The p-Block Elements
Welcome, ASSEB Class 12 students! In this chapter, we explore the rich and diverse chemistry of the p-block elements belonging to Groups 15, 16, 17, and 18 of the periodic table. These elements include nitrogen, phosphorus, oxygen, sulphur, halogens, and the noble gases — each playing crucial roles in chemistry, biology, industry, and the atmosphere. We will study their electronic configurations, periodic trends, and the preparation, properties, and uses of important compounds. This chapter is one of the highest-scoring chapters of inorganic chemistry, so master every reaction, oxidation state, and trend carefully.
Chapter Summary
Group 15 (Nitrogen Family — Pnicogens): Includes N, P, As, Sb, Bi with general electronic configuration ns²np³. The half-filled p-orbital provides extra stability. Down the group, atomic size and metallic character increase, while electronegativity, ionization enthalpy, and non-metallic character decrease. Common oxidation states are −3, +3, and +5, with the stability of +5 decreasing and +3 increasing down the group due to the inert pair effect. Nitrogen shows anomalous behaviour due to its small size, high electronegativity, absence of d-orbitals, and ability to form pπ–pπ multiple bonds. Dinitrogen (N₂) is prepared in the lab by heating NH₄Cl with NaNO₂. Ammonia (NH₃) is prepared by Haber’s process: N₂ + 3H₂ ⇌ 2NH₃. Nitric acid (HNO₃) is prepared industrially by Ostwald’s process — catalytic oxidation of NH₃ over Pt/Rh gauze. Phosphorus exists in allotropes — white (P₄, reactive, toxic), red (polymeric, stable), and black (most stable). PH₃ (phosphine) is prepared by hydrolysis of Ca₃P₂. PCl₃ and PCl₅ are important halides; PCl₅ has trigonal bipyramidal geometry in gas phase and exists as [PCl₄]⁺[PCl₆]⁻ in solid state. Oxoacids of phosphorus include H₃PO₂ (hypophosphorous), H₃PO₃ (phosphorous), H₃PO₄ (orthophosphoric), and H₄P₂O₇ (pyrophosphoric).
Group 16 (Chalcogens — Ore-forming): Includes O, S, Se, Te, Po with configuration ns²np⁴. Down the group, metallic character increases; oxygen is gaseous while others are solids. Common oxidation states are −2, +2, +4, and +6. Oxygen shows anomalous behaviour due to small size and absence of d-orbitals. Allotropes of oxygen are dioxygen (O₂) and ozone (O₃). Sulphur shows several allotropes — rhombic (α-sulphur), monoclinic (β-sulphur), plastic, and colloidal — with S₈ being a puckered ring. Ozone (O₃) is prepared by silent electric discharge through O₂; it is a powerful oxidizing agent and protects life from UV radiation in stratosphere. SO₂ is prepared by burning sulphur in air or roasting sulphide ores; it is acidic, reducing, and used as bleaching agent. H₂SO₄ is manufactured by the Contact Process: S → SO₂ → SO₃ (with V₂O₅ catalyst) → H₂S₂O₇ (oleum) → H₂SO₄. It acts as a strong acid, dehydrating agent, and oxidizing agent. Oxoacids of sulphur include H₂SO₃, H₂SO₄, H₂S₂O₇, H₂S₂O₈ (peroxodisulphuric), and H₂S₂O₃ (thiosulphuric).
Group 17 (Halogens — Salt formers): Includes F, Cl, Br, I, At with configuration ns²np⁵. They are the most electronegative non-metals and exist as diatomic molecules X₂. Down the group, atomic size, melting point, and bond dissociation enthalpy show specific trends; F₂ has unexpectedly low bond enthalpy due to electron–electron repulsion in the small F atom. Chlorine is prepared by oxidation of HCl with MnO₂ or KMnO₄ (Deacon’s process industrially). Hydrogen chloride (HCl) is prepared by heating NaCl with conc. H₂SO₄. Oxoacids of chlorine: HClO (hypochlorous), HClO₂ (chlorous), HClO₃ (chloric), and HClO₄ (perchloric); acidic strength and oxidation state increase from HClO to HClO₄. Interhalogen compounds (XX’, XX’₃, XX’₅, XX’₇) are formed between two different halogens, e.g., ClF, BrF₃, IF₅, IF₇; they are more reactive than halogens (except F₂). Pseudohalogens are polyatomic anions resembling halide ions, e.g., CN⁻, OCN⁻, SCN⁻, N₃⁻; their dimers (CN)₂, (SCN)₂ are pseudohalogen molecules.
Group 18 (Noble gases — Aerogens): Includes He, Ne, Ar, Kr, Xe, Rn with completely filled configuration ns²np⁶ (except He: 1s²). They have very high ionization enthalpies, zero electron affinity, and exist as monatomic gases. Their inertness arises from stable electronic configuration. Xenon forms a number of compounds with fluorine and oxygen due to its large size and low ionization energy. Important xenon compounds: XeF₂ (linear), XeF₄ (square planar), XeF₆ (distorted octahedral), and XeOF₄ (square pyramidal). These are prepared by direct reaction of Xe with F₂ under specific conditions of temperature, pressure, and Xe:F₂ ratio. Noble gases find use in lighting (Ne signs), inert atmosphere (Ar in welding), cryogenics (liquid He), and balloons (He).
1-Mark Questions and Answers
Q1. Write the general electronic configuration of Group 15 elements.
Answer: ns²np³ (a half-filled p-subshell, giving extra stability).
Q2. Why does nitrogen show anomalous behaviour in Group 15?
Answer: Due to its small size, high electronegativity, high ionization enthalpy, and absence of d-orbitals in the valence shell.
Q3. Name the catalyst used in Ostwald’s process for the manufacture of HNO₃.
Answer: Platinum-Rhodium (Pt/Rh) gauze.
Q4. Why is white phosphorus more reactive than red phosphorus?
Answer: Because of the strain in the P₄ tetrahedral structure of white phosphorus (bond angle 60°), which makes the bonds weaker.
Q5. What is the shape of PCl₅ molecule in gaseous state?
Answer: Trigonal bipyramidal (sp³d hybridization).
Q6. Name the catalyst used in the Contact Process for manufacture of H₂SO₄.
Answer: Vanadium pentoxide (V₂O₅).
Q7. What is the geometry of XeF₄?
Answer: Square planar (sp³d² hybridization with two lone pairs on Xe).
Q8. Why is fluorine the strongest oxidizing agent among halogens?
Answer: Because of its low bond dissociation enthalpy, high electronegativity, and high enthalpy of hydration of F⁻.
Q9. Give one example of a pseudohalogen ion.
Answer: Cyanide ion (CN⁻); other examples include OCN⁻, SCN⁻, N₃⁻.
Q10. What is the hybridization of Xe in XeF₆?
Answer: sp³d³ hybridization (distorted octahedral geometry due to one lone pair).
2–3 Marks Questions and Answers
Q11. Discuss the trends in atomic size, ionization enthalpy, and electronegativity in Group 15.
Answer: Down Group 15, atomic size increases due to addition of new shells. Ionization enthalpy decreases because the outermost electron is farther from the nucleus and screening increases. Electronegativity also decreases for the same reason. However, the decrease is not uniform — N has unusually high IE because of its small size and stable half-filled np³ configuration. The reduction in non-metallic character down the group leads to N and P being non-metals, As and Sb metalloids, and Bi a metal.
Q12. How is ammonia prepared in the laboratory? Write its preparation by Haber’s process.
Answer: In the laboratory, NH₃ is prepared by heating ammonium chloride with calcium hydroxide:
2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
Industrially, by Haber’s process:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92 kJ mol⁻¹
Conditions: 200 atm pressure, 700 K, with Fe catalyst and Mo as promoter.
Q13. Why does PCl₅ exist as [PCl₄]⁺[PCl₆]⁻ in the solid state?
Answer: In solid state, PCl₅ ionizes into a tetrahedral cation [PCl₄]⁺ (sp³) and an octahedral anion [PCl₆]⁻ (sp³d²). This ionic structure provides better lattice stability than the molecular trigonal bipyramidal form, which has two longer (axial) and three shorter (equatorial) bonds and is sterically strained.
Q14. Write the preparation and structure of ozone (O₃).
Answer: Ozone is prepared by passing silent electric discharge through pure dry oxygen in an ozonizer:
3O₂ → 2O₃; ΔH = +142 kJ mol⁻¹
Structure: bent (V-shaped), bond angle ~117°, with two equivalent O–O bond lengths (1.28 Å) due to resonance between two structures. Central O is sp² hybridized.
Q15. Why is H₂SO₄ called the “King of Chemicals”?
Answer: H₂SO₄ is called the “King of Chemicals” because of its enormous industrial importance. It is used in the manufacture of fertilizers (ammonium sulphate, superphosphate), petroleum refining, pigments, dyes, detergents, storage batteries, drugs, and explosives. National economic strength is often gauged by sulphuric acid consumption.
Q16. What are interhalogen compounds? Give the types and one example of each.
Answer: Compounds formed between two different halogens are called interhalogen compounds, of general formula XX’ₙ (n = 1, 3, 5, 7). Types: AX (e.g., ClF), AX₃ (e.g., BrF₃), AX₅ (e.g., IF₅), AX₇ (e.g., IF₇). They are generally more reactive than halogens (except F₂) and are diamagnetic covalent molecules.
Q17. Why are noble gases monatomic and chemically inert?
Answer: Noble gases have completely filled valence shells (ns²np⁶, except He which is 1s²), giving stable octet/duplet configuration. Hence they have very high ionization enthalpy and almost zero electron gain enthalpy. They neither lose nor gain electrons easily and do not form bonds, so they exist as single atoms (monatomic) and are chemically inert.
Q17a. Write the preparation and properties of phosphine (PH₃).
Answer: Phosphine is prepared by hydrolysis of calcium phosphide:
Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
Pure PH₃ is prepared by heating white P with conc. NaOH:
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
Properties: Colourless gas with rotten fish smell, slightly soluble in water, weakly basic (much less than NH₃ since lone pair is in larger 3s orbital), forms phosphonium salts (PH₄I) with HI, strong reducing agent.
Q17b. What are the differences between rhombic and monoclinic sulphur?
Answer: Rhombic sulphur (α-sulphur) is yellow, octahedral crystals, m.p. 385.8 K, density 2.06 g/cc, stable below 369 K. Monoclinic sulphur (β-sulphur) is amber-coloured needle-shaped crystals, m.p. 393 K, density 1.98 g/cc, stable above 369 K. Both contain S₈ puckered ring molecules. The transition temperature is 369 K. Above 369 K, rhombic converts to monoclinic; below 369 K, monoclinic converts back to rhombic.
5–7 Marks Questions and Answers
Q18. Describe in detail the manufacture of nitric acid by Ostwald’s process. Write the reactions and conditions involved.
Answer: Ostwald’s process involves catalytic oxidation of ammonia. The steps are:
Step 1 — Catalytic oxidation of NH₃: A 1:8 mixture of NH₃ and air is passed over Pt/Rh gauze at 500 K and 9 bar pressure:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g); ΔH = −907 kJ
Step 2 — Oxidation of NO to NO₂: NO is cooled and oxidized by air:
2NO(g) + O₂(g) → 2NO₂(g)
Step 3 — Absorption of NO₂ in water:
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)
The NO produced is recycled. The acid obtained is about 68% HNO₃, which is concentrated by distillation with conc. H₂SO₄ to get fuming nitric acid (98%). Fuming HNO₃ is yellow due to dissolved NO₂. HNO₃ is stored in dark bottles to prevent decomposition by light.
Q19. Discuss the allotropes of phosphorus, comparing white, red, and black phosphorus.
Answer: Phosphorus exhibits the following allotropes:
White phosphorus (P₄): Translucent waxy solid, soluble in CS₂ but insoluble in water, m.p. 317 K. Consists of discrete tetrahedral P₄ molecules with P–P–P bond angle of 60°. Highly reactive, glows in dark (chemiluminescence), spontaneously catches fire in air; very poisonous; stored under water.
Red phosphorus: Iron-grey crystalline solid, prepared by heating white P at 573 K in inert atmosphere for several days. Polymeric structure with chains of P₄ tetrahedra linked together. Less reactive, non-poisonous, stable, does not glow, used in safety matches.
Black phosphorus: Most stable allotrope, has metallic lustre, layered structure resembling graphite. Two forms: α-black (from red P heated under high pressure) and β-black (from white P at 473 K). It is the least reactive form.
Q20. Explain the manufacture of sulphuric acid by the Contact Process. Mention all reactions, conditions, and the role of catalyst.
Answer: The Contact Process involves the following stages:
Stage 1 — Production of SO₂: Sulphur or pyrites are burned in air:
S + O₂ → SO₂ or 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Stage 2 — Conversion of SO₂ to SO₃: Purified SO₂ and air are passed over V₂O₅ catalyst at 720 K and 2 atm:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH = −196 kJ
Le Chatelier’s principle suggests low temperature and high pressure favour the forward reaction, but optimum temperature ~720 K is used for reasonable rate.
Stage 3 — Absorption of SO₃ in conc. H₂SO₄:
SO₃ + H₂SO₄ → H₂S₂O₇ (oleum)
Direct absorption in water is avoided because it produces a fine mist. Oleum is later diluted with calculated water to give H₂SO₄ of desired concentration:
H₂S₂O₇ + H₂O → 2H₂SO₄
Q21. Discuss the preparation, properties, and structure of XeF₂, XeF₄, XeF₆, and XeOF₄.
Answer: Xenon forms several fluorides and oxofluorides under controlled conditions:
XeF₂: Prepared by Xe + F₂ (2:1) at 673 K and 1 bar in Ni vessel. Linear shape, sp³d hybridization (3 lone pairs on Xe). Colourless crystalline solid; mild fluorinating agent.
XeF₄: Prepared by Xe + F₂ (1:5) at 873 K and 7 bar. Square planar, sp³d² hybridization (2 lone pairs). Colourless crystalline solid.
XeF₆: Prepared by Xe + F₂ (1:20) at 573 K and 60–70 bar. Distorted octahedral, sp³d³ hybridization (1 lone pair). Colourless solid; strong fluorinating agent. On partial hydrolysis it gives XeOF₄ and XeO₂F₂; complete hydrolysis gives XeO₃.
XeOF₄: Prepared by partial hydrolysis of XeF₆: XeF₆ + H₂O → XeOF₄ + 2HF. Square pyramidal shape, sp³d² hybridization with one lone pair and one Xe=O bond. Colourless volatile liquid.
Q22. Compare the oxoacids of chlorine — HClO, HClO₂, HClO₃, and HClO₄ — with respect to oxidation state, structure, and acidic strength.
Answer: The four oxoacids of chlorine are:
HClO (Hypochlorous acid): Cl in +1 state. Structure H–O–Cl (bent at O). Weakest acid; strong oxidizer.
HClO₂ (Chlorous acid): Cl in +3 state. Structure H–O–Cl=O. Stronger than HClO but weaker than HClO₃.
HClO₃ (Chloric acid): Cl in +5 state. Tetrahedral with one O–H, one Cl=O double bond and one resonance pair. Strong acid; strong oxidant.
HClO₄ (Perchloric acid): Cl in +7 state. Regular tetrahedral structure, three Cl=O double bonds and one Cl–O–H. Strongest oxoacid known. Acidic strength order: HClO < HClO₂ < HClO₃ < HClO₄, since increasing oxidation state and number of oxygen atoms stabilize the conjugate base by spreading negative charge.
Multiple Choice Questions (MCQs)
Q1. The general electronic configuration of Group 16 elements is:
(a) ns²np³ (b) ns²np⁴ (c) ns²np⁵ (d) ns²np⁶
Answer: (b) ns²np⁴
Q2. The catalyst used in Haber’s process is:
(a) Pt/Rh (b) V₂O₅ (c) Fe with Mo (d) Ni
Answer: (c) Fe with Mo
Q3. Which allotrope of phosphorus is most stable?
(a) White (b) Yellow (c) Red (d) Black
Answer: (d) Black
Q4. The shape of XeF₂ molecule is:
(a) Bent (b) Linear (c) T-shape (d) Trigonal planar
Answer: (b) Linear
Q5. The strongest oxoacid of chlorine is:
(a) HClO (b) HClO₂ (c) HClO₃ (d) HClO₄
Answer: (d) HClO₄
Q6. Which of the following is a pseudohalogen?
(a) Br⁻ (b) Cl⁻ (c) CN⁻ (d) F⁻
Answer: (c) CN⁻
Q7. The catalyst used in the Contact Process is:
(a) Pt (b) Fe (c) V₂O₅ (d) MnO₂
Answer: (c) V₂O₅
Q8. The shape of XeF₄ is:
(a) Tetrahedral (b) Square planar (c) See-saw (d) Trigonal bipyramidal
Answer: (b) Square planar
Q9. White phosphorus is stored under:
(a) Kerosene (b) Alcohol (c) Water (d) CCl₄
Answer: (c) Water
Q10. Which of the following is an interhalogen compound?
(a) HCl (b) ClF (c) NaCl (d) Cl₂
Answer: (b) ClF
Q11. The basic character of hydrides of Group 15 follows the order:
(a) NH₃ < PH₃ < AsH₃ < SbH₃ (b) NH₃ > PH₃ > AsH₃ > SbH₃ (c) PH₃ > NH₃ > AsH₃ > SbH₃ (d) SbH₃ > AsH₃ > PH₃ > NH₃
Answer: (b) NH₃ > PH₃ > AsH₃ > SbH₃
Q12. The bleaching action of SO₂ is due to:
(a) Oxidation (b) Reduction (c) Hydrolysis (d) Decomposition
Answer: (b) Reduction
Fill in the Blanks
1. The bond angle in P₄ molecule is _______.
Answer: 60°
2. The hybridization of S in SO₃ is _______.
Answer: sp²
3. Ozone is prepared by passing _______ electric discharge through O₂.
Answer: silent
4. The shape of XeF₆ is _______.
Answer: distorted octahedral
5. Phosphine is prepared by hydrolysis of _______.
Answer: calcium phosphide (Ca₃P₂)
True or False
1. Nitrogen forms pπ–pπ multiple bonds easily. — True
2. H₃PO₃ is a tribasic acid. — False (It is dibasic, having two replaceable H atoms.)
3. Ozone is a powerful oxidizing agent. — True
4. XeF₄ has a square planar structure. — True
5. All noble gases form fluorides under normal conditions. — False (Only Xe and Kr form fluorides; lighter ones do not under normal conditions.)
6. Aqua regia dissolves gold and platinum. — True
7. H₂SO₄ acts as a dehydrating agent on sugar. — True
Glossary
| Term | Definition |
|---|---|
| Pnicogens | Group 15 elements (N, P, As, Sb, Bi). |
| Chalcogens | Group 16 elements (O, S, Se, Te, Po) — ore-forming elements. |
| Halogens | Group 17 elements (F, Cl, Br, I, At) — salt-formers. |
| Noble gases | Group 18 elements (He, Ne, Ar, Kr, Xe, Rn) with stable octet/duplet. |
| Inert pair effect | Reluctance of ns² electrons to participate in bonding for heavy p-block elements; +3 state more stable than +5 down Group 15. |
| Allotropy | Existence of an element in two or more different physical forms (e.g., O₂/O₃; white, red, black P). |
| Ostwald’s process | Industrial preparation of HNO₃ by catalytic oxidation of NH₃ over Pt/Rh. |
| Haber’s process | Industrial preparation of NH₃ from N₂ and H₂ using Fe catalyst with Mo promoter. |
| Contact Process | Manufacture of H₂SO₄ by oxidation of SO₂ to SO₃ using V₂O₅, then absorption in H₂SO₄ to form oleum. |
| Oleum | Fuming sulphuric acid; H₂S₂O₇ (pyrosulphuric acid). |
| Interhalogen compound | Compound formed between two different halogens (e.g., ClF, BrF₃, IF₅, IF₇). |
| Pseudohalogen | Polyatomic anion or molecule resembling halides/halogens in chemistry (e.g., CN⁻, SCN⁻, (CN)₂). |
| Disproportionation | Reaction where the same element is simultaneously oxidized and reduced, e.g., 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O. |
| Aerogens | Another name for Group 18 noble gases. |
| Brown ring test | Detection of NO₃⁻ ion using FeSO₄ and conc. H₂SO₄, forming brown [Fe(H₂O)₅NO]²⁺. |
| Ozonizer | Apparatus for producing O₃ by silent electric discharge through O₂ (Siemens’ or Brodie’s ozonizer). |
| Anomalous behaviour | Different chemical behaviour of the first member of a group from the rest, e.g., N differs from P, As, Sb, Bi due to small size and absence of d-orbitals. |
| Aqua regia | Mixture of conc. HNO₃ and conc. HCl in 1:3 ratio; dissolves noble metals like gold and platinum. |
| Tailing of mercury | Mercury loses its meniscus and sticks to the glass surface in presence of ozone due to formation of Hg₂O. |
| Ring test (sulphate) | Used to detect SO₄²⁻ ion using BaCl₂ giving white BaSO₄ precipitate insoluble in dilute HCl. |
| Photographic fixer | Sodium thiosulphate (hypo, Na₂S₂O₃·5H₂O) used to dissolve unreduced AgBr from photographic film. |