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WBB • Class XII • Chemistry • Ch 2
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p-Block Elements (Groups 15, 16, 17 and 18)

The p-block elements of Groups 15, 16, 17, and 18 constitute the most diverse, chemically varied, and industrially critical domain of the periodic table in the WBCHSE Class 12 Chemistry curriculum. Spanning non-metals, metalloids, and reactive non-metallic halogens to noble gas aerogens, these groups exhibit remarkable periodic trends governed by electronic configurations (ns2 np3 to ns2 np6), effective nuclear charge, electronegativity, atomic and ionic radii, oxidation states, and the distinct absence of vacant d-orbitals in period 2 elements. This master study module delivers a rigorous, board-calibrated examination of dinitrogen, ammonia, nitric acid, allotropic forms of phosphorus, phosphine, halides and oxoacids of phosphorus in Group 15; dioxygen, ozone, allotropic sulfur modifications, sulfur dioxide, the Contact Process, and oxoacids of sulfur in Group 16; halogens, hydrogen halides, anomalous fluorine properties, interhalogen compounds, and halogen oxoacids in Group 17; and the landmark synthesis of noble gas compounds by Neil Bartlett, xenon fluorides, oxides, oxyfluorides, and their VSEPR structural geometry in Group 18.

Why This Chapter Matters

p-Block elements drive foundational industrial chemistry, global agriculture, environmental science, and modern technology. Ammonia synthesized via the Haber-Bosch process sustains planetary food production through nitrogenous fertilizers, while nitric acid manufactured via Ostwald's method forms the cornerstone of dye, pharmaceutical, and explosive industries. In Group 16, sulfuric acid is recognized as the chemical barometer of a nation's industrial economy, synthesized via the catalytic Contact Process using vanadium pentoxide. Ozone acts as the planetary ultraviolet shield in the stratosphere, while its laboratory oxidizing power allows quantitative environmental assays. In Group 17, halogens and their interhalogens provide potent oxidizing agents, non-aqueous fluorinating agents, and water disinfection systems. Group 18 noble gases, once deemed entirely chemically inert, reshaped modern valency and chemical bonding theory following Bartlett's 1962 synthesis of xenon hexafluoroplatinate, finding indispensable applications in cryogenic magnetic resonance imaging (MRI), inert atmosphere welding, and deep-sea commercial diving mixtures to prevent decompression sickness.

Chapter Roadmap & Progression

1 Module 1: Group 15 Elements (Pnicto...
2 Module 2: Phosphorus & Its Compound...
3 Module 3: Group 16 Elements (Chalco...
4 Module 4: Oxoacids of Sulfur & Indu...
5 Module 5: Group 17 Elements (Haloge...
6 Module 6: Group 18 Elements (Noble...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Group 15 Elements (Pnictogens — Nitrogen Family)

1. Electronic Configuration & Periodic Trends

Group 15 elements (N, P, As, Sb, Bi) possess the valence shell electronic configuration ns2 np3, featuring a half-filled p-subshell that confers extra thermodynamic stability. Electronegativity, ionization enthalpy, and non-metallic character decrease down the group, whereas atomic radius, metallic character, and density increase progressively. The common oxidation states are -3, +3, and +5. However, due to the inert pair effect (the reluctance of the valence s-electrons to participate in bond formation owing to poor shielding by intervening d and f electrons), the stability of the +5 oxidation state decreases down the group while the stability of the +3 state increases: Bi(V) acts as an exceptionally powerful oxidizing agent (e.g., BiF5).

WBCHSE High-Yield Concept: Anomalous Behavior of Nitrogen
Nitrogen differs significantly from its congeners due to:
  • Extremely small atomic size and high electronegativity: Forms strong pπ-pπ multiple bonds, resulting in a diatomic molecule (:N≡N:) with a colossal bond dissociation enthalpy (945.4 kJ/mol), making N2 chemically inert at room temperature.
  • Absence of vacant d-orbitals: Nitrogen is restricted to a maximum covalency of 4 and cannot expand its octet. Consequently, compounds such as NCl5, NF5, or R3N=O do not exist, whereas P forms PCl5 via sp3d hybridization.
  • Single N-N bond weakness: High interelectronic repulsion between non-bonding lone pairs on the small nitrogen atoms makes the N-N single bond (160 kJ/mol) substantially weaker than the P-P single bond (201 kJ/mol), explaining nitrogen's negligible catenation power.
2. Ammonia (NH₃): Industrial Synthesis, Structure & Basicity

Ammonia is manufactured industrially via Haber's Process by the direct reversible combination of dry dinitrogen and dihydrogen:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   [ΔH° = -92.4 kJ/mol]

According to Le Chatelier's Principle, maximum equilibrium yield of NH3 is favored by: (i) High pressure (~200 atm / 20 MPa) to shift equilibrium toward fewer gaseous moles; (ii) Optimum moderate temperature (~700 K / 450-500°C) balancing exothermic thermodynamics with kinetic activation; (iii) Iron oxide catalyst with promoter additives (K2O and Al2O3).

Molecular Structure: The central nitrogen atom undergoes sp3 hybridization, possessing three bond pairs and one lone pair. The geometry is trigonal pyramidal with a bond angle of 107.8° (reduced from the ideal tetrahedral 109.5° due to lone pair-bond pair repulsion). The localized lone pair renders NH3 a potent Lewis base, forming deep blue complex ions with Cu2+:

Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)   (Tetraamminecopper(II), Deep Blue)
3. Oxides of Nitrogen & Dimerization
Oxide Formula Oxidation State of N Physical State & Acid/Base Magnetic Nature & Key Features
N₂O (Nitrous oxide) +1 Colorless gas, Neutral Diamagnetic, linear (Laughing gas)
NO (Nitric oxide) +2 Colorless gas, Neutral Paramagnetic in gaseous state (odd electron, 11 valence e⁻); dimerizes to diamagnetic N₂O₂ liquid
N₂O₃ (Dinitrogen trioxide) +3 Blue solid, Acidic Diamagnetic, planar, anhydride of HNO₂
NO₂ (Nitrogen dioxide) +4 Brown gas, Acidic Paramagnetic odd-electron molecule; readily dimerizes upon cooling to colorless diamagnetic N₂O₄
N₂O₅ (Dinitrogen pentoxide) +5 Colorless solid, Strongly Acidic Diamagnetic; solid state exists as ionic [NO₂]⁺[NO₃]⁻ (nitronium nitrate), anhydride of HNO₃
4. Nitric Acid (HNO₃): Ostwald Process, Oxidizing Action & Brown Ring Test

Manufactured industrially by Ostwald's Process in three sequential steps:

  • Step 1 (Catalytic Oxidation): 4NH₃(g) + 5O₂(g) xrightarrow[500 K, 9 bar]{Pt/Rh gauge} 4NO(g) + 6H₂O(g)   (Rapid, ~95% conversion).
  • Step 2 (NO Oxidation): 2NO(g) + O₂(g) ⇌ 2NO₂(g).
  • Step 3 (Absorption): 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)   (NO recycled).

Oxidizing Action on Copper: Reaction products depend strictly on acid concentration:

With Dilute HNO₃ (~20%): 3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O
With Concentrated HNO₃ (~70%): Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O

Metal Passivity: Concentrated HNO3 renders metals like Al, Cr, Fe passive due to the immediate formation of an impervious, non-reactive nanoscopic surface oxide film (e.g., Al2O3, Cr2O3).

Brown Ring Test Mechanism: In nitrate solution acidified with dilute H2SO4, freshly prepared FeSO4 reduces NO3- to NO. The NO coordinates to pentaaquairon to yield the brown coordination complex:

[Fe(H₂O)₆]²⁺ + NO + SO₄²⁻ → [Fe(H₂O)₅(NO)]SO₄ + H₂O   (Pentaaquanitrosyliron(I) sulfate)

In this complex, electron transfer occurs from NO to Fe(II), stabilizing the Fe+1 oxidation state (3d7, high-spin, 3 unpaired electrons, magnetic moment μ ≈ 3.87 BM) with coordinated NO+ (nitrosonium ion).

Module 2: Phosphorus & Its Compounds

1. Allotropes of Phosphorus: White, Red & Black

Phosphorus displays multiple allotropic modifications with stark contrasts in stability and chemical reactivity:

Property White Phosphorus (P₄) Red Phosphorus Black Phosphorus (α & β)
Structure Discrete tetrahedral P₄ molecules; P-P-P angle is 60° (severe angular ring strain). Polymeric network of P₄ tetrahedra linked in chains via P-P covalent bridges. Layered puckered hexagonal sheets (graphite-like semiconductor).
Toxicity & Solubility Highly poisonous; insoluble in water, soluble in CS₂. Non-poisonous; insoluble in both water and CS₂. Non-toxic; insoluble in common solvents.
Reactivity & Ignition Highly reactive (ignition temp ~303 K / 30°C); catches fire spontaneously in air yielding dense P₄O₁₀ smoke; exhibits chemiluminescence (green glow in dark). Much less reactive (ignition temp ~533 K / 260°C); does not catch fire at room temperature or glow. Most thermodynamically stable allotrope; completely inert in air at ordinary temperatures.
2. Phosphine (PH₃): Preparation & Vortex Rings

Phosphine is prepared in the laboratory by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2:

P₄ + 3NaOH + 3H₂O → PH₃↑ + 3NaH₂PO₂   (Sodium hypophosphite)

Spontaneous Inflammability: Pure PH3 is non-inflammable. However, laboratory-prepared phosphine contains traces of diphosphine (P2H4) or P4 vapor which spontaneously ignite upon contact with atmospheric oxygen, forming distinctive vortex rings of white P4O10 smoke:

2PH₃ + 4O₂ → P₂O₅ + 3H₂O   (Applications: Holmes' Signals & Smoke Screens)
3. Phosphorus Chlorides: PCl₃ vs PCl₅

Phosphorus Trichloride (PCl₃): Formed by passing dry chlorine over heated white phosphorus: P₄ + 6Cl₂ → 4PCl₃. Hydrolyzes violently with water to form phosphorous acid: PCl₃ + 3H₂O → H₃PO₃ + 3HCl. Geometry is pyramidal (sp3, 3 bp + 1 lp).

Phosphorus Pentachloride (PCl₅): Prepared by reaction with excess chlorine: P₄ + 10Cl₂ → 4PCl₅. In the vapor/gaseous phase, PCl5 exhibits trigonal bipyramidal geometry (sp3d hybridization).

Crucial WBCHSE Examiner Trap: Equivalence of P-Cl Bonds in PCl₅
All five P-Cl bonds in gaseous PCl5 are NOT identical:
  • 3 Equatorial P-Cl bonds: Lie in a plane at 120° angles to each other (shorter and stronger, length = 202 pm).
  • 2 Axial P-Cl bonds: Lie above and below the equatorial plane at 90° to the equatorial bonds (longer and weaker, length = 240 pm). They experience greater repulsion from 3 equatorial pairs, making PCl5 thermally unstable: PCl₅ ⇌ PCl₃ + Cl₂ upon heating.
  • Solid-State Structure: In the solid state, PCl5 does not exist as covalent molecules but as an ionic solid composed of tetrahedral cations and octahedral anions: [PCl₄]⁺[PCl₆]⁻ (hybridizations: sp3 and sp3d2).
4. Oxoacids of Phosphorus: Basicity, Structure & Reducing Action
Name & Formula Oxidation State of P Basicity (Ionizable H) P-H Bonds & Reducing Action
Hypophosphorous acid (H₃PO₂) +1 Monobasic (1 P-OH bond) 2 P-H bonds: Very powerful reducing agent (reduces AgNO₃ to metallic Ag black).
Orthophosphorous acid (H₃PO₃) +3 Dibasic (2 P-OH bonds) 1 P-H bond: Moderately reducing; disproportionates upon heating: 4H₃PO₃ xrightarrow{Δ} 3H₃PO₄ + PH₃.
Orthophosphoric acid (H₃PO₄) +5 Tribasic (3 P-OH bonds) 0 P-H bonds: Non-reducing; stable polyprotic acid.
Pyrophosphoric acid (H₄P₂O₇) +5 Tetrabasic (4 P-OH bonds) 0 P-H bonds: Contains P-O-P bridge; formed by heating H₃PO₄ to ~523 K.

Module 3: Group 16 Elements (Chalcogens — Oxygen Family)

1. Electronic Configuration & Periodic Trends

Group 16 elements (O, S, Se, Te, Po) are termed chalcogens (ore formers), having the valence electronic configuration ns2 np4. Ionization enthalpy is lower than Group 15 counterparts due to extra electron repulsion in the doubly occupied npx orbital. Oxygen exhibits anomalous behavior: it forms pπ-pπ double bonds (O=O), exists as a paramagnetic diatomic gas (explained by Molecular Orbital Theory with 2 unpaired electrons in degenerate π*2p orbitals), has a maximum covalency of 4 (typically 2), and lacks vacant d-orbitals (hence OF6 cannot exist, whereas SF6 is an inert, octahedral gas).

2. Ozone (O₃): Preparation, Thermodynamic Instability & Quantitative Estimation

Ozone is an allotrope of oxygen formed by passing a slow, dry stream of oxygen through a silent electric discharge:

3O₂(g) ⇌ 2O₃(g)   [ΔH° = +142 kJ/mol   (Endothermic)]

A silent electric discharge prevents local overheating which would decompose ozone back to oxygen.

Thermodynamic Instability: Decomposition of ozone is spontaneous and exothermic: 2O₃(g) → 3O₂(g), where ΔH is negative and ΔS is positive (increase in moles). Thus, ΔG = ΔH - TΔS is heavily negative, making high concentrations of ozone potentially explosive.

Potent Oxidizing Action: Ozone releases nascent oxygen readily: O₃ → O₂ + [O]:

  • Oxidizes black lead sulfide to white lead sulfate: PbS + 4O₃ → PbSO₄ + 4O₂.
  • Oxidizes acidified potassium iodide to iodine: 2KI + H₂O + O₃ → 2KOH + I₂ + O₂.
  • Tailing of mercury: Ozone oxidizes mercury to suboxide: 2Hg + O₃ → Hg₂O + O₂, causing mercury to lose its meniscus and stick to glass walls.
Quantitative Estimation of Ozone (Iodometry):
Ozone is reacted with excess potassium iodide buffered with a borate buffer (pH 9.2). The liberated iodine is titrated against standard sodium thiosulfate (Na2S2O3) solution using starch indicator:
I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆   (Sodium tetrathionate)

Structure of Ozone: Ozone has an angular/bent geometry (O-O-O bond angle = 117°) with two resonance hybrid structures where both O-O bond lengths are identical (128 pm), intermediate between single (148 pm) and double (121 pm) bond lengths.

3. Allotropes of Sulfur: Rhombic (α) & Monoclinic (β)

Elemental sulfur forms puckered S8 crown rings with an S-S-S bond angle of 107°:

  • Rhombic Sulfur (α-sulfur): Yellow, stable below 368.6 K (95.6°C), density 2.06 g/cm³, prepared by evaporating roll sulfur solution in CS2.
  • Monoclinic Sulfur (β-sulfur): Needle-shaped, stable above 368.6 K (95.6°C), density 1.98 g/cm³, prepared by melting rhombic sulfur and cooling.
  • Transition Temperature (368.6 K / 95.6°C): The exact temperature at which both rhombic and monoclinic allotropes coexist in dynamic equilibrium.
  • Paramagnetism of S₂ in Vapor Phase: Above 1000 K, sulfur vapor exists predominantly as diatomic S2 molecules, which are paramagnetic with 2 unpaired electrons in antibonding π* orbitals, behaving exactly analogous to O2.
4. Sulfur Dioxide (SO₂): Chemical Nature & Dual Behavior

SO2 is a pungent, choking gas prepared by roasting sulfide ores or burning sulfur: S + O₂ → SO₂. In aqueous solution, it forms sulfurous acid: SO₂ + H₂O ⇌ H₂SO₃.

Reducing Action: In the presence of moisture, SO2 acts as a powerful reducing agent:

1. Decolorizes Acidified KMnO₄: 2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄ (Pink to Colorless)
2. Turns Acidified K₂Cr₂O₇ Green: K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O (Orange to Green Cr³⁺)
3. Temporary Bleaching Action: Bleaches delicate vegetable colors (wool, silk) by reduction: SO₂ + 2H₂O → H₂SO₄ + 2[H]. The bleached article regains color when re-oxidized by atmospheric oxygen.

Module 4: Oxoacids of Sulfur & Industrial Manufacture of Sulfuric Acid

1. Industrial Manufacture of Sulfuric Acid: Contact Process

Sulfuric acid (H2SO4) is manufactured on a massive industrial scale via the Contact Process in four distinct technological stages:

  • Stage 1 (Production of SO₂): Burning sulfur or roasting iron pyrites: S + O₂ → SO₂ or 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂. The SO2 gas is rigorously purified to remove dust, moisture, and arsenic impurities (As2O3 poisons the catalyst).
  • Stage 2 (Catalytic Oxidation of SO₂ to SO₃ — Crucial Step):
    2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   [ΔH° = -196.6 kJ/mol   (Exothermic)]
    Optimum Conditions: (i) Temperature: 720-770 K (450-500°C) — moderate temperature because the reaction is exothermic; (ii) Pressure: 1-2 bar (slight positive pressure to aid gas flow; high pressure favors SO3 but moderate pressure gives ~98% yield economically); (iii) Catalyst: Vanadium pentoxide (V₂O₅), highly resistant to poisoning.
  • Stage 3 (Absorption of SO₃ in conc. H₂SO₄): SO3 is absorbed in 98% H2SO4 to produce Oleum (Pyrosulfuric acid): SO₃ + H₂SO₄ → H₂S₂O₇.
    Note: SO3 is never dissolved directly in water because it forms an unmanageable dense acid mist/fog of H2SO4 that does not easily condense.
  • Stage 4 (Dilution of Oleum): Oleum is diluted with calculated quantities of water to yield desired concentrations (e.g., 98% pure commercial grade): H₂S₂O₇ + H₂O → 2H₂SO₄.
2. Chemical Properties of Concentrated H₂SO₄

Sulfuric acid owes its diverse industrial versatility to four characteristic properties:

  • 1. Low Volatility: Displaces more volatile acids from their salts: 2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl↑.
  • 2. Strong Dibasic Acid Behavior: Ionizes in two steps:
    H₂SO₄(aq) + H₂O(l) → H₃O⁺(aq) + HSO₄⁻(aq)   [K_{a1} > 10   (Very Large)]
    HSO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + SO₄²⁻(aq)   [K_{a2} = 1.2 × 10⁻²]
    Because Ka1 ≫ Ka2, H2SO4 behaves as a strong acid primarily in its first dissociation.
  • 3. Powerful Dehydrating Agent: Removes elements of water with intense affinity:
    Charring of Cane Sugar: C₁₂H₂₂O₁₁ xrightarrow{conc. H₂SO₄} 12C(s) + 11H₂O (leaves black spongy carbon).
    Dehydration of Formic Acid: HCOOH xrightarrow{conc. H₂SO₄} CO↑ + H₂O.
  • 4. Moderately Strong Oxidizing Agent: Concentrated hot H2SO4 oxidizes non-metals and metals:
    C + 2H₂SO₄(conc) → CO₂ + 2SO₂ + 2H₂O
    S + 2H₂SO₄(conc) → 3SO₂ + 2H₂O
    Cu + 2H₂SO₄(conc) xrightarrow{Δ} CuSO₄ + SO₂↑ + 2H₂O
3. Oxoacids of Sulfur: Structures & Oxidation States
Name & Formula Oxidation State of S Key Structural Features
Sulfurous acid (H₂SO₃) +4 Pyramidal; one S=O bond, two S-OH bonds, one lone pair on S.
Sulfuric acid (H₂SO₄) +6 Tetrahedral; two S=O bonds, two S-OH bonds.
Peroxomonosulfuric acid (H₂SO₅ — Caro's acid) +6 (not +8) Contains one peroxo bond (-O-O-H): two S=O, one S-OH, one S-O-O-H.
Peroxodisulfuric acid (H₂S₂O₈ — Marshall's acid) +6 (not +7) Two SO₃H groups joined by a peroxo linkage (HO-SO₂-O-O-SO₂-OH).
Pyrosulfuric acid (H₂S₂O₇ — Oleum) +6 Contains an oxygen bridge (S-O-S): HO-SO₂-O-SO₂-OH.
Thiosulfuric acid (H₂S₂O₃) +2 (average) / +5 & -1 One oxygen atom replaced by sulfur (contains S=S terminal bond and two S-OH).

Module 5: Group 17 Elements (Halogens)

1. Electronic Configuration & Periodic Anomalies

Group 17 elements (F, Cl, Br, I, At) are known as halogens (salt formers), with valence electronic configuration ns2 np5, requiring just one electron to complete an octet. Fluorine shows distinctive anomalies: it exhibits only the -1 oxidation state (highest electronegativity 4.0, lacks vacant d-orbitals), whereas Cl, Br, and I exhibit oxidation states of -1, +1, +3, +5, and +7.

Classic WBCHSE Double Anomaly: Electron Gain Enthalpy vs Bond Dissociation Energy
  • Negative Electron Gain Enthalpy (EA): The order is Cl > F > Br > I (-349, -333, -325, -295 kJ/mol). Fluorine has a smaller value than chlorine because of the extremely compact size of the 2p-orbital in fluorine, which creates intense interelectronic repulsion against the incoming electron.
  • Bond Dissociation Enthalpy: The order is Cl₂ > Br₂ > F₂ > I₂ (242.6 > 192.8 > 158.8 > 151.1 kJ/mol). F2 has an anomalously low bond energy because the three non-bonding lone pairs on each small fluorine atom are packed tightly together, exerting strong lone pair-lone pair electrostatic repulsions that weaken the F-F covalent bond.
  • Why is F₂ the Strongest Oxidizing Agent? Despite lower electron affinity, the standard electrode potential E° for F2/F- (+2.87 V) is greater than Cl2/Cl- (+1.36 V) because the low bond dissociation enthalpy of F2 combined with the immense hydration enthalpy of the tiny F- ion heavily overcompensates.
2. Hydrogen Halides (HX): Trends in Properties
Property Observed Order Scientific Rationale
Thermal Stability HF > HCl > HBr > HI H-X bond dissociation energy decreases as the size of halogen increases (overlap weakens).
Reducing Power HI > HBr > HCl > HF HI decomposes most readily (lowest H-I bond energy), giving off nascent hydrogen easily.
Acidic Strength HI > HBr > HCl > HF Ease of H⁺ release is dictated by decreasing bond dissociation enthalpy (H-I bond is weakest).
Boiling Point HF > HI > HBr > HCl HF has abnormally high b.p. (293 K) due to extensive intermolecular hydrogen bonding; from HCl to HI, b.p. rises due to increasing London dispersion forces.
3. Oxoacids of Halogens & Relative Acidic Strength

Halogens form four series of oxoacids: Hypohalous (HOX), Halous (HXO2), Halic (HXO3), and Perhalic (HXO4). Fluorine forms only one oxoacid, HOF (fluoric(I) acid or hypofluorous acid).

Acidic Strength Order of Chlorine Oxoacids:

HClO₄ > HClO₃ > HClO₂ > HClO   (+7 > +5 > +3 > +1)

Explanation: (1) As the oxidation state of the central halogen increases, electron density is pulled away from the O-H bond, weakening it and facilitating proton release. (2) In the conjugate base series (ClO4-, ClO3-, ClO2-, ClO-), the negative charge in ClO₄⁻ is delocalized over four equivalent electronegative oxygen atoms via resonance, providing maximal stabilization.

4. Interhalogen Compounds: Types, VSEPR Geometries & Enhanced Reactivity

Halogens react among themselves to form binary covalent compounds called interhalogens of types XX', XX'₃, XX'₅, and XX'₇ (where X is larger and more electropositive than X'):

Compound Type Example Central Atom Hybridization VSEPR Molecular Geometry
XX' ClF, BrF, ICl sp³ (1 bp + 3 lp) Linear
XX'₃ ClF₃, BrF₃, ICl₃ sp³d (3 bp + 2 lp) Bent T-shaped (lone pairs occupy equatorial positions to minimize 90° repulsions)
XX'₅ BrF₅, IF₅ sp³d² (5 bp + 1 lp) Square Pyramidal
XX'₇ IF₇ sp³d³ (7 bp + 0 lp) Pentagonal Bipyramidal

Why are Interhalogens More Reactive Than Parent Halogens? In general, interhalogens (XX') are chemically more reactive than dihalogens (X2) except fluorine (F2). This is because the heterogeneous X-X' covalent bond is polarized and significantly weaker than the homogeneous non-polar X-X bond in pure halogens.

Module 6: Group 18 Elements (Noble Gases / Aerogens)

1. Electronic Configuration & Neil Bartlett's Breakthrough (1962)

Group 18 elements (He, Ne, Ar, Kr, Xe, Rn) possess completely filled valence shells (ns2 np6, except He with 1s2). Due to this stable octet configuration, they exhibit exceptionally high first ionization enthalpies and large positive electron gain enthalpies, remaining chemically unreactive under normal conditions.

Neil Bartlett's Historic Discovery (March 1962):
Bartlett synthesized the deep red compound O₂⁺[PtF₆]⁻ by reacting molecular oxygen with platinum hexafluoride (PtF6). He noted two critical physical facts:
  • The first ionization enthalpy of molecular oxygen (O₂ → O₂⁺ + e⁻) is 1175 kJ/mol.
  • The first ionization enthalpy of xenon (Xe → Xe⁺ + e⁻) is 1170 kJ/mol, virtually identical!
Bartlett reasoned that PtF6 should also oxidize xenon. Mixing xenon gas with PtF6 vapor produced an immediate rapid reaction yielding an orange-yellow ionic solid:
Xe(g) + PtF₆(g) → Xe⁺[PtF₆]⁻(s)
This milestone shattered the dogma that noble gases are chemically incapable of forming true chemical compounds.
2. Xenon Fluorides: Preparation & Hydrolysis

Xenon reacts directly with fluorine in nickel vessels under controlled stoichiometry, temperature, and pressure:

  • Xe + F₂ xrightarrow[673 K, 1 bar]{Xe in excess (2:1)} XeF₂ (Xenon difluoride, colorless solid).
  • Xe + 2F₂ xrightarrow[873 K, 7 bar]{1:5 molar ratio} XeF₄ (Xenon tetrafluoride).
  • Xe + 3F₂ xrightarrow[573 K, 60-70 bar]{1:20 molar ratio} XeF₆ (Xenon hexafluoride).

Hydrolysis Reactions (Complete vs Partial):

Hydrolysis of XeF₂: 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂↑
Hydrolysis of XeF₄: 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂↑ (Disproportionation of Xe(IV) to Xe(0) & Xe(VI))
Complete Hydrolysis of XeF₆: XeF₆ + 3H₂O → XeO₃ + 6HF   (Explosive solid XeO₃ formed)
Partial Hydrolysis of XeF₆:
 • XeF₆ + H₂O → XeOF₄ + 2HF   (Xenon oxytetrafluoride, liquid)
 • XeF₆ + 2H₂O → XeO₂F₂ + 4HF   (Xenon dioxydifluoride)
3. VSEPR Geometries of Xenon Compounds
Compound Steric Number (SN) Bond Pairs & Lone Pairs Hybridization & Molecular Shape
XeF₂ SN = 5 2 bp + 3 lp sp³d; Linear (all 3 lone pairs reside in equatorial positions at 120°).
XeF₄ SN = 6 4 bp + 2 lp sp³d²; Square Planar (2 lone pairs occupy axial positions trans to each other at 180°).
XeF₆ SN = 7 6 bp + 1 lp sp³d³; Distorted Octahedral (capped octahedron with fluxional stereochemically active lone pair).
XeO₃ SN = 4 3 bp (σ) + 1 lp sp³; Pyramidal (three Xe=O double bonds and one lone pair, isostructural with SO₃²⁻).
XeOF₄ SN = 6 5 bp + 1 lp sp³d²; Square Pyramidal (Xe=O double bond occupies one axial position, lone pair opposite).
XeO₂F₂ SN = 5 4 bp + 1 lp sp³d; Seesaw (two equatorial Xe=O, one equatorial lone pair, two axial Xe-F bonds).
4. Industrial & Scientific Uses of Noble Gases
  • Helium (He): (1) Used as a diluent for oxygen in deep-sea diving tanks (heliox) because unlike nitrogen, helium has exceptionally low solubility in blood and lipids, preventing the decompression illness known as 'the bends'. (2) Used to fill meteorological balloons due to low density and non-inflammability (unlike H2). (3) Liquid helium (b.p. 4.2 K) serves as an essential cryogenic refrigerant for superconducting magnets in Nuclear Magnetic Resonance (NMR) and Medical Magnetic Resonance Imaging (MRI).
  • Neon (Ne): Filled in discharge tubes and neon glow signs for vibrant orange-red advertising displays; neon lamps produce light that easily penetrates fog.
  • Argon (Ar): Provides an inert gas atmosphere in high-temperature metallurgical processes such as arc welding of aluminum and stainless steel, and filling incandescent light bulbs to inhibit tungsten filament evaporation.
  • Krypton & Xenon (Kr, Xe): Utilized in high-speed photographic flash lamps, airport runway approach beacons, and specialized laser systems.
  • Radon (Rn): Radioactive noble gas employed in radiotherapy for malignant tumor treatment and geological earthquake tracking.

Key Formulas, Reactions & Definitions

Contact Process: Catalytic Oxidation & Equilibrium Yield
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   [ΔH° = -196.6 kJ/mol]
Haber-Bosch Ammonia Synthesis Thermodynamic Equation
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   [ΔH° = -92.4 kJ/mol]
Halogen Electron Gain Enthalpy vs Bond Dissociation Enthalpy Orders
EA: Cl > F > Br > I   |   BDE: Cl₂ > Br₂ > F₂ > I₂
Oxoacid Acidity & Conjugate Base Resonance Polarization
Acidic Strength: HClO₄ > HClO₃ > HClO₂ > HClO
VSEPR Steric Number & Noble Gas Geometry Formula
SN = ½ [V + M - C + A]   (Lone Pairs = SN - Bond Pairs)
Ostwald Process Catalytic Ammonia Oxidation Sequence
4NH₃ + 5O₂ xrightarrow[500 K, 9 bar]{Pt/Rh} 4NO + 6H₂O;   2NO + O₂ → 2NO₂;   3NO₂ + H₂O → 2HNO₃ + NO

Conceptual Solved Examples & Case Studies

Example 1
Explain why PCl₅ is a well-known, stable chemical compound, whereas NCl₅ is completely unknown and does not exist.
Step-by-Step Solution:
Step 1: Write electronic configurations of the central atoms.
• Nitrogen (atomic number Z = 7): 1s² 2s² 2p³ (Principal quantum number n = 2).
• Phosphorus (atomic number Z = 15): 1s² 2s² 2p⁶ 3s² 3p³ 3d⁰ (Principal quantum number n = 3).

Step 2: Compare d-orbital availability and octet expansion.
• In nitrogen, the valence shell (n = 2) contains only s and p subshells (one 2s and three 2p orbitals). There are no vacant 2d-orbitals in existence. Hence, nitrogen cannot unpair its 2s electrons or expand its valence octet beyond 4 (maximum covalency is 4, as in NH₄⁺). Therefore, it cannot form 5 covalent bonds in NCl₅.
• In phosphorus, vacant 3d-orbitals are energetically accessible in the valence shell. One 3s electron can be promoted to an empty 3d-orbital, generating five unpaired electrons to undergo sp³d hybridization, successfully forming PCl₅.

Conclusion: The absence of vacant d-orbitals in nitrogen precludes octet expansion, making NCl₅ non-existent.
Example 2
(a) Account for the fact that fluorine (F₂) is a stronger oxidizing agent than chlorine (Cl₂), even though chlorine has a more negative electron gain enthalpy. (b) Write the balanced chemical equation for the reaction of cold and dilute sodium hydroxide with chlorine gas.
Step-by-Step Solution:
Part (a): Thermodynamic Cycle of Oxidizing Power [2 Marks]
The standard electrode potential E° for the reduction reaction ½ X₂(g) + e⁻ → X⁻(aq) is governed by three sequential thermodynamic enthalpy contributions (Born-Haber cycle):
1. Bond dissociation enthalpy of ½ X₂(g) → X(g)   [Endothermic, ΔH_{diss}]
2. Electron gain enthalpy of X(g) + e⁻ → X⁻(g)   [Exothermic, ΔH_{eg}]
3. Hydration enthalpy of X⁻(g) + aq → X⁻(aq)   [Exothermic, ΔH_{hyd}]

Although chlorine has a slightly more negative electron gain enthalpy (-349 kJ/mol vs -333 kJ/mol for F), fluorine's overall standard reduction potential E° (+2.87 V) is far greater than that of chlorine (+1.36 V) because:
• F₂ has an exceptionally low bond dissociation enthalpy (158.8 kJ/mol) due to severe lone-pair repulsion.
• The resulting tiny fluoride ion (F⁻) has an exceptionally colossal hydration enthalpy (-515 kJ/mol vs -381 kJ/mol for Cl⁻) due to its high charge density.
These two factors overwhelmingly overcompensate for the minor difference in electron gain enthalpy, making F₂ the strongest elemental oxidizing agent.

Part (b): Reaction of Cl₂ with Cold Dilute NaOH [1 Mark]
Chlorine undergoes disproportionation in cold, dilute NaOH yielding sodium chloride and sodium hypochlorite:
2NaOH(cold, dil) + Cl₂ → NaCl + NaOCl + H₂O
(Here, chlorine is simultaneously reduced from 0 to -1 in NaCl and oxidized from 0 to +1 in NaOCl).
Example 3
(a) Explain why hypophosphorous acid (H₃PO₂) acts as a monobasic acid and a powerful reducing agent. (b) What happens when orthophosphorous acid (H₃PO₃) is subjected to thermal heating? Write the balanced chemical equation.
Step-by-Step Solution:
Part (a): Structure and Dual Behavior of H₃PO₂ [2 Marks]
• Basicity: In oxoacids of phosphorus, only those hydrogen atoms that are covalently bound to oxygen in P-OH groups are ionizable in water to furnish H⁺ ions. The hydrogen atoms directly bonded to phosphorus (P-H bonds) are non-ionizable because phosphorus and hydrogen have nearly identical electronegativities (2.1). In hypophosphorous acid (H₃PO₂), the central phosphorus is surrounded by one P=O bond, one P-OH bond, and two P-H bonds.
Since it possesses only one ionizable P-OH group, H₃PO₂ acts strictly as a monobasic acid.
• Reducing Power: The presence of two direct P-H bonds imparts powerful reducing properties, as these bonds readily donate electrons/hydrides. For example, it reduces silver nitrate quantitatively to black metallic silver:
4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag(s)↓ + 4HNO₃ + H₃PO₄

Part (b): Thermal Disproportionation of H₃PO₃ [1 Mark]
Orthophosphorous acid contains phosphorus in the intermediate +3 oxidation state. Upon heating, it undergoes disproportionation into orthophosphoric acid (P in +5) and phosphine gas (P in -3):
4H₃PO₃ xrightarrow{Δ} 3H₃PO₄ + PH₃↑
Example 4
(a) With balanced chemical equations, describe the action of nitric acid on copper metal under: (i) dilute conditions, and (ii) concentrated conditions. (b) Explain the chemical composition and mechanism of the Brown Ring formed during the confirmatory test for nitrate ions, stating the formal oxidation state and magnetic moment of iron.
Step-by-Step Solution:
Part (a): Reactions of Copper with Nitric Acid [2 Marks]
Nitric acid does not evolve hydrogen with copper because copper has a positive standard reduction potential (below hydrogen in the electrochemical series). Instead, HNO₃ acts as a strong oxidizing agent, and the reduced nitrogen product depends strictly on acid concentration:
• (i) With Dilute HNO₃ (~20%):
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O
Nitric oxide (NO, colorless gas) is evolved.
• (ii) With Concentrated HNO₃ (~70%):
Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O
Dense brown pungent fumes of nitrogen dioxide (NO₂) are evolved.

Part (b): Mechanism & State of Fe in the Brown Ring Complex [2 Marks]
• Mechanism: Nitrate solution mixed with freshly prepared FeSO₄ is underlayered with concentrated H₂SO₄ down the side of the test tube. Fe²⁺ reduces NO₃⁻ to NO in acid medium:
NO₃⁻ + 3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O
The liberated NO coordinates with [Fe(H₂O)₆]²⁺ to generate a brown coordination complex ring at the liquid-liquid junction:
[Fe(H₂O)₆]²⁺ + NO + SO₄²⁻ → [Fe(H₂O)₅(NO)]SO₄ + H₂O
• Oxidation State of Iron: Spectroscopic and magnetic studies demonstrate an electron transfer from the NO radical to Fe²⁺, generating a nitrosonium ligand (NO⁺) and reducing iron to the rare Fe⁺¹ oxidation state (3d⁷ configuration).
• Magnetic Moment: With 3 unpaired electrons (t₂g⁵ eg²), the spin-only magnetic moment is calculated as:
μ = √(n(n+2)) = √(3(5)) = √15 ≈ 3.87 BM (Paramagnetic).
Example 5
(a) Deduce the molecular geometry, central atom hybridization, and the number and distribution of lone pairs for: (i) ClF₃, and (ii) XeF₄ using VSEPR theory. (b) Why are interhalogen compounds generally more chemically reactive than the constituent parent halogens (with the exception of F₂)?
Step-by-Step Solution:

Part (a): VSEPR Geometry Deduction [2.5 Marks]
• (i) Chlorine Trifluoride (ClF₃):

  • Valence electrons on central Cl = 7. Monovalent F atoms = 3.
  • Steric Number SN = ½(7 + 3) = 5 ⟹ sp³d hybridization (Trigonal bipyramidal electron geometry).
  • Number of bond pairs = 3; Number of lone pairs = 5 - 3 = 2 lone pairs.
  • Positioning: To minimize intense 90° lone pair-bond pair repulsions, both lone pairs occupy the equatorial positions (at 120°).
  • Resulting Molecular Shape: Bent T-shaped with F(axial)-Cl-F(equatorial) bond angle slightly compressed from 90° to ~87.5°.

• (ii) Xenon Tetrafluoride (XeF₄):

  • Valence electrons on central Xe = 8. Monovalent F atoms = 4.
  • Steric Number SN = ½(8 + 4) = 6 ⟹ sp³d² hybridization (Octahedral electron geometry).
  • Number of bond pairs = 4; Number of lone pairs = 6 - 4 = 2 lone pairs.
  • Positioning: The 2 lone pairs occupy trans-axial positions (180° apart) to achieve maximum spatial separation and zero 90° lone pair-lone pair repulsion.
  • Resulting Molecular Shape: Square Planar with all F-Xe-F bond angles at 90°.

Part (b): Higher Chemical Reactivity of Interhalogens [1.5 Marks]
Interhalogen compounds (XX') are significantly more reactive than elemental halogens (X₂) because:

  1. The X-X' bond is heterogeneous and polar due to the electronegativity difference between the two distinct halogens, whereas the X-X bond in diatomic halogens is purely non-polar.
  2. The polar X-X' covalent bond has lower orbital overlap efficiency, making its bond dissociation enthalpy substantially lower than that of the X-X bond (except F-F). Consequently, interhalogens undergo rapid nucleophilic and electrophilic attack and serve as aggressive halogenating and fluorinating agents.
Example 6
(a) State the physicochemical principles and optimum conditions (temperature, pressure, catalyst) for the Contact Process of sulfuric acid manufacture in terms of Le Chatelier's Principle. (b) Explain why SO₃ is absorbed in concentrated H₂SO₄ rather than directly into water. (c) Write balanced chemical equations for the reaction of hot concentrated H₂SO₄ with: (i) cane sugar (C₁₂H₂₂O₁₁), and (ii) copper metal.
Step-by-Step Solution:
Part (a): Physicochemical Principles of the Contact Process [2 Marks]
The pivotal stage of the Contact Process is the catalytic oxidation of sulfur dioxide:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   [ΔH° = -196.6 kJ/mol]
Applying Le Chatelier's Principle:
1. Temperature: The reaction is exothermic (ΔH < 0). Low temperature favors higher equilibrium conversion to SO₃, but below 700 K the reaction rate becomes imperceptibly slow. Thus, an optimum moderate temperature of 720-770 K (450-500°C) is maintained.
2. Pressure: Forward reaction proceeds with a reduction in gaseous volume (3 moles of reactants → 2 moles of products). High pressure favors SO₃ formation. However, since high conversions (~98%) are already achieved at slight positive pressure without requiring expensive high-pressure towers, an optimum pressure of 1-2 bar is utilized.
3. Catalyst: Divanadium pentoxide (V₂O₅) is used at ~720 K. It is cost-effective and resistant to arsenic poisoning (unlike platinum).

Part (b): Why Direct Dissolution in Water is Avoided [1 Mark]
The direct dissolution of SO₃ in water is violently exothermic: SO₃ + H₂O → H₂SO₄ + Heat. The immense localized heat generated instantly vaporizes the water, producing a dense, unmanageable acid fog/mist of sulfuric acid droplets. This mist is carried away by outgoing gas streams and cannot be condensed in industrial absorption towers. To avert this hazard, SO₃ is absorbed smoothly in 98% H₂SO₄ to yield oleum (H₂S₂O₇), which is then safely diluted with calculated water.

Part (c): Chemical Reactions with Concentrated H₂SO₄ [2 Marks]
• (i) Action on Cane Sugar (Dehydrating Action):
Concentrated H₂SO₄ removes elements of water (H and O in 2:1 ratio) from carbohydrate molecules, charring them into a black spongy mass of elemental carbon ('sugar charcoal'):
C₁₂H₂₂O₁₁(s) xrightarrow{conc. H₂SO₄} 12C(s) + 11H₂O(l)   (Exothermic steam evolution)
• (ii) Action on Copper Metal (Oxidizing Action):
Hot concentrated H₂SO₄ acts as a powerful oxidizing agent, oxidizing copper to cupric sulfate while being reduced to sulfur dioxide gas:
Cu(s) + 2H₂SO₄(conc) xrightarrow{Δ} CuSO₄(aq) + SO₂↑ + 2H₂O(l)

Common Misconceptions & Examiner Traps

Common Misconception

Calculating the oxidation state of sulfur in Marshall's acid (H₂S₂O₈) as +7 by standard algebraic summation.

Scientific Reality & Correction

The maximum possible oxidation state for sulfur cannot exceed its valence electrons (6). H₂S₂O₈ contains a peroxo linkage (-O-O-) where two oxygen atoms have an oxidation state of -1 each. Thus: 2(+1) + 2(S) + 6(-2) + 2(-1) = 0 ⟹ 2S = +12 ⟹ S = +6.

Common Misconception

Assuming hypophosphorous acid (H₃PO₂) is tribasic because its formula contains 3 hydrogen atoms.

Scientific Reality & Correction

Basicity is determined exclusively by the number of ionizable P-OH groups, not total hydrogen atoms. H₃PO₂ contains only one P-OH group and two direct non-ionizable P-H bonds; therefore, it is strictly monobasic (produces only one series of salts, e.g., NaH₂PO₂).

Common Misconception

Assuming iron in the brown ring complex [Fe(H₂O)₅(NO)]²⁺ has an oxidation state of +2 or +3.

Scientific Reality & Correction

The ligand is nitrosonium (NO⁺), which transfers an electron to Fe²⁺ during complex formation. This reduces iron to the rare +1 oxidation state (Fe⁺, 3d⁷ configuration with 3 unpaired electrons, μ ≈ 3.87 BM).

Common Misconception

Believing all five P-Cl bonds in gaseous PCl₅ are identical in length and strength.

Scientific Reality & Correction

In sp³d trigonal bipyramidal geometry, the 2 axial P-Cl bonds (240 pm) suffer greater 90° repulsion from 3 equatorial bonds than the equatorial bonds (202 pm) suffer. Hence, axial bonds are longer, weaker, and cleave first upon heating.

Common Misconception

Ranking fluorine as having the highest electron gain enthalpy because it has the highest electronegativity.

Scientific Reality & Correction

Chlorine has a more negative electron gain enthalpy (-349 kJ/mol) than fluorine (-333 kJ/mol). The small, compact 2p-orbital of fluorine creates intense interelectronic repulsions that hinder the incoming electron.

Visual Learning & Conceptual Map

WBCHSE Class 12 Chemistry: p-Block Elements (Groups 15, 16, 17 & 18) Pnictogens, Chalcogens, Halogens & Noble Gases: Structures, Trends & Industrial Processes 15 Group 15: Pnictogens (N, P) P₄ Tetrahedral & NH₃ Pyramidal P P P P P₄ (60°) N: pπ-pπ multiple bonds, no vacant d-orbitals P₄: 60° bond angle strain, high reactivity Oxoacids: H₃PO₂ (monobasic), H₃PO₄ (tribasic) Brown Ring Test: [Fe(H₂O)₅(NO)]²⁺ (Fe⁺¹) 16 Group 16: Chalcogens (O, S) S₈ Crown Structure & Contact Process S₈ Crown Ring O₃: Ozone resonance (bond angle 117°) S₈: Puckered crown ring, transition at 95.6°C Contact Process: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅ catalyst) Marshall's Acid: H₂S₂O₈ (-O-O- peroxo bridge, S⁺⁶) 17 Group 17: Halogens (F, Cl, Br, I) Bond Enthalpy Anomalies & Interhalogens Cl F F F ClF₃ (T-Shaped) Electron Gain Enthalpy: Cl > F > Br > I Bond Dissociation Energy: Cl₂ > Br₂ > F₂ > I₂ Oxoacid Acidity: HClO₄ > HClO₃ > HClO₂ > HClO Interhalogens: ClF₃ (Bent T-shaped, sp³d) 18 Group 18: Noble Gases (He, Ne, Ar, Kr, Xe) Bartlett Reaction & Xenon Fluorides Xe XeF₄ (Sq. Planar) Bartlett (1962): Xe⁺[PtF₆]⁻ landmark synthesis XeF₂: Linear geometry (sp³d, 3 equatorial lone pairs) XeF₄: Square Planar (sp³d², 2 axial lone pairs) XeF₆: Distorted Octahedral (sp³d³, 1 lone pair) WBCHSE Class 12 Chemistry | p-Block: Electronic Configurations, Oxidation States, VSEPR Geometries & Contact Process

Chapter Summary & 10 Key Takeaways

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Group Key Elements Key Compounds & Processes High-Yield Board Takeaway
Group 15 (Pnictogens) N, P, As, Sb, Bi Haber NH₃, Ostwald HNO₃, PCl₃/PCl₅, H₃PO₂/H₃PO₃/H₃PO₄ N lacks d-orbitals (NCl₅ impossible); H₃PO₂ is monobasic and strongly reducing (2 P-H bonds); Brown ring Fe is in +1 state.
Group 16 (Chalcogens) O, S, Se, Te, Po O₃, S₈ crown ring, Contact Process H₂SO₄, Marshall's acid Contact process optimal conditions: 450-500°C, 1-2 bar, V₂O₅; SO₃ absorbed in 98% H₂SO₄ (oleum); H₂S₂O₈ S is +6 (peroxo bond).
Group 17 (Halogens) F, Cl, Br, I HX acids, ClF₃, BrF₅, IF₇, HClO series EA: Cl > F > Br > I; BDE: Cl₂ > Br₂ > F₂ > I₂; Acid strength: HI > HBr > HCl > HF; ClF₃ is T-shaped (sp³d, 2 eq lp).
Group 18 (Noble Gases) He, Ne, Ar, Kr, Xe, Rn Bartlett Xe⁺[PtF₆]⁻, XeF₂, XeF₄, XeF₆, XeO₃ Bartlett discovery based on IE(Xe) ≈ IE(O₂); XeF₂ is linear (sp³d, 3 lp); XeF₄ is square planar (sp³d², 2 lp); He in diving bends.
Takeaway 38

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
What is the basicity and reducing nature of pyrophosphoric acid (H₄P₂O₇)? Explain based on its molecular structure.
Reveal Answer & Explanation
Answer: Pyrophosphoric acid has four P-OH groups and ZERO direct P-H bonds. Therefore, it is tetrabasic (furnishes 4 H⁺ ions) and possesses no reducing properties.
Draw the structure containing a P-O-P linkage and count the P-OH and direct P-H bonds.
2
Arrange the following hydrogen halides in decreasing order of boiling point and justify the anomalous position of HF: HCl, HBr, HI, HF.
Reveal Answer & Explanation
Answer: Order: HF > HI > HBr > HCl. HF has the highest boiling point (293 K) due to strong intermolecular hydrogen bonding. For HCl, HBr, and HI, boiling point increases down the group with increasing molecular mass and London dispersion forces.
Consider the presence of strong intermolecular hydrogen bonding versus London dispersion forces.
3
Predict the hybridization and molecular geometry of xenon oxytetrafluoride (XeOF₄) using VSEPR theory.
Reveal Answer & Explanation
Answer: SN = ½(8 + 4) = 6 ⟹ sp³d² hybridization. It has 5 bond pairs (4 Xe-F, 1 Xe=O) and 1 lone pair. Molecular geometry is Square Pyramidal.
Calculate the steric number SN = ½ [V + M - C + A] with V=8 for Xe, M=4 for F, and count the axial lone pair.
4
Calculate the formal oxidation state of sulfur in Caro's acid (H₂SO₅) and Marshall's acid (H₂S₂O₈).
Reveal Answer & Explanation
Answer: Both acids contain sulfur in the +6 oxidation state (the group maximum). Caro's acid has one peroxo bond; Marshall's acid has one bridging peroxo bond (-O-O-) linking two sulfur centers.
Both compounds contain peroxo linkages (-O-O-) with oxygen in the -1 oxidation state.
5
Write the products of the complete hydrolysis of xenon hexafluoride (XeF₆) and explain why the reaction is hazardous.
Reveal Answer & Explanation
Answer: XeF₆ + 3H₂O → XeO₃ + 6HF. It is highly hazardous because solid xenon trioxide (XeO₃) is an extremely explosive, sensitive compound that detonates violently upon touch or warming.
Complete replacement of all six fluorine atoms by three oxygen atoms produces a yellow solid oxide.
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