O₃ bent 117°; starch-iodide test.

Common mistake

Calling CO an acidic oxide.

CET trap

CO, NO and N₂O are neutral.

Shortcut

Memory clue: "basic = metal, acidic = non-metal, amphoteric = Al-Zn".

Spot the difference

OxideType
Na₂Obasic
SO₂acidic
ZnOamphoteric
COneutral

Practice questions

Easy
  1. 1Lab preparation of O₂.
  2. 2Types of oxide.
  3. 3What is amphoteric?
Medium
  1. 4Structure of O₃.
  2. 5Preparation of O₃.
  3. 6Use of O₃.
Challenging
  1. 7O₂ from 12.25 g KClO₃.
  2. 8Starch-iodide test.
Show answers (try them yourself first)
  1. 1KClO₃ + MnO₂.
  2. 24 types.
  3. 3Reacts with both.
  4. 4Bent, 117°.
  5. 5Silent discharge.
  6. 6Disinfection.
  7. 74.8 g.
  8. 8Blue colour.

Sulfur dioxide and sulfuric acid

From daily life

Battery acid and fertilisers — sulfuric acid is the ‘king of chemicals’.

Simple definition

Contact process: S → SO₂ → SO₃ (V₂O₅, 720 K, 2 bar) → oleum → H₂SO₄.

ContactContact process for H₂SO₄S + O₂→ SO₂2SO₂ + O₂⇌ 2SO₃SO₃ + H₂SO₄→ H₂S₂O₇H₂S₂O₇ + H₂O→ 2H₂SO₄V₂O₅, 720 K2 bar pressureΔH = –196 kJ/molExothermic: low T favours yield, but slowSO₂: bent, 119.5°H₂SO₄: dehydrating, oxidisingC₁₂H₂₂O₁₁ → 12C + 11H₂O1 mol S ⇒ 1 mol H₂SO₄32 g S ⇒ 98 g H₂SO₄; 1 t S ⇒ 3.06 t

In the figure : Contact process.

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Sulfur dioxide and sulfuric acid

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Rule 1: SO₂ — S + O₂ → SO₂; bent 119.5°, S–O 143 pm; acidic in water; a reducing agent; temporary bleach.

  • Easy : SO₂ + H₂O ⇌ H₂SO₃.
  • Medium : Acid rain.
  • Exam-type : Cr₂O₇²⁻ ⇒ green Cr³⁺.

Rule 2: Contact process — 2SO₂ + O₂ ⇌ 2SO₃ (exothermic, ΔH = –196 kJ/mol), V₂O₅, 720 K, 2 bar; SO₃ + H₂SO₄ → H₂S₂O₇, then dilute.

  • Easy : SO₃ is not passed straight into water (mist).
  • Medium : Le Chatelier's principle.
  • Exam-type : 32 g S ⇒ 98 g H₂SO₄.

Rule 3: Properties of H₂SO₄ — Oily and dense; dehydrating (sugar ⇒ C); oxidising (Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O); dibasic.

  • Easy : Sugar turns black.
  • Medium : Dilute H₂SO₄ + Zn ⇒ H₂.
  • Exam-type : Uses: fertilisers, batteries, dyes.

Calculation: H₂SO₄ (M = 98) from 1 tonne of sulfur (S = 32)

  • Step 1: S → SO₂ → SO₃ → H₂SO₄: 1 mol S ⇒ 1 mol H₂SO₄.
  • Step 2: 32 g S ⇒ 98 g H₂SO₄.
  • Step 3: 1 tonne ⇒ 98/32 = 3.06 tonnes.

CET level (MCQ)

  • CET 1) Catalyst in the contact process? 1) V₂O₅ 2) Fe 3) Pt 4) Ni — Answer: 1).
  • CET 2) Property that blackens sugar? 1) dehydration 2) oxidation 3) reduction 4) hydrolysis — Answer: 1).

Real-life use: Fertilisers, batteries, dyes, chemical industry.

Remember

V₂O₅, 720 K, 2 bar; SO₃ ⇒ oleum; 1 mol S ⇒ 1 mol H₂SO₄; dehydrating agent.

Common mistake

Absorbing SO₃ directly in water.

CET trap

SO₃ is absorbed in H₂SO₄ first (oleum).

Shortcut

Memory clue: "V-five, 720 K, two bar".

Spot the difference

StepCondition
S + O₂→ SO₂
SO₂ → SO₃V₂O₅, 720 K, 2 bar
SO₃in H₂SO₄

Practice questions

Easy
  1. 1Write the contact process.
  2. 2Why these conditions?
  3. 3Oleum.
Medium
  1. 4Properties of H₂SO₄.
  2. 5Properties of SO₂.
  3. 6H₂SO₄ from 1 t of S.
Challenging
  1. 7Acid rain.
  2. 8Cu + H₂SO₄.
Show answers (try them yourself first)
  1. 1Three steps.
  2. 2Exothermic, catalyst.
  3. 3H₂S₂O₇.
  4. 4Dehydrating, oxidising.
  5. 5Reducing, acidic.
  6. 63.06 t.
  7. 7SO₂ and NOx.
  8. 8CuSO₄ + SO₂.

Chlorine, HCl, interhalogens, xenon compounds and noble gases

From daily life

Pool water is cleaned with chlorine; He fills balloons and Ne lights signs.

Simple definition

Cl₂: MnO₂ + 4HCl; HCl: NaCl + H₂SO₄ (420 K); interhalogens: XXʹₙ; Xe: XeF₂, XeF₄, XeF₆.

Cl₂ XeCl₂, HCl, interhalogens and Xe compoundsMnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂ODeacon: 4HCl + O₂ → 2Cl₂ + 2H₂O (CuCl₂, 723 K)NaCl + H₂SO₄ → NaHSO₄ + HCl (420 K)CompoundShapeXeF₂linearXeF₄square planarXeF₆distorted octahedralXeO₃pyramidalXeOF₄square pyramidalInterhalogens XXʹₙ (n = 1,3,5,7): ClF, ClF₃, BrF₅, IF₇XeF₂: Ni tube 400 °C; XeF₄: 1:5, 5–6 atm; XeF₆: electric discharge

In the figure : Cl₂, interhalogens, Xe.

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Chlorine, HCl, interhalogens, xenon compounds and noble gases

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Rule 1: Cl₂ and HCl — MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O; Deacon; electrolysis of brine; HCl: NaCl + H₂SO₄ → NaHSO₄ + HCl.

  • Easy : Cl₂ bleaches moist colours.
  • Medium : Cl₂ + 2NaOH (cold) → NaCl + NaOCl.
  • Exam-type : 8.7 g MnO₂ ⇒ 7.1 g Cl₂.

Rule 2: Interhalogens — XXʹₙ with n = 1, 3, 5, 7 (X larger): ClF, ClF₃, BrF₅, IF₇; more reactive than the halogens (weaker X–Xʹ) and polar.

  • Easy : IF₇ is pentagonal bipyramidal.
  • Medium : ClF₃ is T-shaped.
  • Exam-type : BrF₅ is square pyramidal.

Rule 3: Xenon and noble gases — Xe + F₂: XeF₂ (Ni tube, 400 °C), XeF₄ (1:5, 400 °C, 5–6 atm), XeF₆ (electric discharge, low temperature); XeF₆ + 3H₂O → XeO₃ + 6HF; He balloons, Ne signs, Ar welding.

  • Easy : XeF₂ is linear.
  • Medium : XeF₄ is square planar.
  • Exam-type : XeO₃ is pyramidal.

Calculation: Cl₂ from 8.7 g MnO₂ (M = 87) with excess HCl

  • Step 1: n(MnO₂) = 8.7/87 = 0.1 mol.
  • Step 2: 1 mol MnO₂ ⇒ 1 mol Cl₂ ⇒ 0.1 mol.
  • Step 3: 0.1 × 71 = 7.1 g; at STP 0.1 × 22.4 = 2.24 L.

CET level (MCQ)

  • CET 1) Shape of XeF₄? 1) square planar 2) tetrahedral 3) linear 4) pyramidal — Answer: 1).
  • CET 2) An interhalogen? 1) ClF₃ 2) HCl 3) Cl₂O 4) NaCl — Answer: 1).

Real-life use: Water purification, bleach, balloons, signs.

Remember

Cl₂: MnO₂ + HCl; interhalogens n = 1,3,5,7; XeF₂ linear, XeF₄ square planar.

Common mistake

Thinking XeF₄ is tetrahedral.

CET trap

Xe keeps 2 lone pairs ⇒ square planar.

Shortcut

Memory clue: "2 = line, 4 = square, 6 = distorted".

Spot the difference

CompoundShape
XeF₂linear
XeF₄square planar
XeF₆distorted octahedral
XeO₃pyramidal

Practice questions

Easy
  1. 1Preparation of Cl₂.
  2. 2Preparation of HCl.
  3. 3What are interhalogens?
Medium
  1. 4Making XeF₄.
  2. 5Hydrolysis of XeF₆.
  3. 6Uses of noble gases.
Challenging
  1. 7Cl₂ from 8.7 g MnO₂.
  2. 8Shape of IF₇.
Show answers (try them yourself first)
  1. 1MnO₂ + 4HCl.
  2. 2NaCl + H₂SO₄.
  3. 3Compound of two halogens.
  4. 41:5, Ni tube, 400 °C, 5–6 atm.
  5. 5XeO₃ + 6HF.
  6. 6He, Ne, Ar.
  7. 77.1 g.
  8. 8Pentagonal bipyramidal.

Quick revision of the whole lesson

ConceptKey formula / fact
Group 16ns²np⁴
Group 17ns²np⁵
Group 18ns²np⁶
Ozone3O₂ ⇌ 2O₃
Contact2SO₂ + O₂ ⇌ 2SO₃
Cl₂ prep.MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
S → H₂SO₄1 mol S ⇒ 1 mol H₂SO₄
XeF₆ hydrolysisXeF₆ + 3H₂O → XeO₃ + 6HF
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🎯 Most important points for the exam

  • Configuration.
  • Anomalies.
  • Oxidation states.
  • S allotropes.
  • O₂, O₃.
  • SO₂, H₂SO₄.
  • Cl₂, Xe.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.