Chapter 21: The p-Block Elements (Group 15 to 18)

Introduction

Overview

• Group 13-18 elements form the p-block of the periodic table.
• Properties influenced by atomic size, ionisation enthalpy, electron gain enthalpy, and electronegativity.
• Second period elements lack d-orbitals; heavier elements have d- and/or f-orbitals affecting properties.

Solved Examples

• Ex 1: Explain why ionisation enthalpy decreases down the group.
  Step 1: Atomic size increases down the group.
  Step 2: Outer electrons are farther from nucleus.
  Step 3: Less energy required to remove electron.

Practice Set

• Q1: Which factor primarily influences p-block element properties?
  A. Atomic size
  B. Nuclear spin
  C. Magnetic moment
  D. Isotopic mass
• Q2: Why do second period p-block elements differ from heavier ones?
  A. Presence of f-orbitals
  B. Absence of d-orbitals
  C. Higher electronegativity
  D. Larger atomic radius
• Q3: Which property decreases down the p-block group?
  A. Atomic size
  B. Ionisation enthalpy
  C. Electron gain enthalpy
  D. Electronegativity

Answer Key

• Q1: A - Atomic size affects ionisation energy and electronegativity, key to p-block properties.
• Q2: B - Second period elements lack d-orbitals, affecting their chemical behavior.
• Q3: D - Electronegativity decreases down the group due to increased atomic size.

Memory Aids

• Remember: "No d-orbitals in 2nd period" for unique properties.
• Properties trend: Size ↑, Ionisation energy ↓, Electronegativity ↓ down the group.

Group 15 Elements

General Characteristics

• Elements: N, P, As, Sb, Bi, Moscovium (synthetic, short half-life).
• Known as pnictogens.
• General electronic configuration: ns²np³.
• Atomic radii increase down the group; smaller than group 14 due to higher nuclear charge.
• Melting point increases N to As, then decreases due to inert pair effect.
• Metallic character increases down the group.
• Ionisation enthalpy decreases down the group; higher than group 14 due to half-filled p-orbital stability.
• Electronegativity decreases down the group.
• Allotropy present in all except nitrogen.

Chemical Properties

• Oxidation states: +3 stability increases down the group; +5 decreases due to inert pair effect.
• Halides: General formula EX₃ or EX₅.
• Halide stability: NF₃ > NCl₃ > NBr₃ (due to size differences).
• Lewis acid strength: PCl₃ > NBr₃ > NCl₃ > NF₃ (decreasing electron density on N with increasing halogen electronegativity).
• Halide bond angle: PF₃ < PCl₃ < PBr₃ < PI₃ (due to decreasing bond pair-bond pair repulsions).
• Hydrides (EH₃): Bond angle decreases NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃.
• Basicity: NH₃ > PH₃; AsH₃, SbH₃, BiH₃ are not basic.
• Boiling point order: PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃.
• Melting point order: PH₃ < AsH₃ < SbH₃ < NH₃.
• Thermal stability: NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃.
• Reducing character: NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃.
• Solubility in water: NH₃, PH₃ soluble; AsH₃, SbH₃, BiH₃ insoluble.
• Oxides: General formula E₂O₃ or E₂O₅.
• Acidic strength of trioxides: N₂O₃ > P₄O₆ > As₄O₆ > Sb₄O₆ (due to decreasing electronegativity).
• Nitrogen oxides sequence: N₂O, NO (neutral), N₂O₃, N₂O₄, N₂O₅ (increasing oxidation state).
• Reactivity with metals: Forms binary compounds with -3 oxidation state (e.g., Ca₃N₂, Ca₃P₂, Na₃As).

Solved Examples

• Ex 1: Predict the bond angle trend in group 15 hydrides.
  Step 1: NH₃ has highest bond angle due to lone pair repulsion.
  Step 2: Bond angle decreases down the group due to increasing atomic size.
• Ex 2: Explain the stability trend of +3 and +5 oxidation states.
  Step 1: +3 state stability increases down the group due to inert pair effect.
  Step 2: +5 state stability decreases down the group.

Practice Set

• Q1: Which element is not a pnictogen?
  A. Nitrogen
  B. Phosphorus
  C. Sulfur
  D. Bismuth
• Q2: Which halide is most stable?
  A. NBr₃
  B. NCl₃
  C. NF₃
  D. PCl₃
• Q3: Which hydride has the highest basicity?
  A. PH₃
  B. NH₃
  C. AsH₃
  D. SbH₃

Answer Key

• Q1: C - Sulfur belongs to group 16, not group 15.
• Q2: C - NF₃ is most stable due to size and electronegativity differences.
• Q3: B - NH₃ is the most basic hydride in group 15.

Memory Aids

• Pnictogens: "N P As Sb Bi" - remember Moscovium is synthetic.
• Oxidation state trend: +3 ↑, +5 ↓ down the group.
• Hydrides: Basicity and bond angle decrease down the group.
• Halide stability: NF₃ > NCl₃ > NBr₃.

Dinitrogen

Preparation of Nitrogen Gas

• Ammonium chloride reacts with sodium nitrite:
  NH₄Cl + NaNO₂ → N₂ + 2H₂O + NaCl
• Thermal decomposition of ammonium dichromate:
  (NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + 4H₂O
• Thermal decomposition of barium azide:
  Ba(N₃)₂ → Ba + 3N₂

Chemical Properties

• Reaction with lithium:
  N₂ + 6Li → 2Li₃N (lithium nitride)
• Reaction with magnesium:
  N₂ + Mg → Mg₃N₂ (magnesium nitride)
• Reaction with oxygen at high temperature:
  N₂ + O₂ → 2NO (nitric oxide)
• Haber process (ammonia synthesis):
  N₂ + 3H₂ ⇌ 2NH₃ (high pressure, temperature, catalyst; exothermic)

Solved Examples

• Ex 1: Calculate moles of N₂ produced from 1 mole NH₄Cl and NaNO₂.
  Step 1: Reaction: NH₄Cl + NaNO₂ → N₂ + 2H₂O + NaCl
  Step 2: 1 mole NH₄Cl produces 1 mole N₂.
• Ex 2: Write balanced equation for ammonia synthesis.
  Step 1: N₂ + 3H₂ ⇌ 2NH₃

Practice Set

• Q1: Which reagent is used with ammonium chloride to prepare nitrogen gas?
  A. Sodium hydroxide
  B. Sodium nitrite
  C. Sodium carbonate
  D. Sodium sulfate
• Q2: What is the product when nitrogen reacts with lithium?
  A. LiNO₃
  B. Li₃N
  C. LiNO₂
  D. LiOH
• Q3: In the Haber process, what is the stoichiometric ratio of N₂ to H₂?
  A. 1:1
  B. 1:2
  C. 1:3
  D. 3:1

Answer Key

• Q1: B - Sodium nitrite reacts with ammonium chloride to produce nitrogen gas.
• Q2: B - Lithium reacts with nitrogen to form lithium nitride (Li₃N).
• Q3: C - The Haber process uses 1 mole N₂ and 3 moles H₂.

Memory Aids

• Remember: "NaNO₂ + NH₄Cl → N₂" for nitrogen preparation.
• Haber process ratio: N₂ : H₂ = 1 : 3.
• Nitrogen forms nitrides with active metals like Li and Mg.

Ammonia Preparation

Laboratory Preparation

• Ammonium salts react with alkali to release NH₃.
• Reactions:
  • 2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
  • (NH₄)₂SO₄ + 2NaOH → 2NH₃ + Na₂SO₄ + 2H₂O
  • Mg₃N₂ + 6HCl → 3MgCl₂ + 2NH₃
• Urea hydrolyses in water to form ammonium carbonate, which decomposes to NH₃, CO₂, and H₂O (reversible reaction).

Manufacture

• Haber’s process: N₂ + 3H₂ ⇌ 2NH₃ (ΔH° = -46.1 kJ/mol, exothermic)
• Conditions: 773 K, 200 atm, iron oxide catalyst with K₂O and Al₂O₃ promoters.

Solved Examples

• Ex 1: Calculate moles of NH₃ produced from 1 mole N₂ in Haber’s process.
  • 1 N₂ + 3 H₂ → 2 NH₃
  • So, 1 mole N₂ produces 2 moles NH₃.

Practice Set

• Q1: Which reagent is used in laboratory preparation of NH₃?
  • A. NH₄Cl + Ca(OH)₂
  • B. NH₃ + H₂O
  • C. N₂ + H₂
  • D. NH₄OH + ZnCl₂
• Q2: What is the enthalpy change in Haber’s process?
  • A. +46.1 kJ/mol
  • B. -46.1 kJ/mol
  • C. 0 kJ/mol
  • D. -100 kJ/mol
• Q3: Which catalyst is used in Haber’s process?
  • A. Copper oxide
  • B. Iron oxide
  • C. Zinc oxide
  • D. Aluminium oxide alone

Answer Key

• Q1: A - Ammonia is prepared in lab by reacting ammonium salts with alkali like Ca(OH)₂.
• Q2: B - The reaction is exothermic with ΔH° = -46.1 kJ/mol.
• Q3: B - Iron oxide catalyst is used with promoters K₂O and Al₂O₃.

Memory Aids

• Lab prep: Ammonium salt + alkali → NH₃
• Haber’s process: N₂ + 3H₂ ⇌ 2NH₃, exothermic
• Catalyst: Iron oxide + K₂O + Al₂O₃

Ammonia Structure

Molecular Structure

• NH₃ is covalent with nitrogen using three sp³ hybrid orbitals overlapping with hydrogen 1s orbitals.
• One sp³ orbital holds a lone pair.
• Shape: Trigonal pyramidal, bond angle ≈ 107.8°.
• In liquid/solid, NH₃ molecules are hydrogen bonded.

Solved Examples

• Ex 1: Identify bond angle in NH₃.
  • Answer: 107.8° due to lone pair repulsion.

Practice Set

• Q1: What is the shape of NH₃ molecule?
  • A. Linear
  • B. Trigonal planar
  • C. Trigonal pyramidal
  • D. Tetrahedral
• Q2: Which orbital hybridization is present in nitrogen of NH₃?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d
• Q3: What causes the bond angle in NH₃ to be less than 109.5°?
  • A. Lone pair repulsion
  • B. Hydrogen bonding
  • C. Double bonds
  • D. Ionic character

Answer Key

• Q1: C - NH₃ is trigonal pyramidal due to lone pair on nitrogen.
• Q2: C - Nitrogen is sp³ hybridized in NH₃.
• Q3: A - Lone pair repulsion reduces bond angle from 109.5° to 107.8°.

Memory Aids

• Shape: Trigonal pyramidal = 3 bonds + 1 lone pair
• Hybridization: sp³ for nitrogen
• Lone pair repulsion lowers bond angle

Ammonia Basic Character

Basic Nature

• NH₃ donates lone pair, acts as Lewis base.
• In water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
• Produces alkaline solution due to OH⁻ ions.

Reaction with Heavy Metal Ions

• Forms insoluble metal hydroxide precipitates:
  • FeCl₃ + 3NH₄OH → 3NH₄Cl + Fe(OH)₃ (brown ppt.)
  • ZnCl₂ + 2NH₄OH → 2NH₄Cl + Zn(OH)₂ (white ppt.)

Formation of Complex Compounds

• NH₃ forms stable soluble complexes with Ag⁺, Cu²⁺, Hg²⁺.
  • AgOH + 2NH₄OH → [Ag(NH₃)₂]OH + 2H₂O
  • AgCl + 2NH₄OH → [Ag(NH₃)₂]Cl + 2H₂O
  • Cu(OH)₂ + (NH₄)₂SO₄ + NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O (deep blue solution)

Solved Examples

• Ex 1: Write the reaction of NH₃ with water.
  • NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Practice Set

• Q1: What ion is formed when NH₃ dissolves in water?
  • A. NH₄⁺
  • B. OH⁻
  • C. NH₃
  • D. H₃O⁺
• Q2: What is the color of Fe(OH)₃ precipitate formed with NH₄OH?
  • A. White
  • B. Brown
  • C. Blue
  • D. Green
• Q3: Which complex is formed by NH₃ with Cu²⁺?
  • A. [Cu(NH₃)₂]²⁺
  • B. [Cu(NH₃)₄]²⁺
  • C. [Cu(NH₃)₃]²⁺
  • D. [Cu(NH₃)]²⁺

Answer Key

• Q1: A - NH₃ accepts H⁺ to form NH₄⁺ in water.
• Q2: B - Fe(OH)₃ forms brown precipitate.
• Q3: B - Tetraamminecopper (II) complex [Cu(NH₃)₄]²⁺ is deep blue.

Memory Aids

• NH₃ + H₂O → NH₄⁺ + OH⁻ (basic solution)
• Fe(OH)₃ brown ppt., Zn(OH)₂ white ppt.
• Cu complex deep blue, Ag complex soluble

Nitrogen Hydrides

Types and Nature

• Three hydrides: NH₃ (ammonia), N₂H₄ (hydrazine), N₃H (hydrazoic acid).
• NH₃ and N₂H₄ are basic; N₃H is acidic.

Solved Examples

• Ex 1: Identify acidic hydride among nitrogen hydrides.
  • Answer: N₃H (hydrazoic acid) is acidic.

Practice Set

• Q1: Which nitrogen hydride is acidic?
  • A. NH₃
  • B. N₂H₄
  • C. N₃H
  • D. None
• Q2: Hydrazine formula is?
  • A. NH₃
  • B. N₂H₄
  • C. N₃H
  • D. NH₂OH
• Q3: Which hydride is basic?
  • A. N₃H
  • B. NH₃
  • C. Both NH₃ and N₂H₄
  • D. None

Answer Key

• Q1: C - N₃H is acidic hydride.
• Q2: B - Hydrazine is N₂H₄.
• Q3: C - NH₃ and N₂H₄ are basic.

Memory Aids

• Acidic: N₃H; Basic: NH₃, N₂H₄
• Hydrazine = N₂H₄

Oxides of Nitrogen

Preparation and Properties

• Nitrogen oxides are compounds of nitrogen and oxygen with varying oxidation states.
• Common oxides include N₂O, NO, N₂O₃, NO₂, and N₂O₅.
• Preparation methods vary: thermal decomposition, direct combination, and oxidation reactions.
• Physical states differ: gases (N₂O, NO, NO₂), liquids, or solids.
• Acidic or neutral nature depends on the oxide; e.g., N₂O is neutral, NO₂ is acidic.

Solved Examples

• Ex 1: Preparation of Nitric Oxide (NO)
  Step 1: React NH₄NO₃ (ammonium nitrate) by heating.
  Step 2: NH₄NO₃ → N₂O + 2H₂O (decomposition reaction)
• Ex 2: Formation of Nitrogen Dioxide (NO₂)
  Step 1: Oxidize NO in air.
  Step 2: 2NO + O₂ → 2NO₂

Practice Set

• Q1: Which oxide of nitrogen is a colorless, neutral gas?
  A. NO₂
  B. N₂O
  C. NO
  D. N₂O₅
• Q2: What is the oxidation state of nitrogen in N₂O?
  A. +1
  B. +2
  C. +3
  D. +4
• Q3: Which reaction correctly represents the preparation of NO?
  A. NH₄NO₃ → N₂O + 2H₂O
  B. 2NO + O₂ → 2NO₂
  C. NH₄NO₃ → NO + 2H₂O
  D. N₂ + O₂ → 2NO

Answer Key

• Q1: B - N₂O is a colorless, neutral gas.
• Q2: A - Nitrogen has +1 oxidation state in N₂O.
• Q3: D - Direct combination of N₂ and O₂ forms NO.

Memory Aids

• Remember: N₂O is "laughing gas" (neutral, colorless).
• NO is formed by direct combination of N₂ and O₂ at high temperature.
• NO₂ is brown and acidic; formed by oxidation of NO.

Nitric Acid Preparation

Laboratory Preparation

• React sodium nitrate (NaNO₃) with concentrated sulfuric acid (H₂SO₄):
• NaNO₃ + 5H₂SO₄ → NaHSO₄ + HNO₃

Ostwald’s Process

• Ammonia (NH₃) oxidized to nitric acid using platinum/rhodium catalyst.
• Step 1: NH₃ + O₂ → NO + H₂O (catalyst, high T & P)
• Step 2: NO + O₂ → NO₂
• Step 3: NO₂ + H₂O → HNO₃ + NO
• NO is recycled; HNO₃ concentrated by distillation (~68%) and dehydration (up to 98%).

Solved Examples

• Ex 1: Write the reaction for laboratory preparation of HNO₃.
• Answer: NaNO₃ + 5H₂SO₄ → NaHSO₄ + HNO₃
• Ex 2: Outline the first step of Ostwald’s process.
• Answer: NH₃ + O₂ → NO + H₂O (with catalyst, high temperature and pressure)

Practice Set

• Q1: Which catalyst is used in Ostwald’s process?
• A. Iron B. Platinum/Rhodium C. Copper D. Nickel
• Q2: What is the product when NaNO₃ reacts with H₂SO₄?
• A. NaHSO₄ and HNO₃ B. Na₂SO₄ and HNO₃ C. NaNO₂ and H₂SO₄ D. NaOH and HNO₃
• Q3: What is the maximum concentration of HNO₃ achievable by distillation before dehydration?
• A. 50% B. 68% C. 85% D. 98%

Answer Key

• Q1: B - Platinum/Rhodium catalyst is used for oxidation of NH₃ in Ostwald’s process.
• Q2: A - Reaction produces NaHSO₄ and HNO₃ in laboratory preparation.
• Q3: B - Distillation concentrates HNO₃ up to ~68%; further dehydration needed for 98%.

Memory Aids

• Ostwald’s catalyst: Platinum/Rhodium = "Pt/Rh"
• Lab prep: NaNO₃ + H₂SO₄ → HNO₃ + NaHSO₄
• Distillation max ~68%, dehydration for 98%

Nitric Acid Properties

Physical Properties

• Pure HNO₃ is colourless; impure appears yellow due to nitrogen oxides.
• Gaseous HNO₃ is planar molecule.
• Molecular structure: bond lengths and angles important for shape.

Chemical Properties

• Strong oxidising agent.
• Non-metals oxidised to highest oxyacids; NO₂ gas evolved.
• Examples:
• I₂ + 10HNO₃ → 2HIO₃ + 10NO₂ + 4H₂O
• C + 4HNO₃ → CO₂ + 2H₂O + NO₂
• S₈ + 48HNO₃ → 8H₂SO₄ + 48NO₂ + 16H₂O
• P₄ + 20HNO₃ → 4H₃PO₄ + 20NO₂ + 4H₂O
• Reacts with metals as acid and oxidising agent.

Solved Examples

• Ex 1: Write the reaction of iodine with hot concentrated HNO₃.
• Answer: I₂ + 10HNO₃ → 2HIO₃ + 10NO₂ + 4H₂O
• Ex 2: What gas is evolved when carbon reacts with HNO₃?
• Answer: NO₂ gas

Practice Set

• Q1: What colour is pure nitric acid?
• A. Yellow B. Colourless C. Blue D. Green
• Q2: Which gas is evolved when non-metals react with hot concentrated HNO₃?
• A. NO B. NO₂ C. N₂ D. NH₃
• Q3: Which product forms when sulfur reacts with hot concentrated nitric acid?
• A. H₂SO₄ B. SO₂ C. H₂SO₃ D. S₈

Answer Key

• Q1: B - Pure HNO₃ is colourless; yellow indicates impurities.
• Q2: B - NO₂ gas evolves during oxidation of non-metals.
• Q3: A - Sulfur oxidised to H₂SO₄ by hot concentrated HNO₃.

Memory Aids

• Pure HNO₃ = Colourless; Yellow = NOx impurities
• Non-metals + hot conc. HNO₃ → highest oxyacids + NO₂
• Common gases: NO₂ evolved in oxidation

Nitric Acid Reactions with Metals

Reaction Patterns

• Acts as acid and strong oxidising agent.
• Reactions depend on nitric acid concentration: very dilute, dilute, concentrated.
• Very dilute HNO₃:
• Mg, Mn → H₂ + metal nitrate
• Fe, Zn, Sn → NH₄NO₃ + metal nitrate
• Dilute HNO₃:
• Pb, Cu, Ag, Hg → NO + metal nitrate
• Fe, Zn → N₂O + metal nitrate
• Sn → NH₄NO₃ + Sn nitrate
• Concentrated HNO₃:
• Zn, Fe, Pb, Cu, Ag → NO₂ + metal nitrate
• Sn → NO₂ + H₂SnO₃ (metastannic acid)

Solved Examples

• Ex 1: What gas is evolved when copper reacts with dilute nitric acid?
• Answer: NO gas
• Ex 2: Write the product when tin reacts with concentrated nitric acid.
• Answer: NO₂ and H₂SnO₃ (metastannic acid)

Practice Set

• Q1: Which gas is evolved when iron reacts with dilute nitric acid?
• A. H₂ B. NO C. N₂O D. NO₂
• Q2: What is the product when tin reacts with concentrated nitric acid?
• A. NH₄NO₃ B. NO₂ + H₂SnO₃ C. NO + Sn nitrate D. N₂O
• Q3: Which metal produces hydrogen gas with very dilute nitric acid?
• A. Fe B. Pb C. Mg D. Cu

Answer Key

• Q1: C - Iron with dilute HNO₃ produces N₂O gas.
• Q2: B - Tin with concentrated HNO₃ forms NO₂ and metastannic acid.
• Q3: C - Magnesium produces H₂ with very dilute nitric acid.

Memory Aids

• Very dilute: Mg, Mn → H₂; Fe, Zn, Sn → NH₄NO₃
• Dilute: Pb, Cu, Ag, Hg → NO; Fe, Zn → N₂O; Sn → NH₄NO₃
• Concentrated: Zn, Fe, Pb, Cu, Ag → NO₂; Sn → NO₂ + H₂SnO₃

Brown Ring Test

Test Explanation

• Detects nitrate ion (NO₃⁻).
• Reaction: NO₃⁻ + 3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O
• Complex formation: [Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ (brown ring)

Solved Examples

• Ex 1: Write the brown ring test reaction.
• Answer: NO₃⁻ + 3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O; [Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺

Practice Set

• Q1: What ion does the brown ring test detect?
• A. NO₂⁻ B. NO₃⁻ C. NH₄⁺ D. NO⁺
• Q2: What is the colour of the brown ring?
• A. Blue B. Green C. Brown D. Yellow
• Q3: Which ion reacts with Fe²⁺ in the brown ring test?
• A. NO₃⁻ B. NO⁺ C. NO₂⁻ D. NH₄⁺

Answer Key

• Q1: B - Brown ring test detects nitrate ion (NO₃⁻).
• Q2: C - The ring formed is brown in colour.
• Q3: A - NO₃⁻ ion reacts with Fe²⁺ to form the complex.

Memory Aids

• Brown ring = Nitrate detection
• Fe²⁺ + NO₃⁻ + H⁺ → brown complex

Nitrous Acid

Basic Facts

• Formula: HNO₂
• Unstable; exists only in solution.

Solved Examples

• Ex 1: State the stability of nitrous acid.
• Answer: Nitrous acid is unstable and exists only in solution.

Practice Set

• Q1: What is the formula of nitrous acid?
• A. HNO₃ B. HNO₂ C. H₂SO₄ D. HCl
• Q2: Nitrous acid is stable in which state?
• A. Solid B. Gas C. Solution D. None
• Q3: Which acid is unstable and exists only in solution?
• A. Nitric acid B. Sulfuric acid C. Nitrous acid D. Hydrochloric acid

Answer Key

• Q1: B - Nitrous acid formula is HNO₂.
• Q2: C - Nitrous acid exists only in solution.
• Q3: C - Nitrous acid is unstable and exists only in solution.

Memory Aids

• HNO₂ = Nitrous acid, unstable, solution only

Phosphorus Allotropic Forms

White Phosphorus

• White phosphorus is a white translucent waxy solid, poisonous and insoluble in water but soluble in CS₂.
• It glows in the dark due to chemiluminescence.
• Consists of discrete tetrahedral P₄ molecules with bond angles of 60°, causing angular strain.
• Highly reactive and less stable due to strain in P₄ molecule.
• Reacts with boiling NaOH in inert atmosphere to produce PH₃ and sodium hypophosphite (NaH₂PO₂):
  P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
• Catches fire in air forming dense white fumes of P₄O₁₀:
  P₄ + 5O₂ → P₄O₁₀

Solved Examples

• Ex 1: Write the reaction of white phosphorus with boiling NaOH.
  Solution: P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

Practice Set

• Q1: What is the molecular shape of white phosphorus?
  A. Linear
  B. Tetrahedral
  C. Octahedral
  D. Planar
• Q2: Which gas is produced when white phosphorus reacts with boiling NaOH?
  A. PH₃
  B. NH₃
  C. H₂S
  D. CO₂
• Q3: What is the product when white phosphorus burns in air?
  A. P₄O₆
  B. P₄O₁₀
  C. P₂O₅
  D. P₂O₃

Answer Key

• Q1: B - White phosphorus has a tetrahedral P₄ molecular structure.
• Q2: A - PH₃ (phosphine) is formed when white phosphorus reacts with boiling NaOH.
• Q3: B - Burning white phosphorus produces P₄O₁₀ fumes.

Memory Aids

• White phosphorus = P₄ tetrahedron, 60° bond angle, reactive due to strain.
• Burns in air → P₄O₁₀ (dense white fumes).
• Reacts with NaOH → PH₃ + sodium hypophosphite.

Red Phosphorus

• Obtained by heating white phosphorus at 573 K in an inert atmosphere for several days.
• Has iron grey lustre, odourless, non-poisonous, insoluble in water and CS₂.
• Less reactive than white phosphorus and does not glow in the dark.
• Polymeric structure consisting of chains of P₄ tetrahedra.
• Heating red phosphorus under high pressure forms black phosphorus phases.

Solved Examples

• Ex 1: How is red phosphorus prepared?
  Solution: By heating white phosphorus at 573 K in an inert atmosphere for several days.

Practice Set

• Q1: What is the colour of red phosphorus?
  A. White
  B. Iron grey
  C. Black
  D. Yellow
• Q2: Which property distinguishes red phosphorus from white phosphorus?
  A. Poisonous nature
  B. Glows in dark
  C. Odourless and non-poisonous
  D. Soluble in water
• Q3: What happens when red phosphorus is heated under high pressure?
  A. Converts to white phosphorus
  B. Converts to black phosphorus phases
  C. Burns to P₄O₁₀
  D. Dissolves in CS₂

Answer Key

• Q1: B - Red phosphorus has an iron grey lustre.
• Q2: C - Red phosphorus is odourless and non-poisonous unlike white phosphorus.
• Q3: B - Heating red phosphorus under high pressure forms black phosphorus phases.

Memory Aids

• Red phosphorus = polymeric chains, less reactive, no glow.
• Heated white P at 573 K → red P.
• High pressure + heat on red P → black P phases.

Black Phosphorus

• Exists in two forms: α-black and β-black phosphorus.
• α-black phosphorus: formed by heating red phosphorus at 803 K in a sealed tube.
• Has opaque monoclinic or rhombohedral crystals, can sublime in air, and does not oxidise in air.
• β-black phosphorus: prepared by heating white phosphorus at 473 K under high pressure.
• Does not burn in air up to 673 K and has a layered structure.

Solved Examples

• Ex 1: How is α-black phosphorus prepared?
  Solution: By heating red phosphorus at 803 K in a sealed tube.
• Ex 2: How is β-black phosphorus prepared?
  Solution: By heating white phosphorus at 473 K under high pressure.

Practice Set

• Q1: Which form of black phosphorus is formed by heating red phosphorus at 803 K?
  A. α-black phosphorus
  B. β-black phosphorus
  C. White phosphorus
  D. Red phosphorus
• Q2: What is a key property of β-black phosphorus?
  A. Burns easily in air
  B. Layered structure and does not burn up to 673 K
  C. Glows in dark
  D. Soluble in water
• Q3: Which statement is true about α-black phosphorus?
  A. It oxidises easily in air
  B. It has rhombohedral crystals and can sublime in air
  C. It is highly reactive
  D. It is soluble in CS₂

Answer Key

• Q1: A - α-black phosphorus is formed by heating red phosphorus at 803 K.
• Q2: B - β-black phosphorus has a layered structure and does not burn up to 673 K.
• Q3: B - α-black phosphorus has rhombohedral crystals and can sublime in air without oxidising.

Memory Aids

• α-black P: from red P at 803 K, rhombohedral, stable in air.
• β-black P: from white P at 473 K + high pressure, layered, non-flammable up to 673 K.

Phosphine

Preparation

• Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
• Ca₃P₂ + 6HCl → 3CaCl₂ + 2PH₃
• PH₄I + KOH → KI + H₂O + PH₃
• P₄ + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃

Properties

• Colourless gas with a rotten fish smell
• Highly poisonous
• Reacts with anhydrous HX (HCl, HBr, HI) but not with aqueous HX
• Phosphonium halides formed with aqueous HX decompose to give back PH₃

Reactions

• On heating (Δ), PH₃ decomposes to red phosphorus (P₄) and hydrogen gas (H₂)
• PH₃ + CuSO₄ → Cu₃P₂ + H₂SO₄
• PH₃ + HBr → PH₄Br (phosphonium bromide)

Solved Examples

• Ex 1: Write the reaction of Ca₃P₂ with water.
  • Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
• Ex 2: What happens when PH₃ is heated?
  • PH₃ → P₄ + H₂ (on heating)

Practice Set

• Q1: Which gas is produced when Ca₃P₂ reacts with water?
  • A. NH₃
  • B. PH₃
  • C. H₂S
  • D. CO₂
• Q2: Phosphine reacts with which of the following to form phosphonium halide?
  • A. Aqueous HCl
  • B. Anhydrous HBr
  • C. Water
  • D. NaOH
• Q3: On heating, phosphine decomposes to form:
  • A. White phosphorus and oxygen
  • B. Red phosphorus and hydrogen
  • C. Phosphorus pentoxide and water
  • D. Phosphoric acid and hydrogen

Answer Key

• Q1: B. PH₃ - Ca₃P₂ reacts with water to produce phosphine gas.
• Q2: B. Anhydrous HBr - Phosphine reacts with anhydrous HX to form phosphonium halides.
• Q3: B. Red phosphorus and hydrogen - PH₃ decomposes on heating to red phosphorus and hydrogen gas.

Memory Aids

• PH₃ smells like rotten fish and is highly poisonous.
• Phosphine reacts only with anhydrous HX, not aqueous HX.
• Heating PH₃ gives red phosphorus + H₂.

Phosphorus Halides

Chemical Structures

• PCl₃: Phosphorus bonded to three chlorine atoms, trigonal pyramidal shape.
• PCl₅: Phosphorus bonded to five chlorine atoms, trigonal bipyramidal shape.

Preparation

• PCl₃: Prepared by direct reaction of white phosphorus (P₄) with chlorine gas (Cl₂) in limited amount.
  • P₄ + 6Cl₂ → 4PCl₃
• PCl₅: Prepared by chlorination of PCl₃ or direct reaction of P₄ with excess chlorine.
  • PCl₃ + Cl₂ → PCl₅
  • P₄ + 10Cl₂ → 4PCl₅

Chemical Properties

• Hydrolysis:
  • PCl₃ + 3H₂O → H₃PO₃ + 3HCl
  • PCl₅ + 4H₂O → H₃PO₄ + 5HCl
• Reaction with SO₂:
  • PCl₅ + SO₂ → POCl₃ + SOCl₂
• Reaction with Zn:
  • PCl₅ + 2Zn → 2ZnCl₂ + P
• Organic reactions:
  • PCl₃ used to prepare organophosphorus compounds.

Uses

• PCl₅: Chlorinating and dehydrating agent.
• PCl₃: Used in organic synthesis and preparation of other phosphorus compounds.

Solved Examples

Ex 1: Write the reaction for preparation of PCl₃ from phosphorus and chlorine.

• Step 1: React P₄ with Cl₂ in limited amount.
• Step 2: P₄ + 6Cl₂ → 4PCl₃

Ex 2: Show hydrolysis of PCl₅.

• Step 1: Add water to PCl₅.
• Step 2: PCl₅ + 4H₂O → H₃PO₄ + 5HCl

Practice Set

• Q1: What is the shape of PCl₅ molecule?
  • A. Tetrahedral
  • B. Trigonal bipyramidal
  • C. Octahedral
  • D. Trigonal pyramidal
• Q2: Which reagent is used to prepare PCl₃ from white phosphorus?
  • A. Cl₂ (limited)
  • B. Cl₂ (excess)
  • C. SO₂
  • D. Zn
• Q3: What is the product when PCl₅ reacts with SO₂?
  • A. H₃PO₄
  • B. POCl₃ and SOCl₂
  • C. PCl₃ and Cl₂
  • D. ZnCl₂

Answer Key

• Q1: B - PCl₅ has trigonal bipyramidal shape due to five bonded chlorine atoms.
• Q2: A - Limited Cl₂ reacts with P₄ to form PCl₃.
• Q3: B - PCl₅ reacts with SO₂ to give POCl₃ and SOCl₂.

Memory Aids

• PCl₃ = 3 Cl atoms, trigonal pyramidal shape.
• PCl₅ = 5 Cl atoms, trigonal bipyramidal shape.
• PCl₃ + Cl₂ → PCl₅ (chlorination step).
• PCl₅ hydrolyzes to H₃PO₄, PCl₃ hydrolyzes to H₃PO₃.

Oxoacids of Phosphorus

Overview

• Oxoacids of phosphorus contain phosphorus, oxygen, and hydrogen.
• Phosphorus exhibits different oxidation states in these acids: +3 and +5.
• Common oxoacids include phosphorous acid (H₃PO₃), phosphoric acid (H₃PO₄), and others.
• Structures vary: some have P=O bonds, others have P–OH groups.
• Preparation methods include hydrolysis of phosphorus halides and oxidation of phosphorus compounds.

Solved Examples

• Ex 1: Write the formula and oxidation state of phosphorus in phosphorous acid.
  • Formula: H₃PO₃
  • Oxidation state of P: +3
• Ex 2: How is phosphoric acid prepared from P₄O₁₀?
  • P₄O₁₀ + 6H₂O → 4H₃PO₄

Practice Set

• Q1: What is the oxidation state of phosphorus in H₃PO₄?
  • A. +3
  • B. +4
  • C. +5
  • D. +2
• Q2: Which acid is prepared by hydrolysis of PCl₃?
  • A. H₃PO₄
  • B. H₃PO₃
  • C. H₃PO₂
  • D. H₄P₂O₇
• Q3: Which of the following has phosphorus in +3 oxidation state?
  • A. H₃PO₄
  • B. H₃PO₃
  • C. H₃PO₂
  • D. H₄P₂O₇

Answer Key

• Q1: C. +5 — Phosphorus in phosphoric acid is in +5 oxidation state.
• Q2: B. H₃PO₃ — Hydrolysis of PCl₃ gives phosphorous acid.
• Q3: B. H₃PO₃ — Phosphorous acid has P in +3 oxidation state.

Memory Aids

• Phosphorous acid (H₃PO₃) has P in +3 state; phosphoric acid (H₃PO₄) has P in +5 state.
• Hydrolysis of PCl₃ → H₃PO₃; PCl₅ → H₃PO₄.
• Remember: "-ous" acids have lower oxidation state than "-ic" acids.

Group 16 Elements

Overview

• Elements: Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), Polonium (Po), Livermorium (Lv).
• Also called chalcogens.
• Livermorium is synthetic, radioactive, very short half-life, limited study.
• General electronic configuration: ns²np⁴.

Physical Properties

• Atomic and ionic radii increase down the group.
• Ionization enthalpy decreases down the group; lower than group 15 due to half-filled p-subshell stability in group 15.
• Electronegativity decreases down the group.
• Electron gain enthalpy: O less negative than S due to small size; from S to Po, becomes less negative.
• Melting and boiling points increase down to Te; Po has lower m.pt. and b.pt. than Te.
• Elemental states: O₂ is diatomic (pπ-pπ multiple bonding); S, Se, Te are octa-atomic due to absence of pπ-pπ bonding.
• All elements exhibit allotropy.

Chemical Properties

• Oxidation states: O shows -2, -1, -1/2, 0, +1, +2; others show -2, +2, +4, +6.
• Stability of -2 state decreases down the group (decreasing electronegativity).
• +4 state stability increases; +6 state stability decreases due to inert pair effect.

Hydrides

• General formula: H₂E.
• Physical state: H₂O is liquid; others are gases.
• Boiling points: H₂S < H₂Se < H₂Te < H₂O.
• Bond angles: H₂O > H₂S > H₂Se > H₂Te.
• Reducing power: H₂Te > H₂Se > H₂S > H₂O.
• Acidic character: H₂O < H₂S < H₂Se < H₂Te.
• Thermal stability: H₂O > H₂S > H₂Se > H₂Te.

Oxides

• Form EO₂ and EO₃ types.
• Both types are acidic.
• Reducing property decreases from SO₂ to TeO₂.

Halides

• Form EX₆, EX₄, and EX₂ types.
• Halide stability order: F^- > Cl^- > Br^- > I^-.
• Only hexafluorides are stable with octahedral geometry.
• Tetrafluorides: SF₄ (gas), SeF₄ (liquid), TeF₄ (solid); trigonal bipyramidal structure.
• Dihalides have tetrahedral structure.
• Monohalides exist as dimers.

Solved Examples

• Ex 1: Write electronic configuration of Se.
  Solution: [Ar] 3d¹⁰ 4s² 4p⁴.
• Ex 2: Arrange H₂S, H₂O, H₂Se in order of increasing boiling point.
  Solution: H₂S < H₂Se < H₂O.
• Ex 3: Identify oxidation states of oxygen in H₂O₂ and O₂.
  Solution: H₂O₂: -1; O₂: 0.

Practice Set

• Q1: Which element is not a chalcogen?
  A. Oxygen
  B. Sulphur
  C. Nitrogen
  D. Selenium
• Q2: Which hydride has the highest boiling point?
  A. H₂S
  B. H₂Se
  C. H₂Te
  D. H₂O
• Q3: Which halide of group 16 elements is most stable?
  A. EX₂
  B. EX₄
  C. EX₆ (hexafluoride)
  D. Monohalide

Answer Key

• Q1: C. Nitrogen is in group 15, not group 16.
• Q2: D. H₂O has highest boiling point due to hydrogen bonding.
• Q3: C. Hexafluorides (EX₆) are most stable with octahedral structure.

Memory Aids

• Chalcogens: O, S, Se, Te, Po, Lv (Livermorium synthetic).
• Hydrides boiling point: H₂S < H₂Se < H₂Te < H₂O.
• Halide stability: F^- > Cl^- > Br^- > I^-.
• Oxidation states: O (-2 to +2), others (-2, +2, +4, +6).

Dioxygen Preparation

Methods of Preparation

• Heating oxo salts like chlorates, nitrates, and permanganates in the lab.
• Example: Decomposition of potassium chlorate (KClO₃) with MnO₂ catalyst:
  • KClO₃ 123 K 12B MnO₂ 12B heat 12B KCl + O₂
  • MnO₂ lowers melting point and speeds reaction (ratio 1:4 with KClO₃).
  • MnO₂ must be carbon-free to avoid explosion with KClO₃.
• Thermal decomposition of heavy metal oxides:
  • Metal oxides decompose on heating to metal + O₂ (e.g., HgO, Ag₂O, PbO₂).
• Decomposition of hydrogen peroxide (H₂O₂) using MnO₂ catalyst:
  • 2H₂O₂ 12B MnO₂ 12B 2H₂O + O₂
• Large scale: Electrolysis of water:
  • 2H₂O 12B 2H⁺ + 2OH^-
  • Cathode: 2H⁺ + 2e^- 12B H₂ (byproduct)
  • Anode: 4OH^- 12B 2H₂O + O₂ + 4e^-

Solved Examples

• Ex 1: Write the reaction for oxygen preparation from potassium chlorate.
  • KClO₃ 12B MnO₂ 12B heat 12B KCl + O₂
• Ex 2: Write the reaction for oxygen preparation by decomposition of mercury oxide.
  • 2HgO 12B heat 12B 2Hg + O₂
• Ex 3: Write the reaction for oxygen preparation from hydrogen peroxide using MnO₂.
  • 2H₂O₂ 12B MnO₂ 12B 2H₂O + O₂

Practice Set

• Q1: Which catalyst is used in the decomposition of potassium chlorate to produce oxygen?
  • A. MnO₂
  • B. Fe₂O₃
  • C. CuO
  • D. ZnO
• Q2: What is the product formed when mercury oxide decomposes on heating?
  • A. HgO and O₂
  • B. Hg and O₂
  • C. Hg and H₂O
  • D. HgO and H₂O
• Q3: In the electrolysis of water, what gas is produced at the cathode?
  • A. O₂
  • B. H₂
  • C. CO₂
  • D. Cl₂

Answer Key

• Q1: A. MnO₂ - It acts as a catalyst in KClO₃ decomposition.
• Q2: B. Hg and O₂ - Mercury oxide decomposes to mercury metal and oxygen gas.
• Q3: B. H₂ - Hydrogen gas is produced at the cathode during water electrolysis.

Memory Aids

• MnO₂ lowers KClO₃ melting point and speeds oxygen release.
• Heavy metal oxides decompose to metal + O₂ on heating.
• H₂O₂ decomposes faster with MnO₂ catalyst.
• Electrolysis: Cathode produces H₂, Anode produces O₂.

Dioxygen Physical Properties

Key Physical Properties

• Colourless, odourless, tasteless gas.
• Paramagnetic due to 2 unpaired electrons in antibonding molecular orbitals.

Solved Examples

• Ex 1: Why is O₂ paramagnetic?
  • Because it has 2 unpaired electrons in antibonding orbitals.

Practice Set

• Q1: What is the colour of dioxygen gas?
  • A. Colourless
  • B. Blue
  • C. Green
  • D. Yellow
• Q2: Why is dioxygen paramagnetic?
  • A. No unpaired electrons
  • B. 2 unpaired electrons in antibonding orbitals
  • C. Paired electrons only
  • D. It is diamagnetic
• Q3: Which property is NOT true for dioxygen?
  • A. Odourless
  • B. Tasteless
  • C. Colored
  • D. Paramagnetic

Answer Key

• Q1: A. Colourless - Dioxygen is colourless gas.
• Q2: B. 2 unpaired electrons in antibonding orbitals - This causes paramagnetism.
• Q3: C. Colored - Dioxygen is colourless, so this is false.

Memory Aids

• O₂ is paramagnetic due to 2 unpaired electrons.
• Colourless, odourless, tasteless gas.

Dioxygen Chemical Properties

Reactions and Behavior

• Reacts with all metals except noble metals (Au, Pt).
• Reacts with non-metals and some noble gases.
• Forms oxides:
  • Ca + O₂ 12B CaO (basic oxide)
  • Al + O₂ 12B Al₂O₃ (amphoteric oxide)
  • P + O₂ 12B P₄O₁₀ (acidic oxide)
  • C + O₂ 12B CO₂ (acidic oxide)
• Roasting of zinc sulfide:
  • 2ZnS + 3O₂ 12B 2ZnO + 2SO₂
• Combustion of methane:
  • CH₄ + 2O₂ 12B CO₂ + 2H₂O
• Catalytic oxidation:
  • 2SO₂ + O₂ 12B 2SO₃ (V₂O₅ catalyst)
  • 4HCl + O₂ 12B 2Cl₂ + 2H₂O (CuCl₂ catalyst)

Solved Examples

• Ex 1: Write the reaction of calcium with oxygen.
  • 2Ca + O₂ 12B 2CaO
• Ex 2: Write the catalytic oxidation of sulfur dioxide.
  • 2SO₂ + O₂ 12B 2SO₃ (V₂O₅ catalyst)

Practice Set

• Q1: Which metal does NOT react with oxygen?
  • A. Calcium
  • B. Gold
  • C. Aluminum
  • D. Zinc
• Q2: What type of oxide is Al₂O₃?
  • A. Basic
  • B. Acidic
  • C. Amphoteric
  • D. Neutral
• Q3: Which catalyst is used in the oxidation of SO₂ to SO₃?
  • A. CuCl₂
  • B. V₂O₅
  • C. MnO₂
  • D. Fe₂O₃

Answer Key

• Q1: B. Gold - Noble metals like Au do not react with oxygen.
• Q2: C. Amphoteric - Al₂O₃ reacts with acids and bases.
• Q3: B. V₂O₅ - Catalyst for SO₂ oxidation to SO₃.

Memory Aids

• Oxygen reacts with all metals except Au, Pt.
• Al₂O₃ is amphoteric oxide.
• V₂O₅ catalyzes SO₂ oxidation.

Simple Oxides

Classification of Oxides (Based on Chemical Behaviour)

• Oxides are compounds of oxygen with another element.
• Simple oxides contain one metal + oxygen (e.g., MgO).
• Mixed oxides contain two or more metals + oxygen (e.g., Pb₃O₄).
• Acidic oxides react with water to form acids (e.g., SO₂).
• Basic oxides react with acids to form salts (e.g., Na₂O).
• Amphoteric oxides react with both acids and bases (e.g., Al₂O₃).
• Neutral oxides do not react with acids or bases (e.g., CO).

Acidic Oxides

• Dissolve in water forming acids.
• Examples:
  • P₄O₁₀ + 6H₂O → 4H₃PO₄
  • N₂O₅ + H₂O → 2HNO₃
  • 2NO₂ + H₂O → HNO₂ + HNO₃
  • P₄O₆ + 6H₂O → 4H₃PO₃
• Acidic strength order:
  • Dioxides: SO₂ > SeO₂ > TeO₂ > PoO₂
  • Trioxides: SO₃ > SeO₃ > TeO₃

Basic Oxides

• Dissolve in water forming bases.
• Examples:
  • Na₂O + H₂O → 2NaOH
  • K₂O + H₂O → 2KOH
  • CaO + H₂O → Ca(OH)₂

Amphoteric Oxides

• React with both acids and bases.
• Examples:
  • Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
  • Al₂O₃ + 6NaOH + 3H₂O → 2Na₃[Al(OH)₆]
  • Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (Sodium aluminate)

Solved Examples

• Ex 1: Write the reaction of P₄O₁₀ with water.
  • P₄O₁₀ + 6H₂O → 4H₃PO₄
• Ex 2: Reaction of Al₂O₃ with NaOH.
  • Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Practice Set

• Q1: Which oxide is amphoteric?
  • A. SO₂
  • B. Al₂O₃
  • C. Na₂O
  • D. CO
• Q2: What is the product when Na₂O reacts with water?
  • A. NaOH
  • B. NaCl
  • C. Na₂CO₃
  • D. NaAlO₂
• Q3: Which oxide forms HNO₃ on reaction with water?
  • A. N₂O₅
  • B. CO
  • C. CaO
  • D. Pb₃O₄

Answer Key

• Q1: B - Al₂O₃ is amphoteric, reacting with both acids and bases.
• Q2: A - Na₂O + H₂O forms NaOH, a base.
• Q3: A - N₂O₅ reacts with water to form nitric acid (HNO₃).

Memory Aids

• Acidic oxides form acids with water; basic oxides form bases.
• Amphoteric oxides react with both acids and bases (e.g., Al₂O₃).
• Order of acidic strength: SO₂ > SeO₂ > TeO₂.
• Remember sodium aluminate: Al₂O₃ + NaOH → NaAlO₂.

Ozone

Preparation

• Ozone (O₃) is prepared by electric discharge through oxygen gas (O₂).
• Electric discharge breaks O₂ molecules into oxygen atoms.
• Oxygen atoms recombine to form ozone molecules.
• Chemical equation: 3O₂ \xrightarrow{electric discharge} 2O₃

Properties

• Ozone is pale blue gas, dark blue liquid, and violet-black solid.
• Decomposes to oxygen: O₃ \rightarrow O₂ + O
• Acts as a strong oxidizing agent: PbS + 4O₃ \rightarrow PbSO₄ + 4O₂
• Acts as a reducing agent: 2I^- + H₂O + O₃ \rightarrow 2OH^- + I₂ + O₂
• Used for estimation of ozone by titrating I₂ with sodium thiosulphate.
• Reacts with nitric oxide: NO + O₃ \rightarrow NO₂ + O₂

Structure

• Central oxygen atom in ozone has sp² hybridisation.
• Ozone molecule has an angular shape.
• Ozone is a resonance hybrid of two structures with equal O–O bond lengths (128 pm).

Solved Examples

• Ex 1: Write the reaction for ozone preparation by electric discharge.
  • 3O₂ \xrightarrow{electric discharge} 2O₃
• Ex 2: Show ozone acting as an oxidizing agent with PbS.
  • PbS + 4O₃ \rightarrow PbSO₄ + 4O₂

Practice Set

• Q1: What is the hybridisation of the central oxygen atom in ozone?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d
• Q2: Which of the following reactions shows ozone acting as a reducing agent?
  • A. PbS + 4O₃ \rightarrow PbSO₄ + 4O₂
  • B. 2I^- + H₂O + O₃ \rightarrow 2OH^- + I₂ + O₂
  • C. NO + O₃ \rightarrow NO₂ + O₂
  • D. O₃ \rightarrow O₂ + O
• Q3: How many ozone molecules are formed when 3 molecules of oxygen gas undergo electric discharge?
  • A. 1
  • B. 2
  • C. 3
  • D. 6

Answer Key

• Q1: B. sp² — Central oxygen in ozone is sp² hybridised.
• Q2: B. 2I^- + H₂O + O₃ \rightarrow 2OH^- + I₂ + O₂ — Ozone acts as reducing agent here.
• Q3: B. 2 — 3 O₂ molecules produce 2 O₃ molecules by electric discharge.

Memory Aids

• Ozone preparation: "3 O₂ to 2 O₃" by electric discharge.
• Ozone is both oxidizing and reducing agent.
• Ozone structure: Resonance hybrid with equal bond lengths, angular shape, sp² hybridisation.

Sulphur Allotropic Forms

Rhombic Sulphur (α-sulphur)

• Consists of S₈ molecules, yellow in color.
• Melting point: 385.8 K; Specific gravity: 2.06.
• Crystals formed by evaporating solution of roll sulphur in CS₂.
• Insoluble in water; slightly soluble in benzene, alcohol, and ether.
• Readily soluble in CS₂.

Monoclinic Sulphur (β-sulphur)

• Also consists of S₈ molecules, colorless needle-shaped crystals.
• Melting point: 393 K; Specific gravity: 1.98.
• Soluble in CS₂.

Structural Details

• S₈ ring in both rhombic and monoclinic sulphur is puckered and crown-shaped.
• Cyclo S₆ ring has a chair form.

Hydrogen Sulphide (H₂S)

• Also called sulphurated hydrogen.
• Poisonous and fatal in large amounts.
• Antidote: Dilute chlorine oxidizes H₂S to sulphur.
• Reaction: H₂S + Cl₂ → 2HCl + S

Solved Examples

• Ex 1: Identify the allotrope of sulphur with melting point 385.8 K.
  Answer: Rhombic sulphur (α-sulphur).
• Ex 2: Write the reaction of H₂S with chlorine.
  Answer: H₂S + Cl₂ → 2HCl + S.

Practice Set

• Q1: Which allotrope of sulphur has yellow color and S₈ molecules?
  A. Monoclinic sulphur
  B. Rhombic sulphur
  C. Cyclo S₆
  D. None of these
• Q2: What is the melting point of monoclinic sulphur?
  A. 385.8 K
  B. 393 K
  C. 400 K
  D. 350 K
• Q3: What is the product formed when H₂S reacts with chlorine?
  A. HCl and S
  B. H₂SO₄
  C. SO₂ and H₂O
  D. H₂SO₃

Answer Key

• Q1: B - Rhombic sulphur is yellow and contains S₈ molecules.
• Q2: B - Monoclinic sulphur melts at 393 K.
• Q3: A - H₂S reacts with Cl₂ to form HCl and sulphur.

Memory Aids

• Rhombic sulphur = Yellow, S₈, melts at 385.8 K.
• Monoclinic sulphur = Colorless needles, melts at 393 K.
• H₂S + Cl₂ → 2HCl + S (Dilute chlorine antidote reaction).

Sulphur Dioxide

Preparation

• Direct combination: S + O₂ → SO₂
• Lab method: SO₃^2- + 2H⁺ → H₂O + SO₂
• Roasting of sulphide ores: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Physical Properties

• Colourless gas with pungent smell
• Highly soluble in water
• Liquefies at room temperature at 2 atm pressure
• Boiling point: 263 K

Chemical Properties

• Reacts with water to form sulfurous acid: SO₂ + H₂O → H₂SO₃
• Reacts with NaOH to form sodium sulfite: SO₂ + 2NaOH → Na₂SO₃ + H₂O
• Sodium sulfite reacts with SO₂ and water to form sodium bisulfite: Na₂SO₃ + SO₂ + H₂O → 2NaHSO₃
• Reacts with chlorine to form sulfuryl chloride: SO₂ + Cl₂ → SO₂Cl₂
• Oxidation to sulfur trioxide in presence of V₂O₅ catalyst: 2SO₂ + O₂ → 2SO₃

Bleaching Action

• SO₂ + 2H₂O → H₂SO₄ + 2[H]
• Coloured matter + [H] → Colourless matter (reduction of dyes)

Reducing Nature

• 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₃^2- + 4H⁺
• 5SO₂ + 2MnO₄^- + 2H₂O → 5SO₄^2- + 4H⁺ + 2Mn²⁺

Structure

• SO₂ molecule is angular (bent shape)
• Resonance hybrid of two canonical forms with alternating double and single bonds
• Lone pairs on sulfur and oxygen atoms contribute to resonance

Solved Examples

• Ex 1: Write the reaction for lab preparation of SO₂.
  Solution: SO₃^2- + 2H⁺ → H₂O + SO₂
• Ex 2: Show the reaction of SO₂ with NaOH.
  Solution: SO₂ + 2NaOH → Na₂SO₃ + H₂O

Practice Set

• Q1: Which of the following is a method to prepare SO₂?
  A. S + O₂ → SO₂
  B. H₂ + O₂ → H₂O
  C. CO + O₂ → CO₂
  D. N₂ + O₂ → NO
• Q2: SO₂ reacts with NaOH to form:
  A. NaHSO₃
  B. Na₂SO₃
  C. Na₂SO₄
  D. Na₂SO₂
• Q3: Which catalyst is used for oxidation of SO₂ to SO₃?
  A. V₂O₅
  B. Fe₂O₃
  C. MnO₂
  D. Al₂O₃

Answer Key

• Q1: A - Direct combination of sulfur and oxygen forms SO₂.
• Q2: B - SO₂ reacts with NaOH to form sodium sulfite (Na₂SO₃).
• Q3: A - Vanadium pentoxide (V₂O₅) is the catalyst for SO₂ oxidation.

Memory Aids

• SO₂ preparation: S + O₂ direct combination
• Bleaching action due to reduction by SO₂
• SO₂ is a reducing agent in acidic medium
• Structure: Angular molecule with resonance forms

Oxoacids of Sulphur

Overview

• Oxoacids contain sulfur (S) bonded to oxygen (O) and hydrogen (H).
• Four main oxoacids: Sulphurous acid (H₂SO₃), Sulphuric acid (H₂SO₄), Peroxodisulphuric acid (H₂S₂O₈), Pyrosulphuric acid (Oleum, H₂S₂O₇).
• Sulphurous acid: S bonded to 3 O and 2 H atoms.
• Sulphuric acid: S bonded to 4 O and 2 H atoms.
• Peroxodisulphuric acid: Two S atoms connected by O-O bond, with extra O and H.
• Pyrosulphuric acid: Two S atoms connected by one O atom, plus more O and H.

Solved Examples

• Ex 1: Identify the number of oxygen atoms in H₂SO₄.
  • Count oxygen atoms bonded to S: 4
• Ex 2: Write the formula for Peroxodisulphuric acid.
  • Two S atoms, O-O bond, total 8 O and 2 H: H₂S₂O₈

Practice Set

• Q1: Which acid has sulfur bonded to three oxygen atoms?
  • A. Sulphuric acid
  • B. Sulphurous acid
  • C. Peroxodisulphuric acid
  • D. Pyrosulphuric acid
• Q2: What type of bond connects the two sulfur atoms in Peroxodisulphuric acid?
  • A. Single S-S bond
  • B. Oxygen-oxygen (O-O) bond
  • C. Double S=O bond
  • D. Hydrogen bond
• Q3: Which acid is also known as Oleum?
  • A. Sulphurous acid
  • B. Sulphuric acid
  • C. Pyrosulphuric acid
  • D. Peroxodisulphuric acid

Answer Key

• Q1: B - Sulphurous acid has sulfur bonded to three oxygen atoms.
• Q2: B - Peroxodisulphuric acid has two sulfur atoms connected by an oxygen-oxygen bond.
• Q3: C - Pyrosulphuric acid is also called Oleum.

Memory Aids

• Remember: Sulphurous acid = fewer oxygens (3), Sulphuric acid = more oxygens (4).
• Peroxodisulphuric acid has O-O bond linking two S atoms.
• Oleum = Pyrosulphuric acid with two S atoms linked by one O atom.

Sulphuric Acid

Overview

• Also called oil of vitriol and king of chemicals.
• Strong mineral acid widely used in industry.

Preparation by Contact Process

• Step 1: Sulfur (S) burns in oxygen (O₂) forming sulfur dioxide (SO₂).
• Step 2: SO₂ oxidized to sulfur trioxide (SO₃) using V₂O₅ catalyst.
• Step 3: SO₃ reacts with H₂SO₄ to form oleum (H₂S₂O₇).
• Step 4: Oleum mixed with water to produce concentrated H₂SO₄.

Physical Properties

• Colorless, dense, oily liquid.
• Specific gravity ~1.8 at 298 K.
• Dissolution in water is highly exothermic; always add acid to water with stirring.

Chemical Properties

• Strong diprotic acid: loses two protons stepwise forming H₃O⁺, HSO₄⁻, and SO₄²⁻.
• Reacts with bases like NaOH forming sodium bisulfate or sodium sulfate depending on base amount.
• Reacts with metal halides (MX) producing hydrogen halides (HX) and metal sulfates (M₂SO₄).
• Strong dehydrating agent: removes water from sugars (C₁₂H₂₂O₁₁) leaving carbon.
• Reacts with metals like Cu producing CuSO₄, SO₂, and H₂O.
• Reacts with sulfur producing SO₂ and H₂O.
• Reacts with carbon producing CO₂, SO₂, and H₂O.

Solved Examples

• Ex 1: Write the reaction for SO₂ oxidation in contact process.
  Answer: SO₂ + 1/2O₂ → SO₃
• Ex 2: Reaction of H₂SO₄ with NaOH (1 mole).
  Answer: H₂SO₄ + NaOH → NaHSO₄ + H₂O
• Ex 3: Reaction of concentrated H₂SO₄ with sugar.
  Answer: C₁₂H₂₂O₁₁ → 12C + 11H₂O

Practice Set

• Q1: What is the catalyst used in the contact process?
  A. Fe₂O₃ B. V₂O₅ C. MnO₂ D. CuO
• Q2: Which product forms when SO₃ reacts with H₂SO₄ in contact process?
  A. Sulfuric acid B. Oleum C. Sulfur dioxide D. Sulfur trioxide
• Q3: What happens when concentrated H₂SO₄ reacts with sugar?
  A. Forms CO₂ B. Forms carbon and water C. Forms sulfur dioxide D. Forms copper sulfate

Answer Key

• Q1: B - V₂O₅ is the catalyst for SO₂ oxidation in contact process.
• Q2: B - SO₃ reacts with H₂SO₄ to form oleum (H₂S₂O₇).
• Q3: B - Concentrated H₂SO₄ dehydrates sugar forming carbon and water.

Memory Aids

• Contact Process Steps: S → SO₂ → SO₃ → Oleum → H₂SO₄.
• Always add acid to water to avoid heat splashes.
• H₂SO₄ is diprotic: loses 2 protons stepwise.
• Dehydration of sugar by conc. H₂SO₄ leaves black carbon.

Overview

• Also called oil of vitriol and king of chemicals.
• Strong mineral acid widely used in industry.

Preparation by Contact Process

• Step 1: Sulfur (S) burns in oxygen (O₂) forming sulfur dioxide (SO₂).
• Step 2: SO₂ oxidized to sulfur trioxide (SO₃) using V₂O₅ catalyst.
• Step 3: SO₃ reacts with H₂SO₄ to form oleum (H₂S₂O₇).
• Step 4: Oleum mixed with water to produce concentrated H₂SO₄.

Physical Properties

• Colorless, dense, oily liquid.
• Specific gravity ~1.8 at 298 K.
• Dissolution in water is highly exothermic; always add acid to water with stirring.

Chemical Properties

• Strong diprotic acid: loses two protons stepwise forming H₃O⁺, HSO₄⁻, and SO₄²⁻.
• Reacts with bases like NaOH forming sodium bisulfate or sodium sulfate depending on base amount.
• Reacts with metal halides (MX) producing hydrogen halides (HX) and metal sulfates (M₂SO₄).
• Strong dehydrating agent: removes water from sugars (C₁₂H₂₂O₁₁) leaving carbon.
• Reacts with metals like Cu producing CuSO₄, SO₂, and H₂O.
• Reacts with sulfur producing SO₂ and H₂O.
• Reacts with carbon producing CO₂, SO₂, and H₂O.

Solved Examples

• Ex 1: Write the reaction for SO₂ oxidation in contact process.
  Answer: SO₂ + \(\frac{1}{2}\)O₂ \(\xrightarrow{V_2O_5}\) SO₃
• Ex 2: Reaction of H₂SO₄ with NaOH (1 mole).
  Answer: H₂SO₄ + NaOH \(\rightarrow\) NaHSO₄ + H₂O
• Ex 3: Reaction of concentrated H₂SO₄ with sugar.
  Answer: C₁₂H₂₂O₁₁ \(\xrightarrow{conc. H_2SO_4}\) 12C + 11H₂O

Practice Set

• Q1: What is the catalyst used in the contact process?
  A. Fe₂O₃ B. V₂O₅ C. MnO₂ D. CuO
• Q2: Which product forms when SO₃ reacts with H₂SO₄ in contact process?
  A. Sulfuric acid B. Oleum C. Sulfur dioxide D. Sulfur trioxide
• Q3: What happens when concentrated H₂SO₄ reacts with sugar?
  A. Forms CO₂ B. Forms carbon and water C. Forms sulfur dioxide D. Forms copper sulfate

Answer Key

• Q1: B - V₂O₅ is the catalyst for SO₂ oxidation in contact process.
• Q2: B - SO₃ reacts with H₂SO₄ to form oleum (H₂S₂O₇).
• Q3: B - Concentrated H₂SO₄ dehydrates sugar forming carbon and water.

Memory Aids

• Contact Process Steps: S → SO₂ → SO₃ → Oleum → H₂SO₄.
• Always add acid to water to avoid heat splashes.
• H₂SO₄ is diprotic: loses 2 protons stepwise.
• Dehydration of sugar by conc. H₂SO₄ leaves black carbon.

Group 17 Elements

General Information

• Members: Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), Tennessine (Ts).
• Known as halogens (Greek: halo = salt, genes = born).
• Highly reactive non-metallic elements.
• Astatine and Tennessine are radioactive.
• General electronic configuration: ns²np⁵.

Physical Properties

• Atomic and ionic radii increase down the group; halogens are smallest atoms in a period.
• Ionisation enthalpy is very high but decreases down the group.
• Electronegativity: Fluorine is the most electronegative; decreases down the group.
• Electron gain enthalpy increases from F to Cl, then decreases to I; F has smaller value due to small size and repulsions.
• Bond dissociation enthalpy increases from F₂ to Cl₂, then decreases for Br₂ and I₂; F–F bond weaker due to lone pair repulsions.
• Melting and boiling points increase down the group.

Chemical Properties

• Oxidation states: F only -1; others show -1, +1, +3, +5, +7.
• Higher oxidation states of Cl, Br, I seen in fluorides, oxides, oxoacids.
• Oxidising power decreases down the group: F₂ > Cl₂ > Br₂ > I₂.

Hydrogen Halides

• Reactivity order to form HX: F₂ > Cl₂ > Br₂ > I₂.
• Boiling points: HF > HI > HBr > HCl.
• Melting points: HI > HF > HBr > HCl.
• Bond lengths: HI > HBr > HCl > HF.
• Bond dissociation enthalpy: HF > HCl > HBr > HI.
• Acidic strength: HI > HBr > HCl > HF.
• Thermal stability: HF > HCl > HBr > HI.
• Reducing power: HI > HBr > HCl > HF.

Oxides

• Fluorine forms OF₂ and O₂F₂ (oxygen fluorides).
• Other halogens form oxides with oxidation states from +1 to +7.

Metal Halides

• Ionic character order: MF > MCl > MBr > MI.
• Halides in higher oxidation states are more covalent than in lower states.
• Iodine does not displace Cl or Br from their halides but displaces them from their oxysalts.
• Example reaction: 2KCO₃ + I₂ → 2KIO₃ + Cl₂.

Solved Examples

Ex 1: Write the electronic configuration of Bromine.
Step 1: Bromine atomic number = 35.
Step 2: Configuration = [Ar] 3d¹⁰ 4s² 4p⁵.

Ex 2: Arrange halogens in order of oxidising power.
Answer: F₂ > Cl₂ > Br₂ > I₂.

Practice Set

• Q1: Which halogen has the highest electronegativity?
  A. Chlorine
  B. Fluorine
  C. Bromine
  D. Iodine
• Q2: Which hydrogen halide has the highest boiling point?
  A. HCl
  B. HI
  C. HF
  D. HBr
• Q3: What is the oxidation state of fluorine in its compounds?
  A. +1
  B. -1
  C. 0
  D. +7

Answer Key

• Q1: B. Fluorine is the most electronegative element in group 17.
• Q2: C. HF has the highest boiling point due to strong hydrogen bonding.
• Q3: B. Fluorine shows only -1 oxidation state due to its high electronegativity.

Memory Aids

• Halogen order by oxidising power: F > Cl > Br > I.
• Boiling point order of HX: HF > HI > HBr > HCl.
• Fluorine always -1 oxidation state.
• Bond dissociation enthalpy: F₂ < Cl₂ due to lone pair repulsions.

Chlorine Preparation

Industrial Methods

• MnO₂ or KMnO₄ + HCl → MnCl₂ + H₂O + Cl₂
• NaCl + H₂SO₄ + MnO₂ → Cl₂ + other products
• KMnO₄ + HCl → KCl + MnCl₂ + H₂O + Cl₂
• Deacon’s process: O₂ + HCl (CuCl₂ catalyst) → Cl₂ + H₂O

Electrolytic Process

• Electrolysis of brine (concentrated NaCl solution)
• Chlorine gas is liberated at the anode

Solved Examples

• Ex 1: Write the reaction of MnO₂ with HCl to produce chlorine.
  Answer: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Practice Set

• Q1: Which catalyst is used in Deacon’s process?
  A. CuCl₂
  B. FeCl₃
  C. MnO₂
  D. KMnO₄
• Q2: Chlorine is obtained at which electrode during electrolysis of brine?
  A. Cathode
  B. Anode
  C. Both electrodes
  D. None
• Q3: Which of the following is NOT a method to prepare chlorine?
  A. Reaction of KMnO₄ with HCl
  B. Electrolysis of brine
  C. Reaction of NaCl with H₂SO₄ and MnO₂
  D. Reaction of NaCl with NaOH

Answer Key

• Q1: A - CuCl₂ catalyzes oxygen and HCl to form Cl₂ in Deacon’s process.
• Q2: B - Chlorine is liberated at the anode during brine electrolysis.
• Q3: D - NaCl reacts with H₂SO₄ and MnO₂, not NaOH, to produce chlorine.

Memory Aids

• Deacon’s process uses CuCl₂ catalyst.
• Electrolysis of brine liberates Cl₂ at anode.
• MnO₂ + HCl is a common lab method.

Chlorine Physical Properties

Key Characteristics

• Greenish-yellow gas with pungent, suffocating odor
• 2.5 times heavier than air
• Liquefies to greenish-yellow liquid, boiling point 239 K
• Soluble in water

Solved Examples

• Ex 1: What is the color and odor of chlorine gas?
  Answer: Greenish-yellow with pungent odor.

Practice Set

• Q1: Chlorine gas is:
  A. Colorless
  B. Greenish-yellow
  C. Blue
  D. Red
• Q2: Chlorine is heavier than air by a factor of:
  A. 1.5
  B. 2.5
  C. 3.5
  D. 4.5
• Q3: Boiling point of chlorine is:
  A. 239 K
  B. 273 K
  C. 300 K
  D. 200 K

Answer Key

• Q1: B - Chlorine is greenish-yellow gas.
• Q2: B - Chlorine is about 2.5 times heavier than air.
• Q3: A - Chlorine boils at 239 K.

Memory Aids

• Greenish-yellow gas, pungent smell = Chlorine
• Boils below room temperature (239 K)

Chlorine Chemical Properties

Reactions with Metals and Non-metals

• 2Al + 3Cl₂ → 2AlCl₃
• P₄ + 6Cl₂ → 4PCl₃
• 2Na + Cl₂ → 2NaCl
• S₈ + 4Cl₂ → 4S₂Cl₂
• 2Fe + 3Cl₂ → 2FeCl₃

Reactions with Hydrogen Compounds

• H₂S + Cl₂ → 2HCl + S
• C₁₀H₁₆ + 8Cl₂ → 16HCl + 10C

Reaction with Ammonia

• 8NH₃ + 3Cl₂ → 6NH₄Cl + N₂ (excess NH₃)
• NH₃ + 3Cl₂ → NCl₃ + 3HCl (excess Cl₂)

Reaction with Alkalies

• 2NaOH + Cl₂ → NaCl + NaOCl + H₂O (cold, dilute)
• 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O (hot, concentrated)

Reaction with Slaked Lime

• 2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O

As an Oxidising Agent

• 2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + 2HCl
• SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl
• I₂ + 6H₂O + 5Cl₂ → 2HIO₃ + 10HCl

As a Bleaching Agent

• Cl₂ + H₂O → 2HCl + [O]
• Coloured substance + [O] → Colourless substance

Solved Examples

• Ex 1: Write the reaction of chlorine with sodium hydroxide (cold and dilute).
  Answer: 2NaOH + Cl₂ → NaCl + NaOCl + H₂O

Practice Set

• Q1: Which product is formed when chlorine reacts with hot concentrated NaOH?
  A. NaClO
  B. NaClO₃
  C. NaCl
  D. NaClO₄
• Q2: Reaction of chlorine with ammonia in excess ammonia produces:
  A. NH₄Cl
  B. NCl₃
  C. N₂
  D. NH₃
• Q3: Chlorine acts as an oxidising agent in reaction with:
  A. FeSO₄
  B. NaCl
  C. H₂O
  D. NH₃

Answer Key

• Q1: B - Hot concentrated NaOH with Cl₂ forms NaClO₃.
• Q2: A - Excess NH₃ with Cl₂ forms NH₄Cl and N₂.
• Q3: A - Cl₂ oxidises FeSO₄ to Fe₂(SO₄)₃.

Memory Aids

• Cold dilute NaOH + Cl₂ → NaOCl (bleach)
• Hot concentrated NaOH + Cl₂ → NaClO₃ (chlorate)
• Excess NH₃ + Cl₂ → NH₄Cl + N₂

Hydrogen Chloride

Lab preparation

• Industrial Mannheim process involves two steps:
• At 420 K: NaCl + H₂SO₄ → NaHSO₄ + HCl
• At 823 K: NaHSO₄ + NaCl → Na₂SO₄ + HCl
• HCl gas is dried by passing through concentrated H₂SO₄

Physical properties

• Colourless, pungent smelling gas
• Easily liquefied
• Extremely soluble in water

Chemical properties

• Decomposes salts of weaker acids:
• Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
• Na₂SO₃ + 2HCl → 2NaCl + H₂O + SO₂
• Aqua regia (3 parts conc. HCl + 1 part conc. HNO₃) dissolves noble metals:
• Au + 4H⁺ + NO₃⁻ + 4Cl⁻ → AuCl₄⁻ + NO + 2H₂O
• Pt + 16H⁺ + 4NO₃⁻ + 18Cl⁻ → 3PtCl₆²⁻ + 4NO + 8H₂O

Solved Examples

• Ex 1: Write the reaction for HCl preparation at 420 K in Mannheim process.
  Answer: NaCl + H₂SO₄ → NaHSO₄ + HCl
• Ex 2: Write the reaction of aqua regia with gold.
  Answer: Au + 4H⁺ + NO₃⁻ + 4Cl⁻ → AuCl₄⁻ + NO + 2H₂O

Practice Set

• Q1: Which gas is produced when NaCl reacts with concentrated H₂SO₄ at 420 K?
  A. Cl₂ B. HCl C. SO₂ D. NaHSO₄
• Q2: What is the product formed when NaHSO₄ reacts with NaCl at 823 K?
  A. NaCl B. Na₂SO₄ C. HCl D. SO₂
• Q3: Aqua regia is a mixture of:
  A. 3 parts conc. HCl + 1 part conc. HNO₃
  B. 1 part conc. HCl + 3 parts conc. HNO₃
  C. 2 parts conc. HCl + 2 parts conc. HNO₃
  D. 4 parts conc. HCl + 1 part conc. HNO₃

Answer Key

• Q1: B. HCl - Produced by reaction of NaCl with H₂SO₄ at 420 K.
• Q2: B. Na₂SO₄ - Formed by reaction of NaHSO₄ with NaCl at 823 K.
• Q3: A. 3 parts conc. HCl + 1 part conc. HNO₃ - Correct composition of aqua regia.

Memory Aids

• Mannheim process: NaCl + H₂SO₄ → NaHSO₄ + HCl (420 K), then NaHSO₄ + NaCl → Na₂SO₄ + HCl (823 K)
• Aqua regia = 3 parts HCl + 1 part HNO₃, dissolves Au and Pt
• HCl decomposes salts of weaker acids releasing CO₂ and SO₂

Oxoacids of Halogens

Properties and Trends

• Halogens form oxoacids with oxidation states +1, +3, +5, and +7.
• Thermal stability increases with higher oxidation state.
• Halogen-oxygen bond becomes more covalent as oxidation state increases.
• Oxidizing power decreases with increasing oxidation number.
• Acidity increases with higher oxidation states.
• Electronegativity of halogen atoms remains constant across these acids.
• Opposite trends observed: electronegativity vs. thermal stability, oxidizing power, and acid strength.

Solved Examples

• Ex 1: Identify the oxidation state of chlorine in HClO₃.
  • Cl in HClO₃: Let oxidation state = x
  • Oxygen = -2, Hydrogen = +1
  • Equation: x + 3(-2) + 1 = 0 → x - 6 + 1 = 0 → x = +5
• Ex 2: Compare acidity of HClO and HClO₄.
  • HClO (Cl +1) is less acidic than HClO₄ (Cl +7).
  • Higher oxidation state increases acid strength.

Practice Set

• Q1: Which oxoacid of chlorine has the highest thermal stability?
  • A. HClO
  • B. HClO₂
  • C. HClO₃
  • D. HClO₄
• Q2: In oxoacids of halogens, acidity increases with:
  • A. Decreasing oxidation state
  • B. Increasing oxidation state
  • C. Constant oxidation state
  • D. None of the above
• Q3: Which property decreases as the oxidation state of halogen increases in oxoacids?
  • A. Thermal stability
  • B. Acidity
  • C. Oxidizing power
  • D. Electronegativity

Answer Key

• Q1: D. HClO₄ - Highest oxidation state (+7) gives highest thermal stability.
• Q2: B. Increasing oxidation state - Acidity increases with higher oxidation states.
• Q3: C. Oxidizing power - It decreases as oxidation state increases.

Memory Aids

• Higher oxidation state → Stronger acid, more stable, less oxidizing.
• Remember: HClO < HClO₂ < HClO₃ < HClO₄ in acidity and stability.
• Electronegativity of halogen remains unchanged across oxoacids.

Interhalogen Compounds

General Characteristics

• Formed by combination of two different halogens.
• Types: XX, XX₃, XX₅, XX₇.
• X = larger halogen (more electropositive), X' = smaller halogen.

Preparation

• Cl₂ + F₂ (437 K) → ClF
• I₂ + excess Cl₂ → ICl₃
• Cl₂ + excess F₂ (573 K) → ClF₃
• BrF₃ + F₂ (473 K) → BrF₅
• IF₅ + F₂ (543 K) → IF₇

Structures

• XX' (2 groups): Linear, sp³ hybridization.
• XX'₃ (3 groups): T-shape, sp³d hybridization.
• XX'₅ (5 groups): Square pyramidal, sp³d² hybridization.
• XX'₇ (7 groups): Pentagonal bipyramidal, sp³d³ hybridization.

Pseudohalogens

• Ions with two or more electronegative atoms including nitrogen, similar to halide ions.
• Examples: CN⁻, SCN⁻, OCN⁻.
• Dimeric molecules: (CN)₂, (SCN)₂, (OCN)₂.

Solved Examples

Ex 1: Write the product when Cl₂ reacts with F₂ at 437 K.

Step 1: Cl₂ + F₂ → ClF (Chlorine monofluoride)

Ex 2: Identify the shape and hybridization of ClF₃.

Step 1: ClF₃ has 3 bonded atoms and 2 lone pairs.

Step 2: Shape is T-shaped, hybridization sp³d.

Practice Set

• Q1: Which interhalogen compound is formed by I₂ and excess Cl₂?
  • A. ICl
  • B. ICl₃
  • C. ICl₅
  • D. ICl₇
• Q2: What is the shape of BrF₅ molecule?
  • A. Linear
  • B. T-shaped
  • C. Square pyramidal
  • D. Pentagonal bipyramidal
• Q3: Which hybridization corresponds to XX'₇ interhalogen compounds?
  • A. sp³
  • B. sp³d
  • C. sp³d²
  • D. sp³d³

Answer Key

• Q1: B - ICl₃ is formed by I₂ and excess Cl₂.
• Q2: C - BrF₅ has square pyramidal shape.
• Q3: D - XX'₇ compounds have sp³d³ hybridization.

Memory Aids

• Interhalogen types: 2 (linear), 3 (T-shape), 5 (square pyramidal), 7 (pentagonal bipyramidal).
• Hybridization order: sp³, sp³d, sp³d², sp³d³.
• Preparation temperatures increase with higher fluorine content.
• Pseudohalogens mimic halide ions; remember CN⁻, SCN⁻, OCN⁻.

Group 18 Elements

General Characteristics

• Elements: He, Ne, Ar, Kr, Xe, Rn, Og
• All are monoatomic gases and chemically inert, called noble gases
• General electronic configuration: ns² np⁶
• High ionization enthalpy, decreases down the group
• Atomic radii increase down the group
• Electron gain enthalpy is positive (no tendency to gain electrons)
• Low melting and boiling points due to weak dispersion forces
• Least reactive due to stable electronic configuration

Solved Examples

• Ex 1: Write the electronic configuration of Argon.
  Answer: 1s² 2s² 2p⁶ 3s² 3p⁶

Practice Set

• Q1: Which noble gas has the smallest atomic radius?
  A. Helium
  B. Neon
  C. Argon
  D. Krypton
• Q2: What is the general electronic configuration of noble gases?
  A. ns² np⁵
  B. ns² np⁶
  C. ns¹ np⁶
  D. ns² np⁴
• Q3: Ionization enthalpy trend in noble gases is:
  A. Increases down the group
  B. Decreases down the group
  C. Remains constant
  D. Random

Answer Key

• Q1: A - Helium has the smallest atomic radius due to least number of electron shells.
• Q2: B - Noble gases have completely filled ns² np⁶ configuration.
• Q3: B - Ionization enthalpy decreases down the group as atomic size increases.

Memory Aids

• "He Never Argues, Krypton Xenon Radon Oganesson" for element order
• General config: ns² np⁶ (except He: 1s²)
• High ionization enthalpy = inertness

Xenon Fluorine Compounds

Reactions and Properties

• Xe + F₂ (excess, 673 K, 1 bar) → XeF₂
• Xe + 5F₂ (873 K, 7 bar) → XeF₄
• Xe + 20F₂ (573 K, 60-70 bar) → XeF₆
• XeF₄ + O₂F₂ (143 K) → XeF₆ + O₂
• XeF₂, XeF₄, XeF₆ are colorless crystalline solids, sublime at 298 K
• Powerful fluorinating agents, hydrolyzed by water:
  2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
• React with fluoride ion acceptors/donors to form cations and fluoroanions:
  XeF₂ + PF₅ → [XeF]⁺[PF₆]^-
  XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]^-
  XeF₆ + MF → M⁺[XeF₇]^- (M = Na, K, Rb, Cs)

Solved Examples

• Ex 1: Write the reaction to prepare XeF₆ from XeF₄.
  Answer: XeF₄ + O₂F₂ → XeF₆ + O₂

Practice Set

• Q1: Which xenon fluoride is formed at 673 K and 1 bar?
  A. XeF₂
  B. XeF₄
  C. XeF₆
  D. XeO₃
• Q2: What is the product when XeF₂ reacts with water?
  A. XeF₄
  B. Xe + HF + O₂
  C. XeO₃
  D. XeF₆
• Q3: Which ion is formed when XeF₆ reacts with fluoride ion donors?
  A. [XeF]⁺
  B. [XeF₃]⁺
  C. [XeF₇]^-
  D. [XeF₅]^-

Answer Key

• Q1: A - XeF₂ forms at 673 K and 1 bar with excess fluorine.
• Q2: B - XeF₂ hydrolyzes to Xe, HF, and O₂.
• Q3: C - XeF₆ reacts with fluoride donors to form [XeF₇]^-.

Memory Aids

• XeF₂: linear, formed at low pressure and moderate temperature
• XeF₄: square planar, formed at higher pressure and temperature
• XeF₆: distorted octahedral, formed at very high pressure

Xenon Oxygen Compounds

Reactions and Properties

• 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
• XeF₆ + 3H₂O → XeO₃ + 6HF
• XeF₆ + H₂O → XeOF₄ + 2HF
• XeF₆ + 2H₂O → XeO₂F₂ + 4HF (partial hydrolysis)

Solved Examples

• Ex 1: Write the reaction of XeF₆ with water to form XeO₃.
  Answer: XeF₆ + 3H₂O → XeO₃ + 6HF

Practice Set

• Q1: What is the product when XeF₆ reacts with 3 moles of water?
  A. XeO₃
  B. XeOF₄
  C. XeO₂F₂
  D. XeF₄
• Q2: Partial hydrolysis of XeF₆ produces:
  A. XeO₃
  B. XeOF₄
  C. XeO₂F₂
  D. XeF₂
• Q3: Which compound is formed by complete hydrolysis of XeF₄ with water?
  A. XeO₃
  B. XeF₆
  C. XeOF₄
  D. XeO₂F₂

Answer Key

• Q1: A - XeF₆ + 3H₂O forms XeO₃.
• Q2: C - Partial hydrolysis yields XeO₂F₂.
• Q3: A - Complete hydrolysis of XeF₄ forms XeO₃.

Memory Aids

• Complete hydrolysis → XeO₃
• Partial hydrolysis → XeO₂F₂
• XeOF₄ formed by limited water reaction

Structures of Xenon Compounds

Molecular Shapes and Hybridization

• XeF₂: Linear shape, sp³d hybridization
• XeF₄: Square planar, sp³d² hybridization
• XeF₆: Distorted octahedral, sp³d³ hybridization
• XeO₃: Pyramidal, sp³ hybridization

Solved Examples

• Ex 1: Identify the shape of XeF₄.
  Answer: Square planar

Practice Set

• Q1: What is the hybridization of XeF₂?
  A. sp³d
  B. sp³d²
  C. sp³d³
  D. sp²
• Q2: XeF₆ has which molecular geometry?
  A. Octahedral
  B. Distorted octahedral
  C. Square planar
  D. Linear
• Q3: XeO₃ molecule shape is:
  A. Linear
  B. Tetrahedral
  C. Pyramidal
  D. Trigonal planar

Answer Key

• Q1: A - XeF₂ is sp³d hybridized.
• Q2: B - XeF₆ is distorted octahedral.
• Q3: C - XeO₃ is pyramidal.

Memory Aids

• Remember shapes by number of bonded atoms and lone pairs
• XeF₂: linear, XeF₄: square planar, XeF₆: distorted octahedral
• XeO₃ pyramidal due to lone pair on Xe