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

Group 15 Elements

General Characteristics

• Elements: N, P, As, Sb, Bi, Moscovium (synthetic, chemistry not well-known)

• General electronic configuration: ns2np3

• Known as pnictogens

Physical Properties

• Atomic radii increase down the group; smaller than group 14 due to higher nuclear charge

• Melting point increases from N to As, then decreases due to inert pair effect

• Metallic character increases down the group

• Ionization 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 state stability increases down the group; +5 state stability decreases due to inert pair effect

• Halides: General formula EX3 or EX5

• Halide stability order: NF3 > NCl3 > NBr3 (due to size differences)

• Lewis acid strength: PCl3 > NBr3 > NCl3 > NF3 (decreases with increasing halogen electronegativity)

• Halide bond angle: PF3 < PCl3 < PBr3 < PI3 (due to decreasing bond pair repulsions)

Hydrides (EH3)

• Bond angle trend: 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 (decreases with decreasing electronegativity)

Oxides of Nitrogen

• Sequence by increasing oxidation number: N2O (neutral), NO (neutral), N2O3, N2O4, N2O5

Reactivity with Metals

• Form binary compounds with metals exhibiting -3 oxidation state, e.g., Ca3N2, Ca3P2, Na3As

Solved Examples

• Ex 1: Predict the oxidation state stability trend in group 15 elements.
  - +3 state stability increases down the group due to inert pair effect.
  - +5 state stability decreases down the group.

• Ex 2: Arrange the hydrides NH3, PH3, AsH3 in order of basicity.
  - NH3 > PH3 > AsH3 (AsH3 is weakly basic or non-basic).

• Ex 3: Write the formula of the binary compound formed between calcium and nitrogen.
  - Ca3N2 (calcium nitride).

Practice Set

• Q1: Which element in group 15 has the highest electronegativity?
  A. Nitrogen
  B. Phosphorus
  C. Arsenic
  D. Bismuth

• Q2: Which hydride has the highest boiling point?
  A. PH3
  B. NH3
  C. AsH3
  D. SbH3

• Q3: Which oxide of nitrogen is neutral?
  A. N2O3
  B. NO2
  C. N2O
  D. N2O5

Answer Key

• Q1: A. Nitrogen
  - Highest electronegativity due to smallest atomic size and highest nuclear charge.

• Q2: B. NH3
  - Strong hydrogen bonding leads to highest boiling point.

• Q3: C. N2O
  - N2O and NO are neutral oxides of nitrogen.

Memory Aids

• Oxidation state trend: +3 state stability increases down group due to inert pair effect.

• Hydride boiling point: NH3 highest due to hydrogen bonding.

• Halide stability: NF3 > NCl3 > NBr3 (size difference effect).

Dinitrogen

Preparation of Nitrogen Gas (N2)

• 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 of Nitrogen

• Reacts with lithium to form lithium nitride:
  N2 + 6Li → 2Li3N

• Reacts with magnesium to form magnesium nitride:
  N2 + 3Mg → Mg3N2

• Reacts with oxygen at high temperature to form nitric oxide:
  N2 + O2 → 2NO (at high temperature)

• Haber process for ammonia synthesis:
  N2 + 3H2 ⇌ 2NH3 (high pressure, temperature, catalyst)
  ΔH = -92 kJ/mol (exothermic)

Solved Examples

• Ex 1: Write the balanced equation for the reaction of nitrogen with lithium.
  Solution: N2 + 6Li → 2Li3N

• Ex 2: Identify the product when nitrogen reacts with oxygen at high temperature.
  Solution: Nitric oxide (NO) is formed: N2 + O2 → 2NO

Practice Set

• Q1: Which of the following is a method to prepare nitrogen gas?
  A. Thermal decomposition of ammonium dichromate
  B. Electrolysis of water
  C. Reaction of hydrogen with oxygen
  D. Combustion of methane

• Q2: What is the product when nitrogen reacts with magnesium?
  A. MgN
  B. Mg3N2
  C. Mg2N3
  D. MgNO3

• Q3: In the Haber process, which condition is NOT used?
  A. High pressure
  B. High temperature
  C. Catalyst
  D. Low pressure

Answer Key

• Q1: A - Thermal decomposition of ammonium dichromate produces nitrogen gas.

• Q2: B - Magnesium reacts with nitrogen to form magnesium nitride, Mg3N2.

• Q3: D - Low pressure is not used; high pressure is essential in the Haber process.

Memory Aids

• "Li and Mg form nitrides with N2" - remember alkali and alkaline earth metals react with nitrogen.

• Haber process: N2 + 3H2 → 2NH3, high pressure & temperature, catalyst needed.

• Decomposition of ammonium dichromate gives "Cr2O3 + N2 + H2O" - remember as a classic lab demonstration.

Ammonia Preparation

Laboratory Preparation

• Ammonium salts react with alkali to release NH3 gas.

• Reactions:

  • 2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O

  • (NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O

  • Mg3N2 + 6HCl → 3MgCl2 + 2NH3

• Urea hydrolyzes in water to form ammonium carbonate, which decomposes to NH3, CO2, and H2O.

Manufacture (Haber's Process)

• Large scale synthesis of NH3 from N2 and H2.

• Reaction: N2 + 3H2 ⇌ 2NH3 + heat (ΔH° = -46.1 kJ/mol)

• Conditions: 773 K, 200 atm, iron oxide catalyst with K2O and Al2O3 promoters.

Solved Examples

• Ex 1: Write the reaction for laboratory preparation of NH3 from ammonium chloride and calcium hydroxide.

  • 2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O

• Ex 2: State the conditions for Haber’s process.

  • Temperature: 773 K, Pressure: 200 atm, Catalyst: Fe2O3 with K2O and Al2O3.

Practice Set

• Q1: Which reagent is used with ammonium salts to prepare ammonia in the lab?

  • A. HCl

  • B. NaOH

  • C. H2SO4

  • D. CaCl2

• Q2: In Haber’s process, the catalyst used is:

  • A. Platinum

  • B. Iron oxide

  • C. Copper

  • D. Zinc oxide

• Q3: The reaction N2 + 3H2 ⇌ 2NH3 is:

  • A. Endothermic

  • B. Exothermic

  • C. Neither

  • D. Irreversible

Answer Key

• Q1: B - Alkali (NaOH or Ca(OH)2) reacts with ammonium salts to release NH3.

• Q2: B - Iron oxide catalyst is used in Haber’s process.

• Q3: B - The reaction is exothermic (releases heat).

Memory Aids

• Ammonia lab prep: Ammonium salt + Alkali → NH3

• Haber’s process: N2 + 3H2 ⇌ 2NH3, Fe catalyst, 773 K, 200 atm

• Exothermic reaction: Heat released, equilibrium shifts left if temperature increases

Ammonia Structure

Molecular Geometry

• NH3 is trigonal pyramidal with bond angle ~107.8°.

• Nitrogen uses sp3 hybridization: 3 orbitals overlap with H 1s orbitals; 1 orbital holds lone pair.

• Hydrogen bonding occurs in liquid and solid states.

Solved Examples

• Ex 1: Describe the shape and bond angle of NH3.

  • Trigonal pyramidal shape; bond angle 107.8° due to lone pair repulsion.

Practice Set

• Q1: The hybridization of nitrogen in NH3 is:

  • A. sp

  • B. sp2

  • C. sp3

  • D. dsp2

• Q2: The bond angle in ammonia is approximately:

  • A. 90°

  • B. 104.5°

  • C. 107.8°

  • D. 120°

• Q3: The shape of NH3 molecule is due to:

  • A. Lone pair on nitrogen

  • B. Double bonds

  • C. Hydrogen bonding

  • D. Ionic bonding

Answer Key

• Q1: C - Nitrogen is sp3 hybridized in NH3.

• Q2: C - Bond angle is 107.8° due to lone pair repulsion.

• Q3: A - Lone pair on nitrogen causes trigonal pyramidal shape.

Memory Aids

• NH3 shape: Trigonal pyramidal due to 1 lone pair on N

• Bond angle less than tetrahedral (109.5°) due to lone pair repulsion

• Hydrogen bonding in liquid/solid states increases boiling point

Ammonia Basic Character

Lewis Base Behavior

• NH3 donates lone pair of electrons, acting as a Lewis base.

• In water, NH3 accepts H+ forming NH4+ and OH− ions.

• Reaction: NH3 + H2O ⇌ NH4+ + OH−

Reaction with Heavy Metal Ions

• Forms insoluble metal hydroxide precipitates with Fe3+, Zn2+, etc.

• Examples:

  • FeCl3 + 3NH4OH → Fe(OH)3↓ + 3NH4Cl (brown ppt.)

  • ZnCl2 + 2NH4OH → Zn(OH)2↓ + 2NH4Cl (white ppt.)

Formation of Complex Compounds

• NH3 forms stable soluble complexes with Ag+, Cu2+, Hg2+, etc.

• Examples:

  • AgCl + 2NH4OH → [Ag(NH3)2]Cl + 2H2O (soluble)

  • 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−

• Ex 2: Reaction of FeCl3 with NH4OH.

  • FeCl3 + 3NH4OH → Fe(OH)3↓ + 3NH4Cl (brown ppt.)

Practice Set

• Q1: NH3 acts as a:

  • A. Lewis acid

  • B. Lewis base

  • C. Both

  • D. Neither

• Q2: The precipitate formed when ZnCl2 reacts with NH4OH is:

  • A. ZnCl2

  • B. Zn(OH)2

  • C. ZnO

  • D. Zn(NH3)42+

• Q3: The complex ion formed by NH3 with Ag+ is:

  • A. [Ag(NH3)]+

  • B. [Ag(NH3)2]+

  • C. [Ag(NH3)3]+

  • D. [Ag(NH3)4]+

Answer Key

• Q1: B - NH3 donates lone pair, acts as Lewis base.

• Q2: B - Zn(OH)2 is the white precipitate formed.

• Q3: B - Diamminesilver(I) complex is [Ag(NH3)2]+.

Memory Aids

• NH3 + H2O ⇌ NH4+ + OH−: Basic solution formation

• NH3 forms insoluble metal hydroxides with Fe3+, Zn2+

• Stable complexes: Ag+ forms [Ag(NH3)2]+, Cu2+ forms [Cu(NH3)4]2+

Nitrogen Hydrides

Types and Properties

• Three main hydrides: NH3 (ammonia), N2H4 (hydrazine), N3H (hydrazoic acid).

• NH3 and N2H4 are basic; N3H is distinctly acidic.

Solved Examples

• Ex 1: Identify the acidic nitrogen hydride.

  • Hydrazoic acid (N3H) is acidic.

Practice Set

• Q1: Which nitrogen hydride is acidic?

  • A. NH3

  • B. N2H4

  • C. N3H

  • D. None

• Q2: Hydrazine (N2H4) is:

  • A. Acidic

  • B. Basic

  • C. Neutral

  • D. Amphoteric

• Q3: The formula of hydrazoic acid is:

  • A. NH3

  • B. N2H4

  • C. N3H

  • D. NH2NH2

Answer Key

• Q1: C - N3H (hydrazoic acid) is acidic.

• Q2: B - Hydrazine is basic in nature.

• Q3: C - Hydrazoic acid formula is N3H.

Memory Aids

• NH3 and N2H4: Basic hydrides

• N3H: Acidic hydride

• Hydrazine = N2H4, Hydrazoic acid = N3H

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, oxidation, and reduction reactions.

• Physical states differ: gases (N2O, NO, NO2), liquids, or solids.

• Acidic or neutral nature depends on oxide; e.g., N2O is neutral, NO2 is acidic.

Solved Examples

• Ex 1: Write the preparation reaction of nitric oxide (NO).

  • Reaction: N2 + O2 → 2NO (at high temperature)

• Ex 2: Identify the oxidation state of nitrogen in N2O5.

  • Calculation: Let x = O.S. of N; 2x + 5(-2) = 0 ⇒ 2x -10 = 0 ⇒ x = +5

Practice Set

• Q1: Which oxide of nitrogen is a colorless, neutral gas?

  • A. NO2

  • B. N2O

  • C. N2O5

  • D. NO

• Q2: What is the oxidation state of nitrogen in NO2?

  • A. +2

  • B. +3

  • C. +4

  • D. +5

• Q3: Which reaction correctly represents the preparation of nitrogen dioxide (NO2)?

  • A. 2NO + O2 → 2NO2

  • B. N2 + O2 → 2NO2

  • C. N2O5 → NO2 + O2

  • D. NO + N2O5 → NO2

Answer Key

• Q1: B. N2O is colorless and neutral.

• Q2: C. Nitrogen in NO2 has +4 oxidation state.

• Q3: A. 2NO + O2 → 2NO2 is the correct preparation reaction.

Memory Aids

• Remember: N2O (laughing gas) is neutral and 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:

• 1. 4NH3 + 5O2 → 4NO + 6H2O (catalyst, high T and P)

• 2. 2NO + O2 → 2NO2

• 3. 3NO2 + H2O → 2HNO3 + NO

• NO is recycled; HNO3 concentrated by distillation (~68%) and dehydration with concentrated H2SO4 to 98%

Solved Examples

• Ex 1: Write the reaction for ammonia oxidation in Ostwald’s process.

• Answer: 4NH3 + 5O2 → 4NO + 6H2O (with catalyst)

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 + HNO3 B. Na2SO4 + HNO3 C. NaNO2 + H2SO4 D. NaNO3 + H2O

• Q3: What is the maximum concentration of HNO3 obtained by distillation before dehydration?

• A. 50% B. 68% C. 85% D. 98%

Answer Key

• Q1: B - Platinum/rhodium catalyst is used for ammonia oxidation.

• Q2: A - Reaction produces NaHSO4 and HNO3.

• Q3: B - Distillation concentrates HNO3 up to ~68% before dehydration.

Memory Aids

• Ostwald’s process uses Pt/Rh catalyst for NH3 oxidation.

• NaNO3 + H2SO4 → NaHSO4 + HNO3 (lab prep).

• Distillation max ~68%, dehydration to 98% HNO3.

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 define shape (H–O: 96 pm, angle 102°; N–O: 121 pm, angle 130°).

Chemical Properties

• Strong oxidising agent; converts non-metals to highest oxyacids with evolution of NO and NO2 gases.

• Examples:

• I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O

• C + 4HNO3 → CO2 + 2H2O + NO2

• S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O

• P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O

Reactions with Metals

• Acts as acid and oxidising agent; products depend on metal and acid concentration.

• Very dilute HNO3: Mg, Mn produce H2 + metal nitrate; Fe, Zn, Sn produce NH4NO3 + metal nitrate.

• Dilute HNO3: Pb, Cu, Ag, Hg produce NO + metal nitrate; Fe, Zn produce N2O + metal nitrate; Sn produces NH4NO3 + Sn nitrate.

• Concentrated HNO3: Zn, Fe, Pb, Cu, Ag produce NO2 + metal nitrate; Sn produces NO2 + H2SnO3.

Brown Ring Test

• Detects nitrate ion (NO3−):

• NO3− + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O

• [Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O (brown complex)

Solved Examples

• Ex 1: Write the reaction of iodine with hot concentrated nitric acid.

• Answer: I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O

• Ex 2: What gas is evolved when copper reacts with dilute nitric acid?

• Answer: Nitric oxide (NO) gas is evolved.

Practice Set

• Q1: What is the colour of pure nitric acid?

• A. Yellow B. Colourless C. Brown D. Blue

• Q2: Which gas is evolved when zinc reacts with concentrated nitric acid?

• A. H2 B. NO C. NO2 D. N2O

• Q3: In the brown ring test, the brown complex formed is due to coordination of NO with which ion?

• A. Fe2+ B. Fe3+ C. Cu2+ D. Zn2+

Answer Key

• Q1: B - Pure HNO3 is colourless; yellow colour indicates impurities.

• Q2: C - Concentrated HNO3 with zinc produces NO2 gas.

• Q3: A - Brown ring complex is [Fe(H2O)5(NO)]2+ with Fe2+.

Memory Aids

• Pure HNO3 is colourless; yellow means nitrogen oxides present.

• Brown ring test detects NO3− via Fe2+ complex.

• Metal + HNO3 reaction products depend on acid concentration and metal type.

Phosphorus Allotropic Forms

White Phosphorus

• Appearance: White translucent waxy solid, poisonous.

• Solubility: Insoluble in water, soluble in CS2.

• Structure: Consists of discrete tetrahedral P4 molecules with 60° bond angles causing angular strain.

• Reactivity: Less stable, highly reactive due to strain.

• Chemiluminescence: Glows in dark.

• Reactions:

  • Dissolves in boiling NaOH (inert atmosphere) to give PH3:
    P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2 (Sodium hypophosphite)

  • Burns in air producing dense white fumes of P4O10:
    P4 + 5O2 → P4O10

Red Phosphorus

• Preparation: Heating white phosphorus at 573 K in inert atmosphere for several days.

• Appearance: Iron grey lustre, odourless, non-poisonous.

• Solubility: Insoluble in water and CS2.

• Structure: Polymeric chains of P4 tetrahedra.

• Reactivity: Less reactive than white phosphorus, does not glow in dark.

• Under high pressure, red phosphorus converts to black phosphorus phases.

Black Phosphorus

• Forms:

  • α-Black Phosphorus: Formed by heating red phosphorus at 803 K in sealed tube; has opaque monoclinic or rhombohedral crystals; can sublime in air; does not oxidize in air.

  • β-Black Phosphorus: Prepared by heating white phosphorus at 473 K under high pressure; layered structure; does not burn in air up to 673 K.

Solved Examples

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

• Ex 2: Describe the conversion of white phosphorus to red phosphorus.
  Solution: Heat white phosphorus at 573 K in inert atmosphere for several days to obtain red phosphorus.

Practice Set

• Q1: Which allotrope of phosphorus glows in the dark?
  A. Red phosphorus
  B. White phosphorus
  C. Black phosphorus
  D. None

• Q2: What is the product formed when white phosphorus burns in air?
  A. P4O6
  B. P4O10
  C. PH3
  D. P2O5

• Q3: Which allotrope of phosphorus is obtained by heating white phosphorus at 573 K in inert atmosphere?
  A. Black phosphorus
  B. Red phosphorus
  C. White phosphorus
  D. Violet phosphorus

Answer Key

• Q1: B. White phosphorus glows in the dark due to chemiluminescence.

• Q2: B. White phosphorus burns in air to form P4O10.

• Q3: B. Red phosphorus is formed by heating white phosphorus at 573 K in inert atmosphere.

Memory Aids

• White phosphorus = P4 tetrahedral, reactive, glows.

• Red phosphorus = polymeric chains, less reactive, no glow.

• Black phosphorus = layered structure, stable, two forms α and β.

• White P + NaOH + H2O → PH3 + Sodium hypophosphite.

• White P + O2 → P4O10 (dense white fumes).

Phosphine

Preparation

• From calcium phosphide: Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

• From calcium phosphide and hydrochloric acid: Ca3P2 + 6HCl → 3CaCl2 + 2PH3

• From phosphonium iodide and KOH: PH4I + KOH → KI + H2O + PH3

• From white phosphorus, sodium hydroxide and water: P4 + 3NaOH + 3H2O → 3NaH2PO2 + PH3

Properties

• Colourless gas with a characteristic rotten fish smell

• Highly poisonous

• Decomposes on heating: 4PH3 Δ→ P4 + 6H2

• Reacts with copper sulfate: 2PH3 + 3CuSO4 → Cu3P2 + 3H2SO4

• Reacts with hydrobromic acid to form phosphonium bromide: PH3 + HBr → PH4Br

• Reacts only with anhydrous HX (HCl, HBr, HI), not with aqueous HX because phosphonium halides decompose in water to regenerate PH3

Solved Examples

• Ex 1: Write the reaction for preparation of PH3 from calcium phosphide and water.
  Answer: Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

• Ex 2: What happens when PH3 is heated?
  Answer: 4PH3 Δ→ P4 + 6H2

Practice Set

• Q1: Which of the following is a method to prepare phosphine?
  A. P4 + NaOH + H2O
  B. PH3 + H2O
  C. PH4Br + KOH
  D. PCl3 + H2O

• Q2: Phosphine reacts with which acid only in anhydrous form?
  A. HCl
  B. HBr
  C. HI
  D. All of the above

• Q3: On heating, phosphine decomposes to give:
  A. White phosphorus and hydrogen
  B. Red phosphorus and hydrogen
  C. Phosphorus pentoxide and hydrogen
  D. Phosphoric acid and hydrogen

Answer Key

• Q1: A. P4 + NaOH + H2O; This reaction produces PH3 from white phosphorus.

• Q2: D. All of the above; PH3 reacts with anhydrous HCl, HBr, and HI only.

• Q3: B. Red phosphorus and hydrogen; PH3 decomposes on heating to red phosphorus (P4) and H2.

Memory Aids

• PH3 smells like rotten fish and is highly poisonous.

• Phosphine reacts only with anhydrous HX, aqueous HX regenerates PH3.

• Heating PH3 → red phosphorus + hydrogen gas.

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: React white phosphorus (P4) with limited chlorine (Cl2).

• PCl5: React PCl3 with excess chlorine or directly P4 with excess Cl2.

Chemical Properties

• Hydrolysis: PCl3 + 3H2O → H3PO3 + 3HCl

• PCl5 + H2O → POCl3 + 2HCl (partial hydrolysis)

• Further hydrolysis of POCl3 gives H3PO4.

• React with SO2 and Zn to form other phosphorus compounds.

• React with organic compounds to introduce chlorine atoms.

Uses

• PCl5: Chlorinating agent, dehydrating agent in organic synthesis.

• PCl3: Precursor for organophosphorus compounds, insecticides, and plasticizers.

Solved Examples

• Ex 1: Write the reaction for preparation of PCl3 from white phosphorus.
  Step 1: P4 + 6Cl2 → 4PCl3

• Ex 2: Show hydrolysis of PCl5.
  Step 1: PCl5 + H2O → POCl3 + 2HCl
  Step 2: POCl3 + 3H2O → H3PO4 + 3HCl

Practice Set

• Q1: What is the shape of PCl5?
  A. Tetrahedral
  B. Trigonal bipyramidal
  C. Octahedral
  D. Square planar

• Q2: Which reagent is used to prepare PCl3 from white phosphorus?
  A. Excess Cl2
  B. Limited Cl2
  C. H2O
  D. SO2

• Q3: What is the product when PCl3 reacts with water?
  A. H3PO4
  B. H3PO3
  C. POCl3
  D. PCl5

Answer Key

• Q1: B. Trigonal bipyramidal — PCl5 has five bonded atoms arranged in trigonal bipyramidal geometry.

• Q2: B. Limited Cl2 — Limited chlorine reacts with P4 to form PCl3.

• Q3: B. H3PO3 — PCl3 hydrolyzes to phosphorous acid (H3PO3).

Memory Aids

• PCl3 = 3 chlorines, pyramidal shape, hydrolyzes to H3PO3.

• PCl5 = 5 chlorines, trigonal bipyramidal, partial hydrolysis forms POCl3.

• Preparation: Limited Cl2 → PCl3, Excess Cl2 → PCl5.

• PCl5 used as chlorinating and dehydrating agent; PCl3 used in organic synthesis.

Oxoacids of Phosphorus

Overview

• Oxoacids contain phosphorus bonded to oxygen and hydrogen atoms.

• Phosphorus exhibits different oxidation states: +3 and +5 in oxoacids.

• Common oxoacids include phosphorous acid (H3PO3) and phosphoric acid (H3PO4).

• Structures vary: phosphorous acid has one P=O bond and two P–OH groups; phosphoric acid has one P=O and three P–OH groups.

• Preparation methods involve hydrolysis of phosphorus halides or 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: Describe the preparation of phosphoric acid.
  - React P4O10 with water:
  P4O10 + 6H2O → 4H3PO4

Practice Set

• Q1: What is the oxidation state of phosphorus in H3PO4?
  A. +3
  B. +5
  C. +1
  D. 0

• Q2: Which acid has the formula H3PO3?
  A. Phosphoric acid
  B. Phosphorous acid
  C. Hypophosphorous acid
  D. Perphosphoric acid

• Q3: Which method is used to prepare phosphorous acid?
  A. Hydrolysis of PCl3
  B. Oxidation of PCl5
  C. Reaction of P4O10 with water
  D. Electrolysis of phosphates

Answer Key

• Q1: B. +5
  Phosphorus in H3PO4 is in +5 oxidation state due to four oxygen atoms bonded.

• Q2: B. Phosphorous acid
  H3PO3 is phosphorous acid with P in +3 state.

• Q3: A. Hydrolysis of PCl3
  Phosphorous acid is prepared by hydrolyzing phosphorus trichloride (PCl3).

Memory Aids

• Phosphorous acid (H3PO3) has P in +3 state; remember "ous" acids have lower oxidation states.

• Phosphoric acid (H3PO4) has P in +5 state; "ic" acids have higher oxidation states.

• Preparation: PCl3 + H2O → H3PO3; P4O10 + H2O → H3PO4.

Group 16 Elements

General Characteristics

• Elements: Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), Polonium (Po), Livermorium (Lv).

• Also called chalcogens.

• Livermorium is synthetic, radioactive, with very short half-life; properties not well studied.

• General electronic configuration: ns2np4.

• 

• Atomic number and electronic complexity increase down the group.

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 with pπ–pπ multiple bonding; S, Se, Te form octatomic molecules (S8, Se8, Te8); all show allotropy.

Chemical Properties

• Oxidation states:

  • Oxygen: -2, -1, -1/2, 0, +1, +2.

  • Others: -2, +2, +4, +6.

  • -2 state stability decreases down the group due to 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:

  • Types: EX6, EX4, EX2.

  • 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 the general electronic configuration of group 16 elements.
  Answer: ns2np4.

• Ex 2: Arrange the hydrides H2S, H2O, H2Se, H2Te in order of increasing boiling point.
  Answer: H2S < H2Se < H2Te < H2O.

• Ex 3: Identify the oxidation states exhibited by oxygen.
  Answer: -2, -1, -1/2, 0, +1, +2.

Practice Set

• Q1: Which element is not a member of group 16?
  A. Oxygen
  B. Sulphur
  C. Nitrogen
  D. Selenium

• Q2: Which hydride has the highest thermal stability?
  A. H2S
  B. H2Se
  C. H2Te
  D. H2O

• Q3: Which of the following halides has octahedral geometry?
  A. SF4
  B. SeF4
  C. TeF4
  D. SF6

Answer Key

• Q1: C. Nitrogen is in group 15, not group 16.

• Q2: D. H2O has the highest thermal stability among group 16 hydrides.

• Q3: D. SF6 is a hexafluoride with octahedral geometry; others have different structures.

Memory Aids

• Group 16 elements = Chalcogens ("ore-formers").

• Hydrides formula: H2E; boiling point order: H2S < H2Se < H2Te < H2O.

• Oxidation states: O shows multiple states; others mainly -2, +2, +4, +6.

• Hexafluorides are stable and octahedral; SF6 is a key example.

Dioxygen Preparation

Methods of Preparation

• Heating oxo salts (chlorates, nitrates, permanganates) in lab produces O2.

• Example: Decomposition of potassium chlorate (KClO3) with MnO2 catalyst:

  • 2KClO3 (heat, MnO2) → 2KCl + 3O2

  • MnO2 lowers decomposition temperature from 650 K to 523 K.

  • MnO2 must be carbon-free to avoid explosive mixtures.

• Thermal decomposition of heavy metal oxides:

  • 2HgO (heat) → 2Hg + O2

  • 2Ag2O (heat) → 4Ag + O2

  • 2PbO2 (heat) → 2PbO + O2

• Decomposition of hydrogen peroxide (H2O2) with MnO2 catalyst:

  • 2H2O2 (MnO2) → 2H2O + O2

• Large scale: Electrolysis of water:

  • At cathode: 2H+ + 2e− → H2 (byproduct)

  • At anode: 2OH− → H2O + 1/2 O2 + 2e−

Solved Examples

Ex 1: Write the equation for oxygen preparation from potassium chlorate using MnO2.

Solution:

• 2KClO3 (heat, MnO2) → 2KCl + 3O2

Ex 2: Write the reaction for oxygen release from hydrogen peroxide with MnO2.

Solution:

• 2H2O2 (MnO2) → 2H2O + O2

Practice Set

Q1: Which catalyst is used in the decomposition of KClO3 to produce oxygen?

• A. MnO2

• B. Fe2O3

• C. CuO

• D. ZnO

Q2: What is the product formed when HgO is heated?

• A. Hg + O2

• B. HgO2

• C. Hg2O

• D. Hg + 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 lowering decomposition temperature.

Q2: A. Hg + O2 - Heating HgO decomposes it into mercury and oxygen.

Q3: B. H2 - Hydrogen gas is produced at the cathode during water electrolysis.

Memory Aids

• MnO2 catalyzes KClO3 decomposition, lowers temperature.

• Heavy metal oxides decompose on heating to metal + O2.

• Electrolysis: Cathode → H2, Anode → 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?

Solution:

• O2 has 2 unpaired electrons in antibonding π* orbitals causing paramagnetism.

Practice Set

Q1: What is the magnetic property of dioxygen?

• A. Diamagnetic

• B. Paramagnetic

• C. Ferromagnetic

• D. Non-magnetic

Q2: Which of the following describes dioxygen?

• A. Colourless, odourless, tasteless gas

• B. Coloured, odourless gas

• C. Colourless, odorous liquid

• D. Coloured, odorous solid

Q3: The paramagnetism of O2 is due to:

• A. Paired electrons in bonding orbitals

• B. Unpaired electrons in antibonding orbitals

• C. Absence of electrons

• D. Presence of lone pairs

Answer Key

Q1: B. Paramagnetic - Due to unpaired electrons.

Q2: A. Colourless, odourless, tasteless gas - Standard physical description.

Q3: B. Unpaired electrons in antibonding orbitals - Cause paramagnetism.

Memory Aids

• O2 is paramagnetic: remember "two unpaired electrons".

• Physical state: colourless, odourless, tasteless gas.

Dioxygen Chemical Properties

Reactivity and Reactions

• Reacts with most metals (except noble metals like Au, Pt) to form oxides.

• Reacts with non-metals and some noble gases.

• Examples of reactions:

  • Ca + O2 → CaO (basic oxide)

  • 4Al + 3O2 → 2Al2O3 (amphoteric oxide)

  • 4P + 5O2 → 2P2O5 (acidic oxide)

  • C + O2 → CO2 (acidic oxide)

  • 2ZnS + 3O2 → 2ZnO + 2SO2 (roasting)

  • CH4 + 2O2 → CO2 + 2H2O (combustion)

  • 2SO2 + O2 (V2O5 catalyst) → 2SO3 (catalytic oxidation)

  • 4HCl + O2 (CuCl2 catalyst) → 2Cl2 + 2H2O (catalytic oxidation)

Solved Examples

Ex 1: Write the reaction of aluminum with oxygen.

Solution:

• 4Al + 3O2 → 2Al2O3

Ex 2: Write the reaction for catalytic oxidation of SO2 to SO3.

Solution:

• 2SO2 + O2 (V2O5) → 2SO3

Practice Set

Q1: Which oxide formed by oxygen is amphoteric?

• A. CaO

• B. Al2O3

• C. P2O5

• D. CO2

Q2: What is the product of roasting zinc sulfide?

• A. Zn + SO2

• B. ZnO + SO2

• C. ZnS + O2

• D. ZnSO4

Q3: Which catalyst is used in the oxidation of SO2 to SO3?

• A. CuCl2

• B. V2O5

• C. MnO2

• D. Fe2O3

Answer Key

Q1: B. Al2O3 - Amphoteric oxide reacts with acids and bases.

Q2: B. ZnO + SO2 - Roasting converts ZnS to ZnO and SO2.

Q3: B. V2O5 - Catalyst for SO2 oxidation to SO3.

Memory Aids

• Basic oxide: CaO; Amphoteric: Al2O3; Acidic: P2O5, CO2.

• Roasting: ZnS + O2 → ZnO + SO2.

• V2O5 catalyzes SO2 oxidation; CuCl2 catalyzes HCl 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).

Solved Examples

• Ex 1: Identify the type of oxide: MgO.
  Solution: MgO contains one metal and oxygen; it is a simple basic oxide.

• Ex 2: Write the product when P4O10 reacts with water.
  Solution: P4O10 + 6H2O → 4H3PO4 (phosphoric acid).

Practice Set

• Q1: Which oxide is amphoteric?
  A. Na2O
  B. Al2O3
  C. SO2
  D. CO

• Q2: What is the product of N2O5 reacting with water?
  A. HNO2
  B. HNO3
  C. H2SO4
  D. H3PO4

• Q3: Which oxide is neutral?
  A. CO
  B. SO3
  C. Na2O
  D. Al2O3

Answer Key

• Q1: B. Al2O3 - It reacts with both acids and bases.

• Q2: B. HNO3 - N2O5 + H2O → 2HNO3.

• Q3: A. CO - Neutral oxide; does not react with acids or bases.

Memory Aids

• Acidic oxides + water → acids (e.g., SO2, P4O10).

• Basic oxides + water → bases (e.g., Na2O, CaO).

• 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

Properties and Reactions

• Dissolve in water to form acids.

• Examples and reactions:
  P4O10 + 6H2O → 4H3PO4
  N2O5 + H2O → 2HNO3
  2NO2 + H2O → HNO2 + HNO3
  P4O6 + 6H2O → 4H3PO3

• Acidic strength order for dioxides: SO2 > SeO2 > TeO2 > PoO2

• Acidic strength order for trioxides: SO3 > SeO3 > TeO3

Solved Examples

• Ex 1: Write the acid formed when P4O10 reacts with water.
  Solution: Phosphoric acid (H3PO4).

• Ex 2: Predict the product of N2O5 + H2O.
  Solution: Nitric acid (HNO3).

Practice Set

• Q1: Which oxide forms HNO3 on reaction with water?
  A. N2O5
  B. SO2
  C. CO2
  D. NO

• Q2: Arrange the acidic strength of dioxides in decreasing order.
  A. TeO2 > SO2 > SeO2
  B. SO2 > SeO2 > TeO2
  C. SeO2 > SO2 > TeO2
  D. PoO2 > TeO2 > SO2

• Q3: Which acid is formed from P4O6 + H2O?
  A. H3PO4
  B. H3PO3
  C. HNO3
  D. H2SO4

Answer Key

• Q1: A. N2O5 - Reacts with water to form nitric acid.

• Q2: B. SO2 > SeO2 > TeO2 - Correct acidic strength order.

• Q3: B. H3PO3 - Phosphorous acid formed from P4O6.

Memory Aids

• Acidic oxides + H2O → corresponding acids.

• Stronger acidity: SO > Se > Te > Po in oxides.

• Phosphorus oxides: P4O10 → H3PO4, P4O6 → H3PO3.

Basic Oxides

Properties and Reactions

• Dissolve in water to form bases (alkalis).

• Examples and reactions:
  Na2O + H2O → 2NaOH
  K2O + H2O → 2KOH
  CaO + H2O → Ca(OH)2

• React with acids to form salts and water.

Solved Examples

• Ex 1: Write the base formed when Na2O reacts with water.
  Solution: Sodium hydroxide (NaOH).

• Ex 2: Write the product of CaO + H2O.
  Solution: Calcium hydroxide Ca(OH)2.

Practice Set

• Q1: Which oxide forms KOH on reaction with water?
  A. K2O
  B. CO2
  C. SO3
  D. Al2O3

• Q2: What is the product of CaO + H2O?
  A. CaO
  B. Ca(OH)2
  C. CaH2
  D. CaOH

• Q3: Which of the following is a basic oxide?
  A. SO2
  B. Na2O
  C. CO
  D. Al2O3

Answer Key

• Q1: A. K2O - Reacts with water to form KOH.

• Q2: B. Ca(OH)2 - Calcium hydroxide formed.

• Q3: B. Na2O - Basic oxide.

Memory Aids

• Basic oxides + H2O → alkalis (NaOH, KOH, Ca(OH)2).

• Alkali metals and alkaline earth metals form basic oxides.

Amphoteric Oxides

Properties and Reactions

• React with both acids and bases.

• Example: Aluminum oxide (Al2O3).

• Reactions:
  Al2O3 + 6HCl → 2AlCl3 + 3H2O
  Al2O3 + 6NaOH + 3H2O → 2Na3[Al(OH)6]
  Al2O3 + 2NaOH → 2NaAlO2 + H2O (Sodium aluminate)

Solved Examples

• Ex 1: Write the product when Al2O3 reacts with HCl.
  Solution: Aluminum chloride (AlCl3) and water.

• Ex 2: Write the product when Al2O3 reacts with NaOH.
  Solution: Sodium aluminate (NaAlO2) and water.

Practice Set

• Q1: Which oxide reacts with both acids and bases?
  A. Na2O
  B. SO2
  C. Al2O3
  D. CO

• Q2: What is formed when Al2O3 reacts with NaOH?
  A. NaCl
  B. NaAlO2
  C. AlCl3
  D. NaOH

• Q3: Write the balanced reaction of Al2O3 with HCl.
  A. Al2O3 + 6HCl → 2AlCl3 + 3H2O
  B. Al2O3 + 2HCl → 2AlCl + H2O
  C. Al2O3 + HCl → AlCl + H2O
  D. Al2O3 + 3HCl → AlCl3 + H2O

Answer Key

• Q1: C. Al2O3 - Amphoteric oxide.

• Q2: B. NaAlO2 - Sodium aluminate formed.

• Q3: A. Al2O3 + 6HCl → 2AlCl3 + 3H2O - Correct balanced reaction.

Memory Aids

• Amphoteric oxides react with acids and bases.

• Al2O3 is a classic amphoteric oxide.

• Reactions produce salts like AlCl3 and sodium aluminate.

Ozone

Preparation

• Ozone (O3) is prepared by passing oxygen gas (O2) through an electric discharge.

• Electric discharge breaks O=O bonds in O2 molecules into oxygen atoms.

• Oxygen atoms recombine with O2 molecules to form ozone: 3O2 \( \xrightarrow{electric\ discharge} \) 2O3.

Properties

• Ozone is a pale blue gas, dark blue liquid, and violet-black solid.

• Decomposes to oxygen and atomic oxygen: O3 \( \rightarrow \) O2 + O.

• Strong oxidizing agent: PbS + 4O3 \( \rightarrow \) PbSO4 + 4O2.

• Can act 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 is sp2 hybridized.

• Ozone has an angular (bent) shape.

• Ozone is a resonance hybrid of two structures with alternating single and double bonds.

• O–O bond lengths are equal (~128 pm) due to resonance.

Solved Examples

• Ex 1: Write the reaction for ozone preparation from oxygen.
  Answer: 3O2 \( \xrightarrow{electric\ discharge} \) 2O3.

• Ex 2: Show the reaction of ozone with PbS.
  Answer: PbS + 4O3 \( \rightarrow \) PbSO4 + 4O2.

Practice Set

• Q1: What is the hybridization of the central oxygen atom in ozone?
  A. sp
  B. sp2
  C. sp3
  D. sp3d

• Q2: Which of the following reactions represents ozone acting as an oxidizing agent?
  A. 2I− + H2O + O3 \( \rightarrow \) 2OH− + I2 + O2
  B. PbS + 4O3 \( \rightarrow \) PbSO4 + 4O2
  C. NO + O3 \( \rightarrow \) NO2 + O2
  D. O3 \( \rightarrow \) O2 + O

• Q3: Calculate the number of ozone molecules formed when 9 molecules of oxygen react under electric discharge.
  A. 6
  B. 9
  C. 3
  D. 12

Answer Key

• Q1: B. sp2 hybridization gives ozone its angular shape.

• Q2: B. PbS + 4O3 \( \rightarrow \) PbSO4 + 4O2 shows ozone as oxidizing agent.

• Q3: A. Using 3O2 \( \rightarrow \) 2O3, 9 O2 molecules produce 6 O3 molecules.

Memory Aids

• Ozone preparation: "3 oxygen molecules make 2 ozone molecules with electric spark."

• Ozone structure: "Resonance makes bonds equal and shape bent (sp2 hybrid)."

• Oxidizing action: PbS + O3 \( \rightarrow \) PbSO4 + O2 (remember PbS to PbSO4).

Sulphur Allotropic Forms

Rhombic Sulphur (α-sulphur)

• Consists of S8 molecules; yellow crystals.

• Melting point: 385.8 K; specific gravity: 2.06.

• Formed by evaporating solution of roll sulphur in CS2.

• Insoluble in water; slightly soluble in benzene, alcohol, ether.

• Readily soluble in carbon disulfide (CS2).

Monoclinic Sulphur (β-sulphur)

• Also S8 molecules; colourless 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 Sulfide (H2S)

• Also called sulphurated hydrogen.

• Poisonous; fatal in large amounts.

• Antidote: dilute chlorine oxidizes H2S to sulphur.

• Reaction: H2S + Cl2 → 2HCl + S

Solved Examples

• Ex 1: Identify the form of sulphur with melting point 393 K.
  Answer: Monoclinic sulphur (β-sulphur).

• Ex 2: Write the reaction of H2S with chlorine.
  Answer: H2S + Cl2 → 2HCl + S.

Practice Set

• Q1: Which sulphur allotrope is yellow and has a melting point of 385.8 K?
  A. Monoclinic sulphur
  B. Rhombic sulphur
  C. Cyclo S6
  D. Amorphous sulphur

• Q2: What is the shape of the S8 ring in rhombic sulphur?
  A. Chair form
  B. Planar ring
  C. Puckered crown shape
  D. Linear chain

• Q3: What is the product formed when H2S reacts with chlorine?
  A. HCl and SO2
  B. HCl and S
  C. H2SO4
  D. H2O and S

Answer Key

• Q1: B. Rhombic sulphur - It is yellow and melts at 385.8 K.

• Q2: C. Puckered crown shape - S8 rings in rhombic sulphur have this shape.

• Q3: B. HCl and S - Reaction: H2S + Cl2 → 2HCl + S.

Memory Aids

• Rhombic sulphur: "R" for "Yellow" (think of ripe yellow rhombus).

• S8 ring shape: "Crown" shape for both rhombic and monoclinic.

• H2S antidote: Chlorine "cleans" H2S by oxidation.

Sulphur Dioxide

Preparation

• Direct combustion: S + O2 → SO2

• Laboratory: Sodium sulfite reacts with acid: SO32- + 2H+ → H2O + SO2

• Industrial by-product: Roasting of pyrite ore: 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Physical Properties

• Colourless gas with pungent smell

• Highly soluble in water

• Liquefies at room temperature under 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 acts as a bleaching agent by reducing coloured compounds:

• SO2 + 2H2O → H2SO4 + 2[H]

• Coloured matter + [H] → Colourless matter

Reducing Nature

• Reduces Fe3+ to Fe2+: 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+

• Reduces permanganate ion: 5SO2 + 2MnO4- + 2H2O → 5SO42- + 4H+ + 2Mn2+

Structure

• SO2 molecule is angular (bent shape)

• Resonance hybrid of two canonical forms with alternating double and single bonds between S and O atoms

• Lone pairs on sulfur and oxygen atoms contribute to resonance

• Resonance explains equal bond lengths and molecular properties

Solved Examples

• Ex 1: Write the reaction for laboratory preparation of SO2.

• Solution: Sodium sulfite reacts with acid: SO32- + 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 of SO2 preparation?
  A. S + O2 → SO2
  B. 2SO2 + O2 → 2SO3
  C. SO2 + Cl2 → SO2Cl2
  D. SO3 + H2O → H2SO4

• Q2: SO2 acts as a bleaching agent because it:
  A. Oxidizes coloured matter
  B. Reduces coloured matter
  C. Forms SO3
  D. Reacts with NaOH

• Q3: Which statement about SO2 structure is correct?
  A. Linear molecule
  B. Resonance hybrid with angular shape
  C. No lone pairs on sulfur
  D. Single bond only between S and O

Answer Key

• Q1: A. S + O2 → SO2 is the direct preparation method.

• Q2: B. SO2 reduces coloured matter, causing bleaching.

• Q3: B. SO2 is an angular molecule and a resonance hybrid.

Memory Aids

• SO2 preparation: "Burning Sulfur" (S + O2)

• Bleaching action: SO2 reduces coloured compounds to colourless

• Structure: Remember "Bent SO2 with resonance"

Oxoacids of Sulphur

Structures and Composition

• Sulphurous acid (H2SO3): Sulfur bonded to 3 oxygen atoms and 2 hydrogens.

• Sulphuric acid (H2SO4): Sulfur bonded to 4 oxygen atoms and 2 hydrogens.

• Peroxodisulphuric acid (H2S2O8): Two sulfur atoms connected by an O–O bond, with additional oxygens and hydrogens.

• Pyrosulphuric acid (Oleum, H2S2O7): Two sulfur atoms connected by one oxygen atom, plus more oxygens and hydrogens.

Solved Examples

• Ex 1: Identify the number of sulfur atoms in H2S2O8.
  Step 1: Look at formula H2S2O8.
  Step 2: Count sulfur atoms = 2.

• Ex 2: Differentiate between the oxygen linkage in peroxodisulphuric acid and pyrosulphuric acid.
  Step 1: Peroxodisulphuric acid has O–O (peroxo) bond.
  Step 2: Pyrosulphuric acid has single O bridge between sulfurs.

Practice Set

• Q1: Which acid contains a peroxide (O–O) bond?
  A. Sulphurous acid
  B. Sulphuric acid
  C. Peroxodisulphuric acid
  D. Pyrosulphuric acid

• Q2: What is the molecular formula of Oleum?
  A. H2SO3
  B. H2SO4
  C. H2S2O7
  D. H2S2O8

• Q3: How many oxygen atoms are bonded directly to sulfur in sulphuric acid?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: C. Peroxodisulphuric acid contains the O–O peroxide bond, unique among these acids.

• Q2: C. Oleum is pyrosulphuric acid with formula H2S2O7.

• Q3: C. Sulphuric acid has 4 oxygen atoms bonded directly to sulfur.

Memory Aids

• "Peroxo" means O–O bond: only peroxodisulphuric acid has it.

• Oleum = Pyrosulphuric acid = two sulfurs + one bridging oxygen.

• Sulphuric acid has one more oxygen than sulphurous acid.

Sulphuric Acid

General Information

• Also called oil of vitriol and known as the king of chemicals.

• Chemical formula: H2SO4.

Preparation (Contact Process)

• Step 1: Sulfur (S) burns in oxygen (O2) to form sulfur dioxide (SO2).

• Step 2: SO2 oxidizes to sulfur trioxide (SO3) using V2O5 catalyst.

• Step 3: SO3 reacts with H2SO4 to form oleum (H2S2O7).

• Step 4: Oleum is diluted with water to produce concentrated sulfuric acid.

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 dibasic acid: loses two protons stepwise forming HSO4− and SO42−.

• Reacts with bases (e.g., NaOH) to form normal and acid sulfates.

• Reacts with metal halides (MX) to produce hydrogen halides (HX) and metal sulfates (M2SO4).

• Acts as a strong dehydrating agent, removing water from organic compounds (e.g., sugars) leaving carbon.

• Reacts with metals like Cu and S producing CuSO4, SO2, and H2O.

• Reacts with carbon to produce CO2, SO2, and H2O.

Solved Examples

• Ex 1: Write the reaction of SO2 to SO3 in the contact process.
  Answer: SO2 + 1/2 O2 10 V2O5 catalyst 14 SO3

• Ex 2: Reaction of concentrated H2SO4 with sugar (C12H22O11).
  Answer: C12H22O11 + H2SO4 (conc.) 14 12 C + 11 H2O (dehydration)

Practice Set

• Q1: What is the catalyst used in the contact process?
  A. Fe2O3   B. V2O5   C. MnO2   D. Pt

• Q2: Which product is formed when SO3 reacts with H2SO4?
  A. Sulfuric acid   B. Oleum   C. Sulfur dioxide   D. Sulfur trioxide

• Q3: Concentrated H2SO4 acts as a dehydrating agent with sugar. What is the main product?
  A. CO2   B. Carbon   C. Water   D. Glucose

Answer Key

• Q1: B. V2O5 - It catalyzes SO2 oxidation to SO3.

• Q2: B. Oleum - SO3 reacts with H2SO4 to form oleum (H2S2O7).

• Q3: B. Carbon - H2SO4 removes water from sugar leaving carbon.

Memory Aids

• Contact process catalyst: "V for Vanadium".

• Always add acid to water to avoid splashing.

• Oleum = SO3 + H2SO4.

• H2SO4 is dibasic: loses 2 protons stepwise.

• Dehydration of sugar by H2SO4 leaves black carbon.

Group 17 Elements

Overview

• 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; At and Ts are radioactive.

• General electronic configuration: ns2np5.

Physical Properties

• Atomic and ionic radii increase down the group; halogens are smallest atoms in their periods.

• Ionisation enthalpy is high but decreases down the group.

• Electronegativity decreases down the group; Fluorine is the most electronegative.

• Electron gain enthalpy increases from F to Cl, then decreases to I; F's smaller value due to small size and electron repulsion.

• 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 (+1 to +7) seen in fluorides, oxides, and oxoacids of Cl, Br, I.

• 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 those 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 general electronic configuration of halogens.

Solution: ns2np5

Ex 2

Arrange halogens in order of oxidising power.

Solution: F2 > Cl2 > Br2 > I2

Ex 3

Predict the product of reaction: 2KCO3 + I2 → ?

Solution: 2KIO3 + Cl2

Practice Set

• Q1: Which halogen has the highest electronegativity?
  A. Chlorine
  B. Fluorine
  C. Bromine
  D. Iodine

• Q2: Which hydrogen halide has the highest acidic strength?
  A. HF
  B. HCl
  C. HBr
  D. HI

• Q3: Which halogen shows only -1 oxidation state?
  A. Chlorine
  B. Bromine
  C. Fluorine
  D. Iodine

Answer Key

• Q1: B. Fluorine — It is the most electronegative element in the group.

• Q2: D. HI — Acidic strength increases down the group; HI is strongest acid.

• Q3: C. Fluorine — Only shows -1 oxidation state due to high electronegativity.

Memory Aids

• Halogen order in oxidising power: F2 > Cl2 > Br2 > I2.

• Acidic strength of HX: HI > HBr > HCl > HF.

• General electronic configuration: ns2np5.

• Ionic character decreases down the group: MF > MCl > MBr > MI.

Chlorine Preparation

Industrial Methods

• MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

• KMnO4 + 8HCl → KCl + MnCl2 + 4H2O + 5Cl2

• NaCl + H2SO4 + MnO2 → NaHSO4 + MnCl2 + Cl2

• Deacon’s process: 4HCl + O2 (CuCl2 catalyst) → 2Cl2 + 2H2O

Electrolytic Process

• Electrolysis of brine (concentrated NaCl solution)

• Chlorine gas is liberated at the anode

Solved Examples

• Ex 1: Write the reaction for chlorine production using MnO2 and HCl.
  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 liberated at which electrode during brine electrolysis?
  A. Cathode
  B. Anode
  C. Both electrodes
  D. None

• Q3: Identify the product formed when KMnO4 reacts with HCl.
  A. KCl + MnCl2 + Cl2
  B. KCl + MnO2 + Cl2
  C. KCl + MnCl2 + H2
  D. KCl + MnO2 + H2

Answer Key

• Q1: A - CuCl2 acts as catalyst in Deacon’s process for oxidation of HCl to Cl2.

• Q2: B - Chlorine gas is liberated at the anode during electrolysis of brine.

• Q3: A - KMnO4 reacts with HCl to form KCl, MnCl2, and Cl2.

Memory Aids

• Deacon’s process uses CuCl2 catalyst.

• Electrolysis of brine: Anode → Cl2, Cathode → H2.

• MnO2 + HCl → Cl2 + MnCl2 + H2O (common lab prep).

Chlorine Physical Properties

Key Characteristics

• Greenish-yellow gas with pungent, suffocating odor

• Density about 2.5 times that of air

• Liquefies easily to greenish-yellow liquid

• Boiling point: 239 K (-34 °C)

• Soluble in water forming a pale green solution

Solved Examples

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

Practice Set

• Q1: Chlorine gas is:
  A. Colorless and odorless
  B. Greenish-yellow with pungent odor
  C. Blue with sweet odor
  D. Red with no odor

• Q2: The boiling point of chlorine is:
  A. 273 K
  B. 239 K
  C. 300 K
  D. 200 K

• Q3: Chlorine gas is approximately how many times heavier than air?
  A. 1.5
  B. 2.5
  C. 3.5
  D. 4.5

Answer Key

• Q1: B - Chlorine is greenish-yellow with a pungent odor.

• Q2: B - Chlorine boils at 239 K (-34 °C).

• Q3: B - Chlorine is about 2.5 times heavier than air.

Memory Aids

• Chlorine color: Greenish-yellow (think "green" like leaves).

• Boiling point below room temp: 239 K (~ -34 °C).

• Heavier than air: 2.5 times.

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 Alkalis

• 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 Oxidizing 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 at cold conditions.
  Answer: 2NaOH + Cl2 → NaCl + NaOCl + H2O

• Ex 2: Reaction of chlorine with ammonia in excess ammonia.
  Answer: 8NH3 + 3Cl2 → 6NH4Cl + N2

Practice Set

• Q1: What is the product when chlorine reacts with hot concentrated NaOH?
  A. NaCl + NaOCl + H2O
  B. NaCl + NaClO3 + H2O
  C. NaCl + NaClO + H2O
  D. NaCl + NaOH + HCl

• Q2: Chlorine reacts with ammonia in excess ammonia to form:
  A. NCl3
  B. NH4Cl and N2
  C. NH3Cl
  D. NH4Cl only

• Q3: Which of the following is NOT a product of chlorine reaction with metals?
  A. AlCl3
  B. FeCl3
  C. NaCl
  D. NH4Cl

Answer Key

• Q1: B - Hot concentrated NaOH with Cl2 forms NaCl, NaClO3, and water.

• Q2: B - Excess ammonia reacts with chlorine to give ammonium chloride and nitrogen gas.

• Q3: D - NH4Cl is formed from ammonia and chlorine, not from metals.

Memory Aids

• Cold dilute NaOH + Cl2 → Bleach (NaOCl)

• Hot concentrated NaOH + Cl2 → NaClO3 (chlorate)

• Excess NH3 + Cl2 → NH4Cl + N2

• Excess Cl2 + NH3 → NCl3 (explosive)

Hydrogen Chloride

Lab Preparation

• Industrial Mannheim process involves two steps:

• At 420 K: NaCl + H2SO4 → NaHSO4 + HCl (gas)

• At 823 K: NaHSO4 + NaCl → Na2SO4 + HCl (gas)

• HCl gas is dried by passing through concentrated H2SO4

Physical Properties

• Colourless gas with pungent smell

• Easily liquefied

• Highly soluble in water forming hydrochloric acid

Chemical Properties

• Reacts with salts of weaker acids to decompose them:

• 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: Show reaction of HCl with sodium carbonate.
  Answer: Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Practice Set

• Q1: Which acid is used to dry HCl gas?
  A. HNO3
  B. Conc. H2SO4
  C. H3PO4
  D. Acetic acid

• Q2: In the Mannheim process, at what temperature does NaHSO4 react with NaCl?
  A. 300 K
  B. 420 K
  C. 823 K
  D. 1000 K

• Q3: Aqua regia dissolves gold by forming which complex?
  A. AuCl2−
  B. AuCl3−
  C. AuCl4−
  D. AuCl6−

Answer Key

• Q1: B - Conc. H2SO4 absorbs moisture and dries HCl gas effectively.

• Q2: C - At 823 K, NaHSO4 reacts with NaCl to form Na2SO4 and HCl.

• Q3: C - Gold forms AuCl4− complex in aqua regia.

Memory Aids

• Mannheim process: Two-step reaction at 420 K and 823 K for HCl production.

• Drying HCl: Use concentrated H2SO4.

• Aqua regia ratio: 3 parts HCl + 1 part HNO3 dissolves noble metals.

Oxoacids of Halogens

Properties and Trends

• Halogens form oxoacids with oxidation states +1, +3, +5, and +7.

• Thermal stability increases with higher oxidation states.

• Halogen-oxygen bonds become more covalent as oxidation state rises.

• Oxidizing power decreases with increasing oxidation number.

• Acidity increases with higher oxidation states.

• Electronegativity of halogens remains constant across these acids.

• Opposite trends observed: electronegativity vs. acid strength and oxidizing power.

Solved Examples

• Ex 1: Identify the acid with highest acidity among HClO, HClO2, HClO3, and HClO4.
  Solution: HClO4 has Cl in +7 oxidation state, highest acidity.

• Ex 2: Predict the trend in oxidizing power for HBrO, HBrO3, and HBrO4.
  Solution: Oxidizing power decreases from HBrO (+1) to HBrO4 (+7).

Practice Set

• Q1: Which oxoacid of chlorine is the strongest acid?
  A. HClO
  B. HClO2
  C. HClO3
  D. HClO4

• Q2: As the oxidation state of halogen in oxoacids increases, which property decreases?
  A. Thermal stability
  B. Acidity
  C. Oxidizing power
  D. Covalent character of bond

• Q3: Arrange the following oxoacids in order of increasing thermal stability: HIO, HIO3, HIO4.
  A. HIO < HIO3 < HIO4
  B. HIO4 < HIO3 < HIO
  C. HIO3 < HIO < HIO4
  D. HIO < HIO4 < HIO3

Answer Key

• Q1: D. HClO4 — Highest oxidation state (+7) gives strongest acidity.

• Q2: C. Oxidizing power — It decreases as oxidation state increases.

• Q3: A. HIO < HIO3 < HIO4 — Thermal stability increases with oxidation state.

Memory Aids

• Higher oxidation state → Stronger acid, more stable, less oxidizing.

• Remember: HClO < HClO2 < HClO3 < HClO4 in acidity.

• Oxidizing power trend opposite to acidity and stability.

Interhalogen Compounds

Definition and Types

• Interhalogens are compounds formed by combination of two different halogens.

• General formulas: XX', XX'3, XX'5, XX'7 where X is the larger halogen and X' is the smaller halogen.

• Examples: ClF, ICl3, BrF5, IF7.

Preparation

• Halogen molecules react under controlled temperature and stoichiometric conditions.

• Key reactions:

  • 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

• Molecular shapes depend on hybridization and number of bonded atoms/groups.

• Shapes and hybridizations:

  • XX' (2 groups): Linear, sp3 hybridization

  • XX'3 (3 groups): T-shaped, 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, behaving like halide ions.

• Examples of pseudohalide ions: CN−, SCN−, OCN−.

• Corresponding dimeric molecules (pseudohalogens): (CN)2, (SCN)2, (OCN)2.

Solved Examples

• Ex 1: Write the reaction to prepare ClF3.

• Step 1: React Cl2 with excess F2 at 573 K.

• Reaction: Cl2 + 3F2 → 2ClF3

Practice Set

• Q1: Which interhalogen has a T-shaped molecular geometry?

  • A. ClF

  • B. ICl3

  • C. BrF5

  • D. IF7

• Q2: What is the hybridization of the central atom in IF7?

  • A. sp3

  • B. sp3d

  • C. sp3d2

  • D. sp3d3

• Q3: Which reaction conditions are correct for the formation of BrF5?

  • A. BrF3 + F2 at 473 K

  • B. Br2 + F2 at 437 K

  • C. Br2 + Cl2 at 573 K

  • D. BrF3 + Cl2 at 543 K

Answer Key

• Q1: B. ICl3 - It has T-shaped geometry due to sp3d hybridization with 3 bonded atoms.

• Q2: D. sp3d3 - IF7 has 7 bonded atoms, pentagonal bipyramidal shape.

• Q3: A. BrF3 + F2 at 473 K - Correct condition for BrF5 formation.

Memory Aids

• Interhalogen formulas: Number of X' atoms = 1, 3, 5, or 7 for linear, T-shaped, square pyramidal, pentagonal bipyramidal shapes.

• Higher fluorine excess and temperature increase fluorination level (e.g., ClF → ClF3).

• Pseudohalogens mimic halide behavior; remember CN−, SCN−, OCN−.

Group 18 Elements

General Characteristics

• Elements: He, Ne, Ar, Kr, Xe, Rn, Og

• All are monoatomic gases, 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

• Melting and boiling points are very low due to weak dispersion forces

• Chemically least reactive due to stable electronic configuration

Solved Examples

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

Practice Set

• Q1: Which noble gas has the highest ionization enthalpy?
  A. Helium
  B. Neon
  C. Argon
  D. Xenon

• Q2: Which property increases down the group in noble gases?
  A. Ionization enthalpy
  B. Atomic radius
  C. Electron gain enthalpy
  D. Chemical reactivity

• Q3: Noble gases are chemically inert because:
  A. They have incomplete octet
  B. They have high ionization enthalpy and positive electron gain enthalpy
  C. They have low atomic size
  D. They have low melting points

Answer Key

• Q1: A. Helium - Smallest size and highest nuclear charge leads to highest ionization enthalpy.

• Q2: B. Atomic radius - Increases down the group due to addition of shells.

• Q3: B. High ionization enthalpy and positive electron gain enthalpy make them inert.

Memory Aids

• "He Never Argues Kindly Xenon Radon Oganesson" - Order of noble gases

• Stable ns2 np6 configuration = inertness

• Ionization enthalpy decreases down, atomic size increases down

Xenon-Fluorine Compounds

Reactions and Properties

• Xe + F2 (excess) at 673 K, 1 bar → XeF2

• Xe + 5F2 at 873 K, 7 bar → XeF4

• Xe + 20F2 at 573 K, 60-70 bar → XeF6

• XeF6 also prepared by: XeF4 + O2F2 → XeF6 + O2 (143 K)

• 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 XeF4.
  Answer: Xe + 2F2 (1:5 ratio) at 873 K and 7 bar → XeF4

Practice Set

• Q1: Which xenon fluoride is formed at 673 K and 1 bar?
  A. XeF2
  B. XeF4
  C. XeF6
  D. XeF

• Q2: XeF6 can be prepared by reaction of XeF4 with:
  A. F2
  B. O2F2
  C. H2O
  D. HF

• Q3: Hydrolysis of XeF2 produces:
  A. Xe + HF + O2
  B. XeF4
  C. XeO3
  D. XeF6

Answer Key

• Q1: A. XeF2 - Formed at 673 K, 1 bar with excess fluorine.

• Q2: B. O2F2 - Reacts with XeF4 to form XeF6.

• Q3: A. Xe + HF + O2 - Hydrolysis of XeF2 yields these products.

Memory Aids

• XeF2, XeF4, XeF6 formed by increasing fluorine, temperature, pressure

• Hydrolysis breaks Xe-F bonds producing Xe, HF, and O2

• Fluoride ion acceptors/donors form cations and fluoroanions with xenon fluorides

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 product of complete hydrolysis of XeF6.
  Answer: XeO3 + HF

Practice Set

• Q1: XeF6 reacts with water to form:
  A. XeO3
  B. XeF4
  C. XeO2F2
  D. XeF2

• Q2: Partial hydrolysis of XeF6 produces:
  A. XeO3
  B. XeOF4
  C. XeO2F2
  D. XeF4

• Q3: Which gas is released during hydrolysis of XeF4?
  A. O2
  B. F2
  C. Xe
  D. HF

Answer Key

• Q1: A. XeO3 - Product of complete hydrolysis of XeF6.

• Q2: C. XeO2F2 - Formed by partial hydrolysis of XeF6.

• Q3: A. O2 - Released during hydrolysis of XeF4.

Memory Aids

• Complete hydrolysis of XeF6 → XeO3

• Partial hydrolysis → mixed oxo-fluorides like XeO2F2

• Hydrolysis releases HF and O2

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 shape, sp3 hybridization

• In solid state, XeF6 forms tetrameric and hexameric units with square pyramidal XeF5+ linked by bent F− bridges

Solved Examples

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

Practice Set

• Q1: The hybridization of Xe in XeF2 is:
  A. sp3d
  B. sp3d2
  C. sp3d3
  D. sp2

• Q2: XeF6 has which molecular geometry in gas phase?
  A. Octahedral
  B. Distorted octahedral
  C. Square planar
  D. Linear

• Q3: In solid state, XeF6 contains:
  A. Monomeric units
  B. Tetrameric and hexameric units
  C. Only hexameric units
  D. Only tetrameric units

Answer Key

• Q1: A. sp3d - Hybridization for linear XeF2.

• Q2: B. Distorted octahedral - Gas phase geometry of XeF6.

• Q3: B. Tetrameric and hexameric units - Solid state structure of XeF6.

Memory Aids

• XeF2 linear (sp3d), XeF4 square planar (sp3d2)

• XeF6 distorted octahedral (sp3d3)

• Solid XeF6 forms clusters linked by F− bridges