Chapter 11 — The p-Block Elements (Groups 13 & 14)
Welcome to HSLC Guru! This English-medium guide covers Class 11 Chemistry Chapter 11 — The p-Block Elements (Groups 13 & 14) for the ASSEB syllabus. You will find a clear summary, short and long question answers, multiple-choice questions, fill-in-the-blanks, true/false, and a glossary table to help you master the boron and carbon families.
Summary
The p-block elements occupy Groups 13 to 18 of the periodic table. Their general electronic configuration is ns²np¹⁻⁶. Group 13 is called the boron family and consists of B, Al, Ga, In, and Tl with general configuration ns²np¹. Boron is a non-metal (metalloid), while the rest are metals. Down the group, atomic radii increase, ionisation enthalpy generally decreases (with irregularities due to poor shielding by d and f electrons), and metallic character increases. The common oxidation state is +3, but due to the inert pair effect, the +1 state becomes more stable down the group, making Tl⁺ more stable than Tl³⁺.
Boron shows anomalous behaviour due to its small size, high ionisation enthalpy, and absence of d-orbitals. It forms covalent compounds, exists as a hard solid, and shows a maximum covalency of 4. Important compounds include borax (Na₂B₄O₇·10H₂O) used in glass and as a buffer, orthoboric acid (H₃BO₃) which is a weak monobasic Lewis acid, and diborane (B₂H₆) which has a banana-bond (3-centre 2-electron) structure. Aluminium is extracted from bauxite by the Bayer process followed by the Hall–Héroult electrolytic process using molten cryolite (Na₃AlF₆) and fluorspar. Aluminium is light, ductile, a good conductor, and is used in utensils, transmission lines, foils, and alloys (duralumin, magnalium). Alums are double sulphates of the type M₂SO₄·M₂(SO₄)₃·24H₂O (e.g., potash alum) used as mordants and water purifiers.
Group 14, the carbon family, contains C, Si, Ge, Sn, and Pb with configuration ns²np². Carbon and silicon are non-metals, germanium is a metalloid, while tin and lead are metals. Common oxidation states are +4 and +2; +2 becomes more stable down the group due to the inert pair effect (Pb²⁺ is stable). Carbon shows catenation (self-linking) most strongly because of high C–C bond enthalpy, giving rise to organic chemistry. Allotropes of carbon include diamond (sp³, hardest natural substance), graphite (sp², layered, conducts electricity, lubricant), and fullerenes such as C₆₀ (buckminsterfullerene, soccer-ball shape). Important carbon oxides are CO (neutral, poisonous, reducing agent) and CO₂ (acidic, greenhouse gas).
Silicon is the second most abundant element in the earth’s crust. It forms silicates (orthosilicate, pyrosilicate, cyclic, chain, sheet, three-dimensional) which form rocks, minerals, clays, and gemstones. Silicones are organosilicon polymers with –Si–O–Si– backbone, used as water-repellents, lubricants, and biomedical implants. Zeolites are aluminosilicates used as catalysts (cracking petroleum, ZSM-5), molecular sieves, and ion exchangers. Ultra-pure silicon is the foundation of the semiconductor and silicon-chip industry. Carbides include ionic (CaC₂), covalent (SiC, carborundum), and interstitial types. Together, the boron and carbon families show how trends in size, ionisation, and bonding shape the chemistry of metalloids, light metals, and non-metals.
1-Mark Question Answers
Q1. Write the general electronic configuration of Group 13 elements.
Answer: ns²np¹.
Q2. Name the most abundant metal in the earth’s crust.
Answer: Aluminium.
Q3. What is the formula of borax?
Answer: Na₂B₄O₇·10H₂O.
Q4. Which allotrope of carbon is the hardest natural substance?
Answer: Diamond.
Q5. Which allotrope of carbon is used as a lubricant?
Answer: Graphite.
Q6. Name the process used industrially to extract aluminium from alumina.
Answer: The Hall–Héroult electrolytic process.
Q7. What is the chemical formula of buckminsterfullerene?
Answer: C₆₀.
Q8. Name a poisonous oxide of carbon.
Answer: Carbon monoxide (CO).
Q9. Write the formula of orthoboric acid.
Answer: H₃BO₃.
Q10. What is the hybridisation of carbon in graphite?
Answer: sp² hybridisation.
Q11. Name the most abundant element in the earth’s crust after oxygen.
Answer: Silicon.
Q12. Write the formula of cryolite.
Answer: Na₃AlF₆.
2 to 3-Mark Question Answers
Q1. Why does boron show anomalous behaviour in Group 13?
Answer: Boron differs from the rest of Group 13 because of its (i) very small atomic size, (ii) high ionisation enthalpy and high electronegativity, and (iii) absence of d-orbitals in its valence shell. As a result, boron is a non-metal (metalloid), forms only covalent compounds, has a maximum covalency of 4 (e.g., BF₄⁻), and does not form a stable +1 state. The other members are metals, form ionic +3 (and +1) compounds, and can expand their covalency beyond 4 using d-orbitals.
Q2. What is the inert pair effect? Illustrate with Group 13 and Group 14 examples.
Answer: The reluctance of the outermost ns² electron pair to participate in bonding, due to poor shielding by intervening d- and f-electrons, is called the inert pair effect. Because of this, the lower oxidation state (group number − 2) becomes more stable on going down the group. Examples: in Group 13, Tl⁺ is more stable than Tl³⁺; in Group 14, Pb²⁺ is more stable than Pb⁴⁺ (PbO₂ is a strong oxidising agent).
Q3. Describe the structure of diborane (B₂H₆).
Answer: Diborane has 12 valence electrons but appears to need 14 for normal 2c–2e bonds; it is therefore an electron-deficient molecule. Its structure consists of two BH₂ units in one plane joined by two bridging hydrogens above and below this plane. The four terminal B–H bonds are normal 2-centre 2-electron bonds, while the two B–H–B bridges are 3-centre 2-electron (banana) bonds. Each boron is sp³ hybridised.
Q4. Why is graphite a good conductor of electricity but diamond is not?
Answer: In graphite, each carbon is sp² hybridised and bonded to three others in flat hexagonal layers, leaving one delocalised p-electron per atom that moves freely along the layer; this allows electrical conduction. In diamond, every carbon is sp³ hybridised and uses all four valence electrons in strong, localised C–C σ-bonds in a 3D tetrahedral network; no free electrons are available, so diamond is a non-conductor (but a very good thermal conductor).
Q5. What are silicones? Give two important uses.
Answer: Silicones are synthetic organosilicon polymers having the repeating unit (R₂SiO)ₙ, with an –Si–O–Si– backbone and organic side groups (R = CH₃, C₆H₅, etc.). They are made by hydrolysis of dialkyl- or diaryl-dichlorosilanes followed by condensation. Uses: (i) as water-repellent coatings on textiles and paper, (ii) as high-temperature lubricants and greases, (iii) in electrical insulation, sealants, and biomedical implants.
Q6. Distinguish between CO and CO₂ in two points.
Answer: (i) Nature: CO is a neutral oxide and a powerful reducing agent; CO₂ is an acidic oxide and a weak oxidiser. (ii) Toxicity / use: CO is highly poisonous because it forms carboxyhaemoglobin with blood and is used in metallurgy (e.g., reducing iron ore); CO₂ is non-toxic at normal levels, used in fire extinguishers, fizzy drinks, and is the main greenhouse gas.
Q7. Why is BCl₃ a stronger Lewis acid than AlCl₃ in many reactions, although boron is smaller?
Answer: Both BCl₃ and AlCl₃ have an empty p-orbital on the central atom and can accept a lone pair, making them Lewis acids. However, in BCl₃ there is no back-donation from chlorine into the small B 2p-orbital that fully removes its acceptor character (only weak pπ–pπ overlap); the boron remains strongly electron-deficient. In AlCl₃ the larger Al uses 3p/3d orbitals and is also stabilised by dimerisation to Al₂Cl₆ where each Al is coordinatively saturated, lowering its tendency to accept further electron pairs. Hence in many reactions monomeric BCl₃ behaves as a stronger Lewis acid.
Q8. What are alums? Write the formula of potash alum and give one use.
Answer: Alums are isomorphous double sulphates of the general formula M₂SO₄·M₂(SO₄)₃·24H₂O, where M is a univalent cation (Na⁺, K⁺, NH₄⁺) and M’ is a trivalent cation (Al³⁺, Cr³⁺, Fe³⁺). They form octahedral crystals and contain 24 molecules of water of crystallisation. Potash alum: K₂SO₄·Al₂(SO₄)₃·24H₂O. Use: as a mordant in dyeing, in tanning of leather, in water purification (coagulates suspended impurities), and as a styptic to stop bleeding from minor cuts.
5 to 7-Mark Question Answers
Q1. Discuss the trends in atomic size, ionisation enthalpy, oxidation states, and metallic character in Group 13. Why is ionisation enthalpy of Ga higher than that of Al?
Answer: (i) Atomic size: generally increases from B to Tl, but Ga is almost equal to (slightly smaller than) Al because the 3d¹⁰ electrons in Ga shield the nuclear charge poorly. (ii) Ionisation enthalpy: overall decreases down the group, but the trend is not smooth — Ga has higher first IE than Al, and Tl has higher IE than In, again due to poor shielding by d/f electrons (lanthanide contraction also affects Tl). (iii) Oxidation states: +3 is the principal state; +1 becomes increasingly stable down the group due to the inert pair effect, so Tl⁺ > Tl³⁺ in stability. (iv) Metallic character: increases from B (metalloid, non-metal) through Al, Ga, In to Tl (typical metals); this is because ionisation enthalpies generally decrease and electropositive character grows. Boron forms only covalent compounds, while the rest form mainly ionic compounds in +3 and +1 states.
Q2. Describe the preparation, properties, and uses of (a) borax and (b) orthoboric acid.
Answer: (a) Borax (Na₂B₄O₇·10H₂O): obtained from the mineral tincal by recrystallisation, or by treating colemanite with Na₂CO₃. It is a white crystalline solid, soluble in water, giving an alkaline solution due to hydrolysis: Na₂B₄O₇ + 7H₂O → 2NaOH + 4H₃BO₃. On heating, it loses water and finally yields a transparent glassy bead of sodium metaborate and boric anhydride (NaBO₂ + B₂O₃) — the basis of the borax-bead test for coloured metal ions. Uses: in glass and enamel industry, as a flux in metallurgy, in pharmaceutical preparations, and as a buffer.
(b) Orthoboric acid (H₃BO₃): prepared by acidifying an aqueous solution of borax: Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃. It is a soft, white, soapy crystalline solid with layered structure (planar BO₃ units linked by H-bonds). It is a weak monobasic Lewis acid that accepts OH⁻ from water: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺. On heating, it loses water successively to form metaboric acid (HBO₂) and finally boric anhydride (B₂O₃). Uses: antiseptic, eye-wash, in glass, glaze, and enamels, and as a food preservative.
Q3. Explain the extraction of aluminium by the Hall–Héroult process. Give two important properties and three uses of aluminium.
Answer: Aluminium is extracted from purified alumina (Al₂O₃) obtained from bauxite by the Bayer process. In the Hall–Héroult process, alumina is dissolved in molten cryolite (Na₃AlF₆) along with fluorspar (CaF₂) to lower the melting point (from 2050°C to about 950°C) and increase electrical conductivity. The molten mixture is electrolysed in a steel tank lined with carbon (cathode) using carbon anodes. Reactions: at cathode, Al³⁺ + 3e⁻ → Al (l); at anode, 2O²⁻ → O₂ + 4e⁻, and the liberated oxygen reacts with the carbon anode to form CO and CO₂, so the anodes must be replaced periodically. Molten aluminium settles at the bottom and is tapped out (about 99% pure). Properties: light silvery-white metal; high tensile strength; excellent conductor of heat and electricity; reacts with both acids and alkalis (amphoteric); develops a protective Al₂O₃ layer in air. Uses: (i) household utensils, foils, and packaging; (ii) overhead electrical transmission lines (high conductivity per unit weight); (iii) construction, aircraft, and automobile bodies (alloys like duralumin and magnalium).
Q4. Compare the structure and properties of diamond, graphite, and fullerene.
Answer: (i) Diamond: each C is sp³ hybridised and bonded tetrahedrally to four other carbons by strong σ-bonds, giving a giant 3D covalent network. It is the hardest natural substance, has a very high melting point, is a non-conductor of electricity (no free electrons) but an excellent thermal conductor, and is used in cutting tools and jewellery. (ii) Graphite: each C is sp² hybridised, bonded to three others to form planar hexagonal layers; the fourth electron is delocalised in a π-cloud above and below the layer. Layers are held together by weak van der Waals forces, so they slide easily — graphite is soft, slippery, and used as a lubricant and pencil lead. The delocalised electrons make graphite a good conductor of electricity along the layers; it is used in electrodes. (iii) Fullerenes: closed-cage molecules of carbon, the most famous being C₆₀ (buckminsterfullerene), shaped like a soccer ball with 20 hexagons and 12 pentagons. Each C is sp² hybridised. Fullerenes are molecular solids, soluble in organic solvents, and find use in nanotechnology, superconductors, and drug delivery research.
Q5. What are silicates and zeolites? Give the classification of silicates and two uses of zeolites.
Answer: Silicates are compounds containing the basic tetrahedral SiO₄⁴⁻ unit, in which silicon is sp³ hybridised. They are the chief constituents of rocks, sand, clay, and minerals. Depending on how the SiO₄ tetrahedra are linked through corner-sharing of oxygen, silicates are classified as: (i) Orthosilicates (single tetrahedron, e.g., Mg₂SiO₄ — olivine, zircon ZrSiO₄). (ii) Pyrosilicates (two tetrahedra sharing one O, Si₂O₇⁶⁻; e.g., thortveitite). (iii) Cyclic silicates (closed rings, e.g., beryl Be₃Al₂Si₆O₁₈). (iv) Chain silicates (single chains like pyroxenes, double chains like amphiboles/asbestos). (v) Sheet silicates (layered, e.g., mica, talc, kaolin). (vi) Three-dimensional silicates (all four oxygens shared, e.g., quartz). Zeolites are 3D aluminosilicates with a porous framework where some Si is replaced by Al; the resulting negative charge is balanced by Na⁺, K⁺, or Ca²⁺ ions. Uses of zeolites: (i) as catalysts (ZSM-5 in petroleum cracking and conversion of alcohol to gasoline); (ii) as molecular sieves for separating molecules and softening hard water by ion exchange.
Multiple Choice Questions
Q1. The general electronic configuration of Group 14 elements is —
(a) ns²np¹ (b) ns²np² (c) ns²np³ (d) ns²np⁴
Answer: (b) ns²np².
Q2. Which of the following is a metalloid?
(a) Aluminium (b) Boron (c) Indium (d) Thallium
Answer: (b) Boron.
Q3. Diborane has —
(a) only 2c–2e bonds (b) only 3c–2e bonds (c) both 2c–2e and 3c–2e bonds (d) ionic bonds
Answer: (c) both 2c–2e and 3c–2e bonds.
Q4. The hybridisation of carbon in diamond is —
(a) sp (b) sp² (c) sp³ (d) sp³d
Answer: (c) sp³.
Q5. The chief ore of aluminium is —
(a) Cryolite (b) Bauxite (c) Corundum (d) Feldspar
Answer: (b) Bauxite.
Q6. Cryolite used in Hall–Héroult process has the formula —
(a) Na₃AlF₆ (b) NaAlF₄ (c) Na₂AlF₅ (d) NaAlO₂
Answer: (a) Na₃AlF₆.
Q7. The shape of buckminsterfullerene C₆₀ resembles —
(a) a cube (b) a soccer ball (c) a tetrahedron (d) a sheet
Answer: (b) a soccer ball.
Q8. Which carbide is also called carborundum?
(a) CaC₂ (b) Al₄C₃ (c) SiC (d) Be₂C
Answer: (c) SiC.
Q9. Which of the following is a Lewis acid?
(a) BCl₃ (b) NH₃ (c) H₂O (d) NaOH
Answer: (a) BCl₃.
Q10. Which oxidation state of Pb is more stable?
(a) +1 (b) +2 (c) +3 (d) +4
Answer: (b) +2 (due to inert pair effect).
Fill in the Blanks
Q1. The most abundant element in the earth’s crust is __________.
Answer: oxygen (silicon is the most abundant after oxygen).
Q2. The phenomenon of self-linking of carbon atoms is called __________.
Answer: catenation.
Q3. Diborane contains __________ banana bonds.
Answer: two.
Q4. Potash alum has the formula K₂SO₄·Al₂(SO₄)₃·__________ H₂O.
Answer: 24.
Q5. Zeolites are crystalline __________ used as catalysts and molecular sieves.
Answer: aluminosilicates.
True / False
Q1. Boron forms only covalent compounds.
Answer: True.
Q2. Aluminium is amphoteric in nature.
Answer: True.
Q3. Carbon monoxide is an acidic oxide.
Answer: False (it is a neutral oxide).
Q4. Graphite is a good conductor of electricity.
Answer: True.
Q5. Pb⁴⁺ is more stable than Pb²⁺.
Answer: False (Pb²⁺ is more stable due to inert pair effect).
Glossary
| Term | Meaning |
|---|---|
| p-Block elements | Elements of Groups 13 to 18 in which the last electron enters a p-orbital. |
| Inert pair effect | Reluctance of the ns² electron pair to take part in bonding, stabilising lower oxidation states down a group. |
| Borax | Sodium tetraborate decahydrate, Na₂B₄O₇·10H₂O; used in glass, flux, and bead test. |
| Orthoboric acid | H₃BO₃; a weak monobasic Lewis acid with a layered planar structure. |
| Diborane | B₂H₆; an electron-deficient hydride containing 3-centre 2-electron banana bonds. |
| Hall–Héroult process | Industrial extraction of aluminium by electrolysis of alumina dissolved in molten cryolite. |
| Alum | Double sulphate of the type M₂SO₄·M₂(SO₄)₃·24H₂O, e.g., potash alum. |
| Catenation | Property of an element to form long chains or rings of its own atoms; strongest in carbon. |
| Allotropes of carbon | Different forms of carbon — diamond (sp³), graphite (sp²), fullerenes (e.g., C₆₀). |
| Carbide | Binary compound of carbon with a less electronegative element; ionic, covalent, or interstitial. |
| Silicate | Compound containing the SiO₄⁴⁻ tetrahedral unit; basis of rocks and minerals. |
| Silicone | Synthetic organosilicon polymer with –Si–O–Si– backbone and organic side groups. |
| Zeolite | Porous 3D aluminosilicate used as catalyst, molecular sieve, and ion exchanger. |
| Lewis acid | An electron-pair acceptor; e.g., BF₃, BCl₃, AlCl₃. |
| Amphoteric oxide | An oxide that reacts with both acids and bases; e.g., Al₂O₃, SnO, PbO. |