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State Board · Class 11 · Chemistry

Lesson 9: Elements of Groups 13, 14 and 15

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

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Groups 13, 14, 15: configuration and trends

From daily life

Aluminium utensils (group 13), pencil graphite (group 14) and N/P in fertiliser (group 15) — all in daily life.

Simple definition

Group 13: ns²np¹ (B, Al, Ga, In, Tl); group 14: ns²np² (C, Si, Ge, Sn, Pb); group 15: ns²np³ (N, P, As, Sb, Bi). All are p-block; down a group radius grows and ionisation enthalpy generally falls.

Ionisation enthalpyFirst ionisation enthalpy IE₁ (kJ/mol)BNPAsSbBiperiod 23456Group 13Group 14Group 15IE falls down a group; Ga ≈ Al (poor d-shielding)

In the figure : IE₁ trends.

Rule 1: Atomic radius — Increases down a group. Covalent radius (pm): B 85, Al 143, Ga 135, In 167, Tl 170. Ga < Al because 3d electrons shield poorly.

  • Easy : C 77 < Si 118.
  • Medium : N 70 < P 110 < As 120.
  • Exam-type : Ga is smaller than Al.

Rule 2: Ionisation enthalpy — IE₁ (kJ/mol): group 13 – B 801, Al 577, Ga 579, In 558, Tl 589; group 14 – C 1086 → Pb 715; group 15 – N 1402 → Bi 703. N is high due to its half-filled 2p³.

  • Easy : Ga ≈ Al (exception).
  • Medium : N > P > As > Sb > Bi.
  • Exam-type : Tl > In (poor d and f shielding).

Rule 3: Electronegativity and metallic character — Group 13: B (2.0) is a non-metal, others are metals; group 14: C non-metal, Si/Ge metalloids, Sn/Pb metals; group 15: N, P non-metals, As/Sb metalloids, Bi metal.

  • Easy : B is more electronegative than Al.
  • Medium : Si and Ge are semiconductors.
  • Exam-type : Bi is the most metallic of group 15.

Reasoning: why is IE of Ga not lower than Al?

  • Step 1: Ga has a filled 3d¹⁰; d electrons shield poorly.
  • Step 2: Effective nuclear charge rises.
  • Step 3: The outer electron is held tighter ⇒ IE ≈ Al and a smaller radius.

CET level (MCQ)

  • CET 1) Valence configuration of group 14? 1) ns²np¹ 2) ns²np² 3) ns²np³ 4) ns²np⁴ — Answer: 2).
  • CET 2) Smallest covalent radius? 1) Al 2) Ga 3) In 4) Tl — Answer: 2).

Real-life use: Aluminium utensils, silicon chips, fertilisers.

Remember

IE: N 1402 (highest); Ga ≈ Al; Ga smaller than Al.

Common mistake

Assuming IE always falls down the group; Ga and Tl are exceptions.

CET trap

Ga < Al in radius and IE ≈ Al is a CET favourite exception.

Shortcut

Memory clue: "groups 13-14-15 = np¹, np², np³".

Spot the difference

GroupConfigurationFirst element
13ns²np¹B
14ns²np²C
15ns²np³N

Practice questions

Easy
  1. 1Write the group 13 configuration.
  2. 2Why is Ga smaller than Al?
  3. 3Why is N's IE high?
Medium
  1. 4Name a semiconductor.
  2. 5How does radius change down a group?
  3. 6Why is Tl's IE above In?
Challenging
  1. 7The metal in group 15?
  2. 8Why is B a non-metal?
Show answers (try them yourself first)
  1. 1ns²np¹.
  2. 2Poor d shielding.
  3. 3Half-filled 2p³.
  4. 4Si or Ge.
  5. 5It increases.
  6. 6Poor f and d shielding.
  7. 7Bi.
  8. 8Small, high IE, covalent bonding.

Oxidation states, inert pair effect and bonding

From daily life

Lead in batteries prefers +2, while nitrogen in fertilisers shows states from −3 to +5.

Simple definition

Inert pair effect: in heavy elements the ns² pair stays out of bonding ⇒ the state 2 below the group valence is stable (Tl⁺, Pb²⁺, Bi³⁺).

Inert pairInert pair effectGroup 13B · Al · Ga · InStable: +3Tl : +1ns² pair inertGroup 14C · Si · GeStable: +4Pb : +2ns² pair inertGroup 15N · P · As · SbStable: +5Bi : +3ns² pair inertDown the group the ns² pair stays out of bonding ⇒ lower state is stable

In the figure : Inert pair.

Rule 1: Common states — Group 13: +3 (Tl: +1); group 14: +4, +2; group 15: −3, +3, +5. Compounds of B and C are almost entirely covalent.

  • Easy : AlCl₃: Al +3.
  • Medium : PbCl₂: Pb +2.
  • Exam-type : NH₃: N −3; HNO₃: N +5.

Rule 2: Inert pair effect — Tl⁺ is more stable than Tl³⁺; Pb²⁺ more than Pb⁴⁺; Bi³⁺ more than Bi⁵⁺. Pb⁴⁺ and Tl³⁺ are strong oxidising agents.

  • Easy : PbO₂ is an oxidant.
  • Medium : SnCl₂ is a reductant (Sn²⁺ → Sn⁴⁺).
  • Exam-type : TlCl₃ is unstable, TlCl stable.

Rule 3: Bonding and valence shell — N has a maximum covalency of 4 (no d orbitals); P reaches 5/6 (PCl₅, PCl₆⁻). B is electron-deficient (BF₃, 6 e⁻); so is Al.

  • Easy : NCl₅ does not exist; PCl₅ does.
  • Medium : BF₃ + :NH₃ → F₃B←NH₃.
  • Exam-type : AlCl₃ dimerises to Al₂Cl₆.

Reasoning: why is Pb⁴⁺ an oxidant?

  • Step 1: Pb⁴⁺ + 2e⁻ → Pb²⁺.
  • Step 2: The inert 6s² pair makes Pb²⁺ stable.
  • Step 3: So Pb⁴⁺ gains electrons to give Pb²⁺ ⇒ oxidant.

CET level (MCQ)

  • CET 1) Stable state of Tl? 1) +1 2) +3 3) +2 4) +5 — Answer: 1).
  • CET 2) Why no NCl₅? 1) no d orbitals 2) high IE 3) large size 4) inert — Answer: 1).

Real-life use: Lead-acid batteries, tin plating.

Remember

Tl +1, Pb +2, Bi +3; N maximum covalency 4.

Common mistake

Writing covalency 5 for N; its maximum covalency is 4.

CET trap

NCl₅ is wrong, yet N is +5 in HNO₃ (oxidation state).

Shortcut

Memory clue: "heavy = 2 less" (Tl +1, Pb +2, Bi +3).

Spot the difference

GroupGroup valenceStable in heaviest
13+3Tl +1
14+4Pb +2
15+5Bi +3

Practice questions

Easy
  1. 1What is the inert pair effect?
  2. 2Stable state of Tl?
  3. 3Why is SnCl₂ a reductant?
Medium
  1. 4Why is NCl₅ absent?
  2. 5Why is BF₃ electron-deficient?
  3. 6Role of PbO₂?
Challenging
  1. 7Stable state of Bi?
  2. 8Oxidation state of N in HNO₃?
Show answers (try them yourself first)
  1. 1The ns² pair not taking part in bonding.
  2. 2+1.
  3. 3Sn²⁺ easily goes to Sn⁴⁺.
  4. 4N has no d orbitals.
  5. 5Only 6 electrons.
  6. 6Oxidising agent.
  7. 7+3.
  8. 8+5.

Reactivity with air, water, halogens; nature of oxides

From daily life

A thin oxide layer on aluminium utensils protects the metal beneath.

Simple definition

Oxides of groups 13–15 go from acidic to amphoteric to basic down a group (metallic character rises).

Oxide natureOxide nature: acidic → basic down the group13B₂O₃Al₂O₃Ga₂O₃In₂O₃Tl₂O14CO₂SiO₂GeO₂SnO₂PbO₂15N₂O₅P₄O₁₀As₂O₃Sb₂O₃Bi₂O₃acidicamphotericbasic

In the figure : Nature of oxides.

Rule 1: Air and water — B gives B₂O₃ + BN on heating; Al forms a protective Al₂O₃ layer and, hot, 2Al + N₂ → 2AlN. C and Si do not react with water readily; Sn and Pb react slowly in air.

  • Easy : 4Al + 3O₂ → 2Al₂O₃.
  • Medium : 2B + 3O₂ → 2B₂O₃ (hot).
  • Exam-type : Al + NaOH + H₂O → Na[Al(OH)₄] + H₂.

Rule 2: With halogens — Group 13: MX₃ (BCl₃, AlCl₃). Group 14: MX₄ (CCl₄, SiCl₄), Pb/Sn also give MX₂. Group 15: NX₃, PX₃, PX₅. SiCl₄ + 2H₂O → SiO₂ + 4HCl (hydrolysis); CCl₄ does not hydrolyse.

  • Easy : BCl₃ + 3H₂O → H₃BO₃ + 3HCl.
  • Medium : C has no d orbitals ⇒ CCl₄ does not hydrolyse.
  • Exam-type : PCl₃ + 3H₂O → H₃PO₃ + 3HCl.

Rule 3: Nature of oxides — B₂O₃ is acidic; Al₂O₃ and Ga₂O₃ amphoteric; In₂O₃ and Tl₂O basic. CO₂ and SiO₂ acidic; SnO₂ and PbO₂ amphoteric. N₂O₅ and P₄O₁₀ acidic; Bi₂O₃ basic. CO is neutral.

  • Easy : Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O.
  • Medium : Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄].
  • Exam-type : CO, N₂O and NO are neutral.

Calculation: H₂ from 2 mol Al with NaOH

  • Step 1: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂.
  • Step 2: 2 mol Al ⇒ 3 mol H₂.
  • Step 3: At STP: 3 × 22.4 = 67.2 L.

CET level (MCQ)

  • CET 1) Amphoteric oxide? 1) B₂O₃ 2) Al₂O₃ 3) Tl₂O 4) CO₂ — Answer: 2).
  • CET 2) Which does not hydrolyse? 1) SiCl₄ 2) CCl₄ 3) PCl₃ 4) BCl₃ — Answer: 2).

Real-life use: Anodising aluminium, acid/base cleaning.

Remember

B₂O₃ acidic, Al₂O₃ amphoteric, Tl₂O basic; CCl₄ does not hydrolyse.

Common mistake

Calling Al₂O₃ only basic.

CET trap

CO and N₂O are neutral, not acidic.

Shortcut

Memory clue: "lower = basic".

Spot the difference

OxideNature
B₂O₃acidic
Al₂O₃amphoteric
Tl₂Obasic
COneutral

Practice questions

Easy
  1. 1Protective layer on Al?
  2. 2Why is Al₂O₃ amphoteric?
  3. 3Why no hydrolysis of CCl₄?
Medium
  1. 4BCl₃ + H₂O equation.
  2. 5Name a neutral oxide.
  3. 6H₂ at STP from 2 mol Al?
Challenging
  1. 7Hydrolysis of SiCl₄.
  2. 8Nature of Bi₂O₃?
Show answers (try them yourself first)
  1. 1Al₂O₃.
  2. 2Reacts with both acid and base.
  3. 3C has no d orbitals.
  4. 4BCl₃ + 3H₂O → H₃BO₃ + 3HCl.
  5. 5CO.
  6. 667.2 L.
  7. 7SiCl₄ + 2H₂O → SiO₂ + 4HCl.
  8. 8Basic.

Allotropy and catenation: carbon and phosphorus

From daily life

Pencil graphite and a diamond ring are both pure carbon; red phosphorus is on a matchbox.

Simple definition

Allotropy: different physical forms of one element. Catenation: bonding with own atoms to form chains (C > Si > S ...).

AllotropesAllotropes of carbon and phosphorusDiamond sp³hard, insulatorGraphite sp²layers, conducts335 pm between layersFullerene C₆₀60 C, 32 faces12 pentagons + 20 hexagonsWhite P₄: 60° strain, very reactiveRed: chains, less reactiveBlack: layered, most stable

In the figure : Allotropes of C and P.

Rule 1: Allotropes of carbon — Diamond: sp³, C–C 154 pm, hard insulator. Graphite: sp², layered, conducts, lubricant. Fullerene C₆₀: 12 pentagons + 20 hexagons, football-like.

  • Easy : Diamond is the hardest.
  • Medium : Graphite conducts (free π electrons).
  • Exam-type : C₆₀ has 60 carbons.

Rule 2: Allotropes of phosphorus — White P₄ (60° strain, ignites in air, stored under water, dissolves in CS₂), red (chains, stable), black (layered, most stable).

  • Easy : P₄ + 5O₂ → P₄O₁₀.
  • Medium : 12.4 g P₄ → 28.4 g P₄O₁₀.
  • Exam-type : White is poisonous, red is not.

Rule 3: Catenation — C–C bond energy 348 kJ/mol; Si–Si about 297; so carbon catenates most. N is poor (weak N–N), P moderate.

  • Easy : Countless hydrocarbons.
  • Medium : Few silanes SiₙH₂ₙ₊₂.
  • Exam-type : P–P bonds in P₄.

Calculation: P₄O₁₀ from burning 12.4 g P₄

  • Step 1: M(P₄) = 124 ⇒ n = 12.4 ÷ 124 = 0.1 mol.
  • Step 2: P₄ + 5O₂ → P₄O₁₀ (1 : 1).
  • Step 3: M(P₄O₁₀) = 284 ⇒ 0.1 × 284 = 28.4 g.

CET level (MCQ)

  • CET 1) Hybridisation in graphite? 1) sp 2) sp² 3) sp³ 4) sp³d — Answer: 2).
  • CET 2) White phosphorus is stored? 1) in air 2) under water 3) in oil 4) in CS₂ — Answer: 2).

Real-life use: Pencils, diamond drills, matches, carbon nanotubes.

Remember

Diamond sp³; graphite sp² + free electrons; C₆₀ = 60 C; P₄ strain 60°.

Common mistake

Calling graphite hard; it is soft.

CET trap

Diamond and graphite differ by hybridisation, not element.

Shortcut

Memory clue: "diamond = 3 (sp³), graphite = 2 (sp²)".

Spot the difference

ItemDiamondGraphite
Hybridisationsp³sp²
Conductivitynoyes
Hardnesshighlow

Practice questions

Easy
  1. 1What is allotropy?
  2. 2Why does graphite conduct?
  3. 3Structure of C₆₀.
Medium
  1. 4Why store white P under water?
  2. 5P₄O₁₀ from 12.4 g P₄?
  3. 6What is catenation?
Challenging
  1. 7Why is red P stable?
  2. 8Why does C lead in catenation?
Show answers (try them yourself first)
  1. 1Different forms of one element.
  2. 2Free π electrons.
  3. 312 pentagons + 20 hexagons.
  4. 4It ignites in air.
  5. 528.4 g.
  6. 6Chain formation by own atoms.
  7. 7Chain structure, less strain.
  8. 8Strong C–C bond.

Molecular structures of important compounds

From daily life

Boric acid eye drops, silica sand and ammonia cleaner each have a different shape.

Simple definition

Shapes follow VSEPR and hybridisation: BF₃ trigonal planar (sp²), NH₃ pyramidal (sp³), SiCl₄ tetrahedral, PCl₅ trigonal bipyramidal (sp³d), B₂H₆ bridged.

StructuresShapes of important compoundsB₂H₆BB2 bridges 3c–2e (red)BF₃ vs NH₃planar, 120°pyramidal, 107°SiCl₄: tetrahedral (sp³) · PCl₅: trigonal bipyramidal (sp³d)BF₃ is electron-deficient ⇒ forms the BF₃·NH₃ adductC₂H₆: sp³ · C₂H₄: sp² · C₂H₂: sp

In the figure : Structures.

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Molecular structures of important compounds

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Rule 1: Diborane B₂H₆ — 12 valence electrons, too few for 8 normal pairs. 4 terminal B–H (2c–2e) plus 2 bridges B–H–B (3c–2e). B is sp³. B–H–B angle ≈ 97°.

  • Easy : Terminal H: 4; bridging H: 2.
  • Medium : Six bonding positions but only 12 electrons.
  • Exam-type : B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂.

Rule 2: BF₃, NH₃ and hybridisation — BF₃: sp², 120°, electron-deficient (Lewis acid). NH₃: sp³, one lone pair, 107°. SiO₂: tetrahedral Si in a network.

  • Easy : BF₃ + NH₃ → F₃B←NH₃.
  • Medium : [BF₄]⁻ is tetrahedral.
  • Exam-type : NH₄⁺ is tetrahedral, 109.5°.

Rule 3: Boric acid and silicates — H₃BO₃: layered through H-bonds; a Lewis acid: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺. Silicates: SiO₄⁴⁻ units forming chains, sheets or 3-D networks.

  • Easy : H₃BO₃ is monobasic (Lewis).
  • Medium : Zeolites: Si–O–Al framework.
  • Exam-type : Quartz: SiO₂ network.

Reasoning: why 3c–2e bonds in B₂H₆?

  • Step 1: Total electrons = 2×3 + 6×1 = 12.
  • Step 2: Eight normal bonds would need 16 electrons.
  • Step 3: 12 electrons ⇒ 4 terminal (8 e⁻) + 2 bridges (4 e⁻) as 3c–2e.

CET level (MCQ)

  • CET 1) Shape of NH₃? 1) planar 2) linear 3) pyramidal 4) tetrahedral — Answer: 3).
  • CET 2) 3c–2e bonds in B₂H₆? 1) 1 2) 2 3) 4 4) 6 — Answer: 2).

Real-life use: Boric acid antiseptic, silica gel, zeolites.

Remember

BF₃ 120° sp²; NH₃ 107° sp³; B₂H₆: 4 terminal + 2 bridges.

Common mistake

Treating all B₂H₆ bonds as normal.

CET trap

H₃BO₃ does not donate H⁺; it accepts OH⁻ (Lewis acid).

Shortcut

Memory clue: "BF₃ flat, NH₃ pyramid".

Spot the difference

CompoundHybridisationShape
BF₃sp²trigonal planar
NH₃sp³pyramidal
SiCl₄sp³tetrahedral
PCl₅sp³dbipyramidal

Practice questions

Easy
  1. 1Electrons in B₂H₆?
  2. 2What is a 3c–2e bond?
  3. 3Why is BF₃ a Lewis acid?
Medium
  1. 4Why is NH₃ 107°?
  2. 5How is H₃BO₃ acidic?
  3. 6Shape of PCl₅?
Challenging
  1. 7Structure of SiO₂?
  2. 8Shape of [BF₄]⁻?
Show answers (try them yourself first)
  1. 112.
  2. 2A bond of 2 electrons over 3 atoms.
  3. 3Incomplete octet.
  4. 4Lone-pair repulsion.
  5. 5By accepting OH⁻.
  6. 6Trigonal bipyramidal.
  7. 7A 3-D network.
  8. 8Tetrahedral.

Borax and silicones

From daily life

Borax in glass and jewellery work; silicones in waterproof sprays, sealants and baking moulds.

Simple definition

Borax Na₂B₄O₇·10H₂O (M ≈ 381). Silicones: –[R₂Si–O]ₙ– polymers, R = CH₃.

Borax, siliconesBorax and siliconesNa₂B₄O₇·10H₂OboraxNa₂B₄O₇anhydrous2NaBO₂ + B₂O₃glassy beadBorax-bead test: Cu blue · Cr green · Co deep blue · Mn violet · Fe yellowSilicone: (CH₃)₂SiCl₂ + 2H₂O → (CH₃)₂Si(OH)₂ → –[Si(CH₃)₂–O]ₙ–SiCH₃CH₃SiCH₃CH₃SiCH₃CH₃SiCH₃CH₃SiCH₃CH₃(CH₃)₃SiCl stops the chain · heat- and water-resistant, lubricants

In the figure : Borax and silicones.

Rule 1: Heating borax — Na₂B₄O₇·10H₂O → (heat) Na₂B₄O₇ → 2NaBO₂ + B₂O₃ (glassy bead). Ion [B₄O₅(OH)₄]²⁻; aqueous solution is alkaline (Na₂B₄O₇ + 7H₂O → 2NaOH + 4H₃BO₃).

  • Easy : 38.1 g borax = 0.1 mol.
  • Medium : Anhydrous 20.1 g; B₂O₃ 6.96 g.
  • Exam-type : Water lost: 18 g.

Rule 2: Borax bead test — B₂O₃ + CuO → Cu(BO₂)₂ (blue); Cr → green; Co → deep blue; Mn → violet; Fe → yellow-brown; Ni → brown.

  • Easy : Cu: blue.
  • Medium : Co: deep blue.
  • Exam-type : Cr: green.

Rule 3: Silicones — (CH₃)₂SiCl₂ + 2H₂O → (CH₃)₂Si(OH)₂ → –[Si(CH₃)₂O]ₙ– + H₂O. (CH₃)₃SiCl stops the chain; CH₃SiCl₃ makes cross-links. Water-repellent, heat-stable and inert.

  • Easy : Lubricants, sealants.
  • Medium : Medical implants.
  • Exam-type : Si–O–Si chain.

Calculation: B₂O₃ from heating 38.1 g borax

  • Step 1: M ≈ 381 ⇒ n = 38.1 ÷ 381 = 0.1 mol.
  • Step 2: 1 mol borax ⇒ 1 mol B₂O₃.
  • Step 3: M(B₂O₃) = 69.6 ⇒ 0.1 × 69.6 = 6.96 g.

CET level (MCQ)

  • CET 1) Formula of borax? 1) Na₂B₄O₇·10H₂O 2) Na₂BO₃ 3) H₃BO₃ 4) B₂O₃ — Answer: 1).
  • CET 2) Monomer of silicone? 1) (CH₃)₂SiCl₂ 2) SiCl₄ 3) CH₃SiH₃ 4) SiO₂ — Answer: 1).

Real-life use: Glass, jewellery flux, waterproofing.

Remember

Borax M ≈ 381; bead: Cu blue; silicone = Si–O chain.

Common mistake

Calling silicone SiO₂; it is an organosilicon polymer.

CET trap

Borax solution is alkaline, not acidic.

Shortcut

Memory clue: "Cu-blue, Cr-green, Co-deep blue".

Spot the difference

ItemBoraxSilicone
Typesaltpolymer
Propertyalkaline solutionwater-repellent

Practice questions

Easy
  1. 1Formula of borax?
  2. 2What is the borax bead test for?
  3. 3How are silicones made?
Medium
  1. 4Role of (CH₃)₃SiCl?
  2. 5B₂O₃ from 38.1 g borax?
  3. 6Why is borax solution alkaline?
Challenging
  1. 7Bead colour of Co?
  2. 8Two uses of silicones.
Show answers (try them yourself first)
  1. 1Na₂B₄O₇·10H₂O.
  2. 2To identify metal radicals.
  3. 3Hydrolysis and condensation of (CH₃)₂SiCl₂.
  4. 4Chain stopper.
  5. 56.96 g.
  6. 6Hydrolysis gives NaOH.
  7. 7Deep blue.
  8. 8Lubricants and sealants.

Ammonia: Haber process and properties

From daily life

Urea fertiliser and household cleaner both come from ammonia.

Simple definition

Ammonia NH₃: pungent, very soluble gas, weak base; N₂ + 3H₂ ⇌ 2NH₃ (Haber), ΔH = −92 kJ.

Haber processAmmonia: Haber processN₂1 mol3H₂3 molReactorN₂ + 3H₂ ⇌ 2NH₃ΔH = −92 kJ~700 K, 200 atmiron oxide + K₂O, Al₂O₃2NH₃liquefiedOstwald: 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt, 1100 K) → HNO₃NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ · deep blue [Cu(NH₃)₄]²⁺ with Cu²⁺2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃ (laboratory)Dry with CaO (not H₂SO₄ or CaCl₂)

In the figure : Haber process.

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Ammonia: Haber process and properties

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Rule 1: Haber process — About 700 K, 200 atm, iron oxide catalyst (with a little K₂O and Al₂O₃). Le Chatelier: high pressure and moderate temperature; the catalyst speeds up but does not shift equilibrium.

  • Easy : 5.6 g N₂ ⇒ 6.8 g NH₃.
  • Medium : 56 g N₂ ⇒ 68 g NH₃.
  • Exam-type : 1 mol N₂ + 3 mol H₂ ⇒ 2 mol NH₃.

Rule 2: Properties — NH₃ + H₂O ⇌ NH₄⁺ + OH⁻; Lewis base (lone pair); with Cu²⁺ gives deep-blue [Cu(NH₃)₄]²⁺; dissolves AgCl; excess Cl₂ ⇒ NCl₃, limited Cl₂ ⇒ N₂.

  • Easy : Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺.
  • Medium : AgCl + 2NH₃ → [Ag(NH₃)₂]Cl.
  • Exam-type : HCl gives white fumes of NH₄Cl.

Rule 3: Laboratory and uses — 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃; dry with CaO (not H₂SO₄ or CaCl₂). Uses: fertilisers, HNO₃ (Ostwald), refrigerant.

  • Easy : 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt).
  • Medium : 17 g NH₃ ⇒ 30 g NO.
  • Exam-type : Urea fertiliser.

Calculation: NH₃ from 5.6 g N₂ (100 % yield)

  • Step 1: n(N₂) = 5.6 ÷ 28 = 0.2 mol.
  • Step 2: 1 : 2 ⇒ NH₃ = 0.4 mol.
  • Step 3: 0.4 × 17 = 6.8 g.

CET level (MCQ)

  • CET 1) Catalyst in the Haber process? 1) Pt 2) Fe 3) Ni 4) V₂O₅ — Answer: 2).
  • CET 2) To dry NH₃? 1) H₂SO₄ 2) CaCl₂ 3) CaO 4) P₄O₁₀ — Answer: 3).

Real-life use: Fertilisers, nitric acid, refrigeration, cleaners.

Remember

700 K, 200 atm, iron oxide; dry with CaO; [Cu(NH₃)₄]²⁺ deep blue.

Common mistake

Drying ammonia with H₂SO₄.

CET trap

High temperature lowers yield because the reaction is exothermic.

Shortcut

Memory clue: "high pressure, moderate temperature".

Spot the difference

ItemHaberOstwald
ProductNH₃HNO₃
CatalystFePt

Practice questions

Easy
  1. 1Conditions of Haber process.
  2. 2Why high pressure?
  3. 3How is NH₃ dried?
Medium
  1. 4NH₃ from 5.6 g N₂?
  2. 5Why is NH₃ a base?
  3. 6Colour with Cu²⁺?
Challenging
  1. 7NO from 17 g NH₃?
  2. 8Uses of NH₃.
Show answers (try them yourself first)
  1. 1700 K, 200 atm, Fe.
  2. 2Fewer moles of gas.
  3. 3With CaO.
  4. 46.8 g.
  5. 5A lone pair.
  6. 6Deep blue.
  7. 730 g.
  8. 8Fertilisers, HNO₃, refrigeration.

Quick revision of the whole lesson

ConceptKey formula / fact
Groups 13/14/15np¹ / np² / np³
Highest IE₁N 1402 kJ/mol
Inert pairTl +1, Pb +2, Bi +3
Oxidesacidic → basic
Carbondiamond sp³, graphite sp²
B₂H₆4 terminal + 2 bridges
BoraxM ≈ 381
Haber700 K, 200 atm, Fe
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🎯 Most important points for the exam

  • p-block trends.
  • Inert pair.
  • Oxide nature.
  • Allotropy.
  • Structures.
  • Borax and silicones.
  • Ammonia.
Study mantra

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