Chapter 20: The p-Block Elements (Group 13 and 14)

Introduction to P-Block Elements

General Electronic Configuration

• Last electron enters outermost p-orbital.

• General configuration: ns2 np1-6 (except He).

• Maximum oxidation state = total valence electrons (s + p).

• Inert pair effect causes oxidation states two less than max.

• First period elements show anomalous behavior due to smaller size and no d-orbitals.

Solved Examples

• Ex 1: Determine max oxidation state of group 15 p-block element.
  Solution: ns2 np3 → max oxidation state = 5.

Practice Set

• Q1: What is the general electronic configuration of p-block elements?
  A. ns2 np1-6 B. ns2 nd1-10 C. ns1 np0 D. ns2 nd0

• Q2: The inert pair effect causes oxidation states to be lower by how many units?
  A. 1 B. 2 C. 3 D. 4

• Q3: Why do first period p-block elements show anomalous behavior?
  A. Larger size B. Smaller size and no d-orbitals C. High electronegativity D. High atomic mass

Answer Key

• Q1: A - General configuration is ns2 np1-6.

• Q2: B - Oxidation states are two less due to inert pair effect.

• Q3: B - Smaller size and absence of d-orbitals cause anomalies.

Memory Aids

• p-block = last electron in p-orbital.

• Max oxidation state = group number (valence electrons).

• Inert pair effect lowers oxidation state by 2.

Group 13 Elements: Boron Family

Elements and Properties

• Elements: B, Al, Ga, In, Tl, Nh.

• Boron: semi-metal, high ionisation energy, closer to non-metals.

• Aluminium: metal, chemically similar to boron.

• Ga, In, Tl, Nh: metallic character increases down the group.

• Nh is radioactive, chemistry not well established.

Solved Examples

• Ex 1: Identify the element with highest ionisation energy in group 13.
  Solution: Boron (B) has highest ionisation energy.

Practice Set

• Q1: Which element in group 13 is a semi-metal?
  A. Al B. Ga C. B D. Tl

• Q2: Which element is radioactive with a half-life of 20 seconds?
  A. In B. Nh C. Tl D. Ga

• Q3: Which element is the most abundant metal in earth's crust?
  A. B B. Al C. Ga D. In

Answer Key

• Q1: C - Boron is semi-metal.

• Q2: B - Nihonium (Nh) is radioactive.

• Q3: B - Aluminium is most abundant metal.

Memory Aids

• Boron = semi-metal, high IE.

• Aluminium = abundant metal.

• Nh = radioactive, short half-life.

Atomic and Ionic Radii

Trends and Exceptions

• Group 13 elements smaller than group 2 due to higher nuclear charge.

• Atomic and ionic radii increase down the group except Ga.

• Ga atomic radius less than Al due to poor d-electron shielding.

Solved Examples

• Ex 1: Compare atomic radius of Ga and Al.
  Solution: Ga < Al due to d-electron poor shielding.

Practice Set

• Q1: Which element has smaller atomic radius?
  A. Al B. Ga C. In D. Tl

• Q2: Why does Ga have smaller radius than Al?
  A. Better shielding B. Poor d-electron shielding C. More electrons D. Less nuclear charge

• Q3: Ionic radius of M3+ is generally ______ metallic radius.
  A. Larger than B. Equal to C. Smaller than D. Unrelated

Answer Key

• Q1: B - Ga has smaller radius than Al.

• Q2: B - Poor d-electron shielding in Ga.

• Q3: C - Ionic radius smaller than metallic radius for M3+.

Memory Aids

• Ga radius < Al due to d-electrons.

• Ionic radius < metallic radius for cations.

Ionisation Enthalpy

Trends and Observations

• IE values do not decrease smoothly down group 13.

• Irregularities in Ga and Tl due to d- and f-electron poor shielding.

• Order: B > Tl > Ga > Al > In for first IE.

• IEs less than corresponding group 2 elements due to p-electron shielding.

• IE1 < IE2 < IE3; sum of first three IE very high.

Solved Examples

• Ex 1: Arrange B, Al, Ga in order of IE.
  Solution: B > Tl > Ga > Al > In.

Practice Set

• Q1: Which element has highest first ionisation enthalpy?
  A. Al B. Ga C. B D. In

• Q2: Why is IE of group 13 less than group 2?
  A. More shielding B. Less penetrating p-electrons C. More protons D. Smaller size

• Q3: Sum of first three IE corresponds to formation of which ion?
  A. M+ B. M2+ C. M3+ D. M4+

Answer Key

• Q1: C - Boron has highest IE.

• Q2: B - p-electrons less penetrating than s-electrons.

• Q3: C - Sum of IE1 to IE3 forms M3+.

Memory Aids

• IE irregular due to d- and f-electron shielding.

• IE sum for M3+ very high.

Electronegativity

Trends

• Electronegativity decreases from B to Al, then increases due to poor d- and f-electron shielding.

• Group 13 elements more electronegative than groups 1 and 2.

Solved Examples

• Ex 1: Which element has highest electronegativity?
  Solution: Boron (2.0) highest among group 13.

Practice Set

• Q1: Electronegativity order in group 13 is?
  A. B > Al > Ga B. Al > B > Ga C. B > Ga > Al D. Ga > Al > B

• Q2: Why does electronegativity increase after Al?
  A. Better shielding B. Poor d- and f-electron shielding C. Larger size D. More protons

• Q3: Group 13 elements are more electronegative than?
  A. Group 1 B. Group 2 C. Both A and B D. None

Answer Key

• Q1: C - B > Ga > Al due to shielding effects.

• Q2: B - Poor shielding increases effective nuclear charge.

• Q3: C - More electronegative than groups 1 and 2.

Memory Aids

• Electronegativity dips at Al, then rises.

• Group 13 > groups 1 and 2 electronegativity.

Physical Properties

Key Points

• Boron: hard, black solid, many allotropes, very high melting point.

• Other members: soft metals, low melting points, high electrical conductivity.

• Gallium: low melting point (303 K), liquid near room temp, high boiling point (2676 K).

• Density increases down the group from B to Tl.

Solved Examples

• Ex 1: Why is gallium used in high temperature thermometry?
  Solution: High boiling point and low melting point.

Practice Set

• Q1: Which element is hardest in group 13?
  A. B B. Al C. Ga D. Tl

• Q2: Gallium melts at?
  A. 303 K B. 500 K C. 1000 K D. 2676 K

• Q3: Density trend down group 13 is?
  A. Decreases B. Increases C. Constant D. Random

Answer Key

• Q1: A - Boron is hardest.

• Q2: A - Gallium melts at 303 K.

• Q3: B - Density increases down the group.

Memory Aids

• Boron = hard, high MP.

• Gallium = melts near room temp.

• Density increases down group.

Chemical Properties

Oxidation States and Reactivity

• Valence config: ns2 np1 → +1 and +3 oxidation states.

• +1 state stability increases down group due to inert pair effect.

• Stability order +3: B3+ > Al3+ > Ga3+ > In3+ > Tl3+.

• Stability order +1: Tl+ > In+ > Ga+ > Al+ > B+.

• +1 compounds more ionic than +3 compounds.

• Boron halides (BX3) are electron deficient, act as Lewis acids.

Solved Examples

• Ex 1: Write reaction of BX3 with NH3.
  Solution: BX3 + NH3 → {H3N → BX3} (Lewis acid-base complex).

Practice Set

• Q1: Which oxidation state becomes more stable down group 13?
  A. +3 B. +1 C. 0 D. -1

• Q2: Order of Lewis acid strength of boron halides?
  A. BF3 < BCl3 < BBr3 < BI3 B. BI3 < BBr3 < BCl3 < BF3 C. BCl3 < BF3 < BI3 < BBr3 D. BBr3 < BI3 < BF3 < BCl3

• Q3: Which compound forms a dimer?
  A. BCl3 B. AlCl3 C. BF3 D. BI3

Answer Key

• Q1: B - +1 oxidation state more stable down group.

• Q2: A - Lewis acid strength increases BF3 < BCl3 < BBr3 < BI3.

• Q3: B - AlCl3 forms dimer Al2Cl6.

Memory Aids

• Inert pair effect stabilizes +1 state down group.

• Boron halides are Lewis acids.

• AlCl3 forms dimers; BCl3 does not.

Reactivity towards Air, Acids, Alkalis, and Halogens

Reactivity towards Air

• Group 13 elements form trioxides (E2O3) on heating in oxygen.

• Tl also forms some Tl2O.

Reactivity towards Acids and Alkalis

• Boron unaffected by non-oxidising acids (dil. HCl, H2SO4).

• Other elements dissolve in acids forming M3+ salts and liberate H2.

• Oxidising acids (conc. H2SO4, HNO3) react with all; conc. HNO3 passivates Al.

• B, Al, Ga react with alkalis liberating H2.

Reactivity towards Halogens

• Group 13 elements react with halogens to form trihalides (except TlI3).

• Reaction: 2E(s) + 3X2(g) → 2EX3(s) (X = F, Cl, Br, I).

Solved Examples

• Ex 1: Write reaction of Al with HCl.
  Solution: 2Al + 6HCl → 2AlCl3 + 3H2.

• Ex 2: Write reaction of Al with NaOH.
  Solution: 2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2.

Practice Set

• Q1: Which oxide is acidic?
  A. B2O3 B. Al2O3 C. In2O3 D. Tl2O

• Q2: Which element is not affected by dilute HCl?
  A. B B. Al C. Ga D. Tl

• Q3: Reaction of group 13 element with halogen forms?
  A. EX B. EX2 C. EX3 D. EX4

Answer Key

• Q1: A - B2O3 is acidic oxide.

• Q2: A - Boron unaffected by dilute HCl.

• Q3: C - Trihalides EX3 formed.

Memory Aids

• Oxides: B2O3 acidic, Al2O3 amphoteric, others basic.

• Boron inert to dilute acids.

• Group 13 + halogens → trihalides.

Important Trends and Anomalous Properties of Boron

Key Characteristics

• Boron is the first element of group 13 and shows unique behaviour compared to aluminium and others.

• It has the smallest atomic size and highest ionisation energy in the group.

• Highest electronegativity among group 13 elements.

• Maximum covalency of 4 due to absence of d-orbitals; others can have covalency up to 6.

• Exhibits only +3 oxidation state; others show +1 and +3.

• Boron halides are monomeric; others form dimeric halides.

• Does not exhibit inert pair effect unlike heavier group members.

• Forms volatile, electron-deficient hydrides; others form polymeric hydrides; thallium does not form hydrides.

• Combines with metals to form borides; others form alloys instead.

Solved Examples

Ex 1: Explain why boron shows only +3 oxidation state while thallium shows +1 and +3.

• Boron lacks d-orbitals, so it cannot expand its octet; stable oxidation state is +3.

• Thallium has inert pair effect; s-electrons are less involved, allowing +1 and +3 states.

Ex 2: Why are boron halides monomeric but aluminium halides dimeric?

• Boron halides have small size and high charge density, forming strong covalent bonds as monomers.

• Aluminium halides form dimers to satisfy electron deficiency via coordinate bonds.

Practice Set

Q1: Which property of boron is responsible for its anomalous behaviour in group 13?

• A. Large atomic size

• B. Low ionisation energy

• C. Absence of d-orbitals

• D. Presence of inert pair effect

Q2: Boron forms volatile hydrides because:

• A. They are electron-rich compounds

• B. Boron has d-orbitals to expand octet

• C. They are electron-deficient compounds

• D. Boron forms ionic hydrides

Q3: Which of the following statements is correct?

• A. Boron halides are dimeric

• B. Thallium forms hydrides

• C. Boron forms borides with metals

• D. Aluminium shows only +3 oxidation state

Answer Key

Q1: C. Absence of d-orbitals — Boron's lack of d-orbitals limits its covalency and oxidation states.

Q2: C. They are electron-deficient compounds — Boron's hydrides lack sufficient electrons, making them volatile.

Q3: C. Boron forms borides with metals — Boron forms borides; thallium does not form hydrides; boron halides are monomeric; aluminium shows +1 and +3 oxidation states.

Memory Aids

• Boron’s small size and no d-orbitals = max covalency 4, only +3 state.

• Halides: Boron monomeric, others dimeric.

• Hydrides: Boron volatile & electron-deficient; others polymeric; Tl no hydrides.

• Boron forms borides; others form alloys.

Some Important Compounds of Boron

Borax

• White crystalline solid: Na2[B4O5(OH)4].8H2O (commonly Na2B4O7.10H2O)

• Dissolves in water to give alkaline solution: Na2B4O7 + 7H2O → 2NaOH + 4H3BO3

• On heating, loses water, swells, melts to transparent liquid, then solidifies to glass-like borax bead

• Borax bead test identifies transition metals by characteristic colours of their metaborates

Solved Examples

• Ex 1: Write the reaction when borax is heated.

  • Na2B4O7.10H2O (heat) → Na2B4O7 + 10H2O

  • Further heating: Na2B4O7 → 2NaBO2 + B2O3

Practice Set

• Q1: Borax formula is:

  • A. Na2B4O7.10H2O

  • B. NaBO2

  • C. B2O3

  • D. NaOH

• Q2: Borax bead test is used to identify:

  • A. Alkali metals

  • B. Transition metals

  • C. Halogens

  • D. Noble gases

• Q3: On heating borax, which compound is NOT formed?

  • A. NaBO2

  • B. B2O3

  • C. NaOH

  • D. Borax bead

Answer Key

• Q1: A - Borax is Na2B4O7.10H2O

• Q2: B - Borax bead test identifies transition metals by their coloured metaborates

• Q3: C - NaOH is formed in aqueous solution, not on heating borax

Memory Aids

• Borax bead test: Transition metals give coloured beads

• Borax formula: Think Na2B4O7.10H2O

Orthoboric Acid

Properties and Preparation

• White crystalline solid with soapy touch

• Sparingly soluble in cold water, highly soluble in hot water

• Prepared by acidifying borax solution: Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3

• Layered structure with planar BO3 units joined by hydrogen bonds

• Weak monobasic acid; acts as Lewis acid by accepting OH− electrons: B(OH)3 + 2H2O → [B(OH)4]− + H3O+

Thermal Decomposition

• On heating above 370 K: H3BO3 → HBO2 (metaboric acid)

• Further heating: HBO2 → B2O3 (boric oxide)

Detection Tests

• Treated with HF or conc. H2SO4 + CaF2 to form BF3 gas, which gives green flame in Bunsen burner

• Reaction: H3BO3 + 3HF → 2BF3 + 3H2O

• With conc. H2SO4 and methyl alcohol, warmed and ignited, gives green flame

  • B(OH)3 + 3CH3OH → B(OCH3)3 + 3H2O

Solved Examples

• Ex 1: Write the reaction for preparation of orthoboric acid from borax.

  • Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3

Practice Set

• Q1: Orthoboric acid acts as:

  • A. Strong acid

  • B. Lewis acid

  • C. Base

  • D. Salt

• Q2: On heating orthoboric acid, the first product formed is:

  • A. Boric oxide

  • B. Metaboric acid

  • C. Borax

  • D. Borazine

• Q3: BF3 gas formed during detection test imparts which flame colour?

  • A. Red

  • B. Blue

  • C. Green

  • D. Yellow

Answer Key

• Q1: B - Orthoboric acid is a Lewis acid accepting electrons from OH−

• Q2: B - Heating orthoboric acid first forms metaboric acid (HBO2)

• Q3: C - BF3 gas gives green flame in Bunsen burner

Memory Aids

• Orthoboric acid = Lewis acid, not proton donor

• Heating sequence: H3BO3 → HBO2 → B2O3

• BF3 gas → green flame

Diborane

Properties and Preparation

• Simplest boron hydride: B2H6

  Colourless, highly toxic gas; boiling point 180 K

• Catches fire spontaneously in air: B2H6 + 3O2 → B2O3 + 3H2O; ΔH° = -1967 kJ/mol

• Prepared by:

  • BF3 + LiAlH4 → B2H6 + LiF + AlF3

  • NaBH4 + I2 → B2H6 + NaI + H2

  • BF3 + NaH (450 K) → B2H6 + NaF

  

• Undergoes hydrolysis: B2H6 + 6H2O → 2H3BO3 + 6H2

• Cleavage with Lewis bases forms borane adducts: B2H6 + 2NMe3 → 2BH3NMe3

Structure

• Non-planar; B is sp3 hybridised

• Four normal 2c–2e B–H bonds in plane

• Two 3c–2e bridge bonds (banana bonds) above and below plane

Reaction with Ammonia

• Diborane + NH3 forms addition product

• On heating, forms borazine (B3N3H6), called inorganic benzene

• 

  Borazine has alternating BH and NH groups in cyclic structure

Borohydrides

• Tetrahydridoborates of metals prepared by metal hydrides + B2H6 in diethyl ether

• LiBH4 and NaBH4 are important reducing agents in organic synthesis

Solved Examples

• Ex 1: Write the reaction of NaBH4 with I2 to form diborane.

  • NaBH4 + I2 → B2H6 + NaI + H2

Practice Set

• Q1: Diborane is:

  • A. Solid

  • B. Colourless gas

  • C. Coloured liquid

  • D. Metal

• Q2: Borazine is called inorganic benzene because:

  • A. It is aromatic and cyclic with alternating BH and NH groups

  • B. It is a hydrocarbon

  • C. It is a metal complex

  • D. It is a salt

• Q3: Which bond type is present in diborane?

  • A. Only 2c–2e bonds

  • B. Only ionic bonds

  • C. 3c–2e bridge bonds (banana bonds)

  • D. Metallic bonds

Answer Key

• Q1: B - Diborane is a colourless gas

• Q2: A - Borazine is aromatic with alternating BH and NH groups

• Q3: C - Diborane has 3-center 2-electron bridge bonds

Memory Aids

• Diborane = B2H6, sp3 hybridised B, banana bonds

• Borazine = inorganic benzene, alternating BH and NH

• LiBH4, NaBH4 = reducing agents

Uses of Boron and Aluminium

Applications of Boron and Its Compounds

• Boron fibres: Used in bullet-proof vests and lightweight composite materials for aircraft.

• Boron-10 isotope (\(^{10}B\)): High neutron absorption; metal borides used in nuclear reactors as control rods and protective shields.

• Borax and boric acid: Manufacture of heat-resistant glass (e.g., Pyrex), glass-wool, and fibre glass.

• Borax: Acts as a flux in soldering metals; provides heat, scratch, and stain-resistant glazed coatings on earthenware.

• Orthoboric acid solution: Used as a mild antiseptic.

Applications of Aluminium

• Aluminium: Bright silvery-white metal with high tensile strength.

• Electrical conductivity: Aluminium's conductivity is twice that of copper.

• Thermal conductivity: High thermal conductivity makes it useful in heat exchangers.

• Extensive industrial and everyday use due to lightweight, strength, and conductivity.

Solved Examples

Ex 1: Why is boron-10 used in nuclear reactors?

• Because \(^{10}B\) has a high neutron absorption cross-section, it effectively controls neutron flux.

Ex 2: Calculate the electrical conductivity ratio of aluminium to copper if aluminium's conductivity is twice that of copper.

• Given: \(\sigma_{Al} = 2 \times \sigma_{Cu}\)

• Ratio \(= \frac{\sigma_{Al}}{\sigma_{Cu}} = 2\)

Practice Set

Q1: Which isotope of boron is used in nuclear reactors for neutron absorption?

• A. \(^{11}B\)

• B. \(^{10}B\)

• C. \(^{9}B\)

• D. \(^{12}B\)

Q2: Borax is commonly used as:

• A. A fuel

• B. A flux in soldering metals

• C. A reducing agent

• D. An oxidizing agent

Q3: Aluminium is preferred over copper in electrical wiring because:

• A. It has higher tensile strength and twice the electrical conductivity of copper

• B. It is cheaper and has lower conductivity

• C. It is heavier and more durable

• D. It has lower thermal conductivity

Answer Key

Q1: B. \(^{10}B\) - It has a high neutron absorption capacity, making it ideal for nuclear control rods.

Q2: B. A flux in soldering metals - Borax lowers melting points and removes oxides during soldering.

Q3: A. It has higher tensile strength and twice the electrical conductivity of copper - Aluminium is lightweight and conducts electricity efficiently.

Memory Aids

• Boron-10: "10 absorbs neutrons" - remember isotope 10 for neutron absorption.

• Borax uses: Flux and glass - think "Borax fixes and glazes".

• Aluminium conductivity: "Al is twice Cu" - aluminium's conductivity is double copper's.

Group 14 Elements

Overview

• Group 14 contains C, Si, Ge, Sn, Pb, and synthetic Fl (Flerovium).

• Carbon, silicon, tin, and lead are well-known; germanium is rare; flerovium is radioactive with short half-life.

• Silicon is the second most abundant element in Earth's crust after oxygen.

• Electronic configurations follow noble gas shorthand with outer electrons in s and p orbitals.

Solved Examples

• Ex 1: Write the electronic configuration of Tin (Sn, Z=50).
  Answer: [Kr] 4d10 5s2 5p2

Practice Set

• Q1: Which element in group 14 is synthetic and radioactive?
  A. Carbon
  B. Silicon
  C. Flerovium
  D. Lead

• Q2: Which element is the second most abundant in Earth's crust?
  A. Carbon
  B. Silicon
  C. Germanium
  D. Tin

• Q3: The electronic configuration of lead (Pb) ends with:
  A. 6s2 6p2
  B. 5d10 6s2 6p2
  C. 4f14 5d10 6s2 6p2
  D. 5s2 5p2

Answer Key

• Q1: C - Flerovium is synthetic and radioactive.

• Q2: B - Silicon is second most abundant after oxygen.

• Q3: C - Lead's configuration includes 4f14 and 5d10 filled orbitals.

Memory Aids

• Group 14 elements: C, Si, Ge, Sn, Pb, Fl (Flerovium).

• Silicon abundant in crust; Flerovium synthetic and unstable.

Covalent Radius

Trends

• Covalent radius increases down the group.

• Increase from Si downwards is small due to poor shielding by d- and f-electrons.

• Group 14 elements have smaller covalent radii than group 13 elements.

Solved Examples

• Ex 1: Compare covalent radius of Si and Sn.
  Answer: Sn has larger radius than Si but increase is small due to ineffective shielding.

Practice Set

• Q1: Covalent radius trend in group 14 is:
  A. Decreases down the group
  B. Increases down the group
  C. Remains constant
  D. Random

• Q2: Compared to group 13, group 14 elements have:
  A. Larger covalent radii
  B. Smaller covalent radii
  C. Same covalent radii
  D. No trend

• Q3: The small increase in covalent radius from Si to Sn is due to:
  A. Effective shielding
  B. Ineffective shielding by d- and f-electrons
  C. Increase in nuclear charge
  D. Decrease in atomic size

Answer Key

• Q1: B - Covalent radius increases down the group.

• Q2: B - Group 14 elements have smaller covalent radii than group 13.

• Q3: B - Ineffective shielding by d- and f-electrons causes small increase.

Memory Aids

• Radius increases down group 14 but less after Si due to poor shielding.

• Group 14 smaller radius than group 13 due to higher nuclear charge.

Ionisation Enthalpy

Trends

• Ionisation enthalpy decreases down the group, but not regularly.

• Order: C > Si > Ge > Sn < Pb (Pb has slightly higher than Sn).

• Small decrease from Si to Sn; slight increase from Sn to Pb due to poor shielding and atomic size.

• Group 14 elements have higher ionisation enthalpy than group 13 due to greater nuclear charge and smaller size.

Solved Examples

• Ex 1: Explain why Pb has higher ionisation enthalpy than Sn.
  Answer: Poor shielding by d- and f-electrons and increased atomic size cause slight increase.

Practice Set

• Q1: Ionisation enthalpy trend in group 14 is:
  A. Increases down the group
  B. Decreases down the group irregularly
  C. Constant
  D. Random

• Q2: Which element has the highest ionisation enthalpy?
  A. Carbon
  B. Silicon
  C. Tin
  D. Lead

• Q3: Ionisation enthalpy of group 14 elements compared to group 13 is:
  A. Lower
  B. Higher
  C. Same
  D. No relation

Answer Key

• Q1: B - Ionisation enthalpy decreases irregularly down the group.

• Q2: A - Carbon has highest ionisation enthalpy.

• Q3: B - Group 14 elements have higher ionisation enthalpy than group 13.

Memory Aids

• Ionisation enthalpy decreases down group 14 but Pb shows slight increase due to inert pair effect.

• Group 14 > Group 13 in ionisation enthalpy due to higher nuclear charge.

Electronegativity

Trends

• Group 14 elements are slightly more electronegative than group 13 due to smaller size.

• Electronegativity decreases from C to Si.

• From Si to Pb, electronegativity values are almost constant.

Solved Examples

• Ex 1: Arrange C, Si, Pb in order of decreasing electronegativity.
  Answer: C > Si > Pb (Pb ~ Si)

Practice Set

• Q1: Electronegativity trend in group 14 is:
  A. Increases down the group
  B. Decreases down the group
  C. Constant
  D. Random

• Q2: Compared to group 13, group 14 elements are:
  A. Less electronegative
  B. More electronegative
  C. Same electronegativity
  D. No trend

• Q3: Electronegativity values from Si to Pb are:
  A. Increasing
  B. Decreasing
  C. Almost same
  D. Random

Answer Key

• Q1: B - Electronegativity decreases down the group.

• Q2: B - Group 14 elements are more electronegative than group 13.

• Q3: C - Values from Si to Pb are almost the same.

Memory Aids

• Electronegativity decreases from C to Si, then remains steady down to Pb.

• Group 14 > Group 13 in electronegativity due to smaller atomic size.

Physical Properties

Key Points

• Melting and boiling points of group 14 elements are higher than group 13 due to four covalent bonds.

• Strong binding forces exist in solid and liquid states.

• Melting and boiling points decrease down the group from C to Pb due to weaker M–M bonds from increased atomic size.

Solved Examples

• Ex 1: Why does carbon have a higher melting point than lead?
  Answer: Carbon forms strong covalent bonds; lead has larger atoms with weaker bonds.

Practice Set

• Q1: Melting points of group 14 elements:
  A. Increase down the group
  B. Decrease down the group
  C. Remain constant
  D. Random

• Q2: Compared to group 13, group 14 elements have:
  A. Lower melting points
  B. Higher melting points
  C. Same melting points
  D. No trend

• Q3: The reason for high melting points in group 14 is:
  A. Ionic bonding
  B. Metallic bonding
  C. Four covalent bonds per atom
  D. Van der Waals forces

Answer Key

• Q1: B - Melting points decrease down the group.

• Q2: B - Group 14 elements have higher melting points than group 13.

• Q3: C - Due to four covalent bonds per atom.

Memory Aids

• Group 14 elements have high melting points due to strong covalent bonding.

• Melting points decrease down the group as atomic size increases.

Chemical Properties

Oxidation States and Reactivity

• Group 14 elements show +4 and +2 oxidation states.

• C⁴⁺ does not exist due to high ionisation enthalpy and small size; carbon forms negative oxidation states in carbides.

• Silicon shows similar trends to carbon.

• Ge, Sn, Pb show +2 and +4 states; +2 state stability increases down the group due to inert pair effect.

• Compounds with +4 state are covalent; +2 state compounds are ionic (e.g., SnCl4 covalent, SnCl2 ionic).

• Tendency to form covalent compounds decreases down the group; ionic compounds increase.

Reactivity Towards Oxygen

• Elements form oxides on heating in air.

• Reactions: C + O2 → CO2, Si + O2 → SiO2.

• Monoxides CO, SiO, GeO exist; SiO is unstable.

• Higher oxidation state oxides are more acidic.

• PbO exists as red litharge and yellow massicot.

• Pb3O4 (red lead) is a mixed oxide: 2PbO·PbO3.

Reactivity Towards Water

• C, Si, Ge are unaffected by water but decompose steam on red heating.

• Reactions: C + H2O → CO + H2, Si + 2H2O → SiO2 + 2H2.

• Lead is unaffected by water due to protective oxide film.

Reactivity Towards Halogens

• Group 14 elements form MX4 and MX2 halides.

• Tetrahalides are covalent except SnF4 and PbF4 which are ionic.

• Heavier elements (Ge, Sn, Pb) form stable dihalides MX2 due to inert pair effect.

• CCl4 does not hydrolyze (no d-orbitals); SiCl4 and heavier halides hydrolyze due to vacant d-orbitals.

• SiCl4 + 4H2O → Si(OH)4 + 4HCl (silicic acid formation).

• Thermal stability of tetrahalides decreases down group: CX4 > SiX4 > GeX4 > SnX4 > PbX4.

• Thermal stability decreases with heavier halogen: MF4 > MCl4 > MBr4 > MI4.

Solved Examples

• Ex 1: Write the reaction of SiCl4 with water.
  Answer: SiCl4 + 4H2O → Si(OH)4 + 4HCl

Practice Set

• Q1: Which oxidation states are common in group 14?
  A. +1 and +3
  B. +2 and +4
  C. +3 and +5
  D. +1 and +4

• Q2: PbO exists in which forms?
  A. Litharge and massicot
  B. Red lead and litharge
  C. Massicot and red lead
  D. Only litharge

• Q3: Which halide does not hydrolyze?
  A. CCl4
  B. SiCl4
  C. SnCl4
  D. PbCl4

Answer Key

• Q1: B - +2 and +4 oxidation states are common.

• Q2: A - PbO exists as litharge (red) and massicot (yellow).

• Q3: A - CCl4 does not hydrolyze due to lack of d-orbitals.

Memory Aids

• Inert pair effect increases down group 14, stabilizing +2 state.

• PbO forms litharge (red) and massicot (yellow).

• CCl4 is resistant to hydrolysis; SiCl4 hydrolyzes to silicic acid.

Allotropes of Carbon

Overview

• Carbon exists in different structural forms called allotropes.

• Main allotropes: Diamond, Graphite, Fullerene (Crystalline); Carbon black, Coke, Charcoal (Amorphous).

• Allotropes differ in bonding, structure, and properties.

Solved Examples

• Ex 1: Identify the allotrope of carbon used in jewellery.
  Answer: Diamond, due to its brilliant shine and hardness.

Practice Set

• Q1: Which allotrope of carbon is the hardest?
  A. Graphite
  B. Diamond
  C. Fullerene
  D. Carbon black

• Q2: Which allotrope conducts electricity?
  A. Diamond
  B. Graphite
  C. Charcoal
  D. Coke

• Q3: Fullerene has a structure similar to:
  A. Cube
  B. Soccer ball
  C. Sheet
  D. Chain

Answer Key

• Q1: B - Diamond is the hardest due to strong 3D covalent bonding.

• Q2: B - Graphite conducts electricity because of delocalized electrons in layers.

• Q3: B - Fullerene resembles a soccer ball with carbon atoms arranged in rings.

Memory Aids

• Diamond = Hardest, 3D network.

• Graphite = Conducts electricity, layered structure.

• Fullerene = Soccer ball shape.

Crystalline Allotropes of Carbon

Properties and Structure

• Diamond: Each carbon atom bonded tetrahedrally to 4 others; very hard; electrical insulator.

• Graphite: Carbon atoms in layers with weak forces between layers; soft, slippery; conducts electricity.

• Fullerene: Carbon atoms arranged in spherical shape (C60); resembles soccer ball; unique molecular structure.

Solved Examples

• Ex 1: Explain why graphite is a good lubricant.
  Answer: Layers slide over each other easily due to weak interlayer forces.

Practice Set

• Q1: Which allotrope has a 3D network structure?
  A. Graphite
  B. Diamond
  C. Fullerene
  D. Carbon black

• Q2: Why does graphite conduct electricity?
  A. Free electrons between layers
  B. Ionic bonding
  C. Metallic bonding
  D. None of the above

• Q3: Fullerene is also called:
  A. Graphitic acid
  B. Buckminsterfullerene
  C. Carbon black
  D. Coke

Answer Key

• Q1: B - Diamond has a strong 3D covalent network.

• Q2: A - Delocalized electrons between layers allow conductivity.

• Q3: B - Fullerene is known as Buckminsterfullerene.

Memory Aids

• Diamond = 3D network, hardest.

• Graphite = Layers slide, conducts electricity.

• Fullerene = Buckminsterfullerene, soccer ball shape.

Amorphous Allotropes of Carbon

Characteristics and Preparation

• Amorphous carbon forms: Carbon black, Coke, Charcoal.

• Carbon black: Produced by incomplete combustion of hydrocarbons.

• Charcoal: Obtained by heating wood in absence of air.

• Coke: Obtained by heating coal in absence of air.

• These are impure forms of graphite or fullerenes.

Solved Examples

• Ex 1: How is carbon black prepared?
  Answer: By burning hydrocarbons with limited air supply.

Practice Set

• Q1: Charcoal is prepared by:
  A. Burning wood in air
  B. Heating wood without air
  C. Burning coal in air
  D. Heating coal with air

• Q2: Carbon black is used as:
  A. Fuel
  B. Lubricant
  C. Black pigment
  D. Reducing agent

• Q3: Coke is mainly used as:
  A. Fuel
  B. Reducing agent in metallurgy
  C. Lubricant
  D. Pigment

Answer Key

• Q1: B - Charcoal is made by heating wood in absence of air.

• Q2: C - Carbon black is used as black pigment in ink and tyres.

• Q3: B - Coke is used as a reducing agent in metallurgy.

Memory Aids

• Carbon black = pigment, incomplete combustion.

• Charcoal = heated wood, no air.

• Coke = heated coal, reducing agent.

Uses of Carbon

Applications of Different Allotropes

• Graphite: Lubricant at high temperatures; refractory material for crucibles and electrodes.

• Carbon black: Black pigment in ink; filler in automobile tyres.

• Coal and Charcoal: Used as fuels.

• Coal: Used to manufacture producer gas and water gas.

• Coke: Used as reducing agent in metallurgy.

• Diamond: Used in jewellery for its brilliant shine.

Solved Examples

• Ex 1: Why is graphite used in electrodes?
  Answer: It withstands high temperatures and conducts electricity.

Practice Set

• Q1: Which allotrope is used as a lubricant?
  A. Diamond
  B. Graphite
  C. Coke
  D. Charcoal

• Q2: Producer gas is made from:
  A. Charcoal
  B. Coal
  C. Carbon black
  D. Fullerene

• Q3: Diamond is mainly used in:
  A. Fuel
  B. Jewellery
  C. Pigment
  D. Reducing agent

Answer Key

• Q1: B - Graphite acts as a lubricant due to its layered structure.

• Q2: B - Producer gas is manufactured from coal.

• Q3: B - Diamond is used in jewellery for its brilliance.

Memory Aids

• Graphite = lubricant, electrodes.

• Carbon black = pigment, tyres.

• Diamond = jewellery.

Chemical Reactions of Graphite

Oxidation Reactions

• Graphite + Conc. HNO3 → Graphitic acid (C11H4O5).

• Graphite + Alkaline KMnO4 → Mellitic acid (C6(COOH)6) + Oxalic acid ((COOH)2).

• These reactions show oxidation of graphite to different acids.

Solved Examples

• Ex 1: Write the product when graphite is oxidized with conc. HNO3.
  Answer: Graphitic acid (C11H4O5).

Practice Set

• Q1: Graphite reacts with conc. HNO3 to form:
  A. Oxalic acid
  B. Graphitic acid
  C. Mellitic acid
  D. Carbon black

• Q2: Alkaline KMnO4 oxidizes graphite to:
  A. Graphitic acid
  B. Oxalic acid and Mellitic acid
  C. Carbon black
  D. Charcoal

• Q3: Which acid is benzene hexacarboxylic acid?
  A. Graphitic acid
  B. Oxalic acid
  C. Mellitic acid
  D. Citric acid

Answer Key

• Q1: B - Conc. HNO3 produces graphitic acid.

• Q2: B - Alkaline KMnO4 yields mellitic and oxalic acids.

• Q3: C - Mellitic acid is benzene hexacarboxylic acid.

Memory Aids

• Conc. HNO3 → Graphitic acid.

• Alkaline KMnO4 → Mellitic + Oxalic acids.

• Mellitic acid = benzene ring with 6 COOH groups.

Carbon Monoxide

Properties and Preparation

• Colourless, odourless, poisonous, almost water-insoluble gas.

• Prepared by:

  • Incomplete combustion of carbon: C + \(\frac{1}{2}\)O2 \(\rightarrow\) CO

  • Thermal decomposition of formic acid: HCOOH \(\xrightarrow{\Delta, H_2SO_4}\) CO + H2O

  • Reaction of carbon with steam (water gas): C + H2O \(\xrightarrow{\Delta}\) CO + H2

  • Reaction of carbon with air (producer gas): 2C + O2 + N2 \(\xrightarrow{\Delta}\) 2CO + N2

• Strong reducing agent; reduces many metal oxides except alkali, alkaline earth, Al, and some transition metals.

• Forms metal carbonyls by acting as a ligand (donor) to metals.

• Toxic due to stable complex formation with haemoglobin, preventing oxygen transport.

Solved Examples

• Ex 1: Write the reaction for preparation of CO from formic acid.

  • HCOOH \(\xrightarrow{\Delta, H_2SO_4}\) CO + H2O

• Ex 2: Show the reaction of CO with iron oxide.

  • Fe2O3 + 3CO \(\rightarrow\) 2Fe + 3CO2

Practice Set

• Q1: Which gas is formed by incomplete combustion of carbon?

  • A. CO2

  • B. CO

  • C. CH4

  • D. H2

• Q2: CO acts as a reducing agent for which of the following?

  • A. Na2O

  • B. Fe2O3

  • C. CaO

  • D. Al2O3

• Q3: Why is CO toxic?

  • A. It is acidic

  • B. It forms stable complex with haemoglobin

  • C. It is flammable

  • D. It is water soluble

Answer Key

• Q1: B - CO is formed by incomplete combustion of carbon.

• Q2: B - CO reduces Fe2O3 but not alkali or alkaline earth metal oxides.

• Q3: B - CO binds strongly with haemoglobin, preventing oxygen transport.

Memory Aids

• CO = Incomplete combustion product, strong reducing agent.

• CO toxicity due to haemoglobin binding stronger than O2.

• Metal carbonyls formed by CO acting as ligand.

Carbon Dioxide

Properties and Preparation

• Colourless, odourless gas; molecular structure: O=C=O with resonance.

• Prepared by:

  • Complete combustion of carbon or methane: C + O2 \(\rightarrow\) CO2

  • Heating limestone (CaCO3): CaCO3 \(\xrightarrow{\Delta}\) CaO + CO2

  • Reaction of calcium carbonate with HCl: CaCO3 + 2HCl \(\rightarrow\) CaCl2 + CO2 + H2O

• Acidic oxide; reacts with bases to form salts.

• Solid CO2 is dry ice; used as refrigerant.

• Non-poisonous but excess causes greenhouse effect.

• Removed from air by photosynthesis: 6CO2 + 12H2O \(\xrightarrow{light, chlorophyll}\) C6H12O6 + 6O2 + 6H2O

Solved Examples

• Ex 1: Write the reaction for preparation of CO2 by heating limestone.

  • CaCO3 \(\xrightarrow{\Delta}\) CaO + CO2

• Ex 2: Write the photosynthesis equation.

  • 6CO2 + 12H2O \(\xrightarrow{light, chlorophyll}\) C6H12O6 + 6O2 + 6H2O

Practice Set

• Q1: Which is an acidic oxide?

  • A. CO

  • B. CO2

  • C. SO2

  • D. NO

• Q2: Dry ice is the solid form of:

  • A. CO

  • B. CO2

  • C. SO2

  • D. N2

• Q3: Photosynthesis removes CO2 from air by producing:

  • A. Oxygen and glucose

  • B. Carbon monoxide

  • C. Methane

  • D. Water

Answer Key

• Q1: B - CO2 is an acidic oxide.

• Q2: B - Dry ice is solid CO2.

• Q3: A - Photosynthesis produces oxygen and glucose.

Memory Aids

• CO2 = Acidic oxide, dry ice = solid CO2.

• Photosynthesis: 6CO2 + 12H2O \(\rightarrow\) glucose + O2.

• CO2 not poisonous but causes greenhouse effect.

Silicon Dioxide

Properties and Reactions

• Also called silica; found as sand and quartz.

• Covalent, 3D network solid with high Si–O bond enthalpy.

• Almost inert; attacked by HF and NaOH only.

• Reactions:

  • SiO2 + 2NaOH \(\rightarrow\) Na2SiO3 + H2O

  • SiO2 + 4HF \(\rightarrow\) SiF4 + 2H2O

Uses

• Quartz used in piezoelectric devices: clocks, radios, TVs, mobile communications.

• Silica gel used as drying agent and catalyst support.

• Kieselghur (amorphous silica) used in filtration.

Solved Examples

• Ex 1: Write reaction of SiO2 with NaOH.

  • SiO2 + 2NaOH \(\rightarrow\) Na2SiO3 + H2O

• Ex 2: Reaction of SiO2 with HF.

  • SiO2 + 4HF \(\rightarrow\) SiF4 + 2H2O

Practice Set

• Q1: Silicon dioxide is commonly known as:

  • A. Quartz

  • B. Silica

  • C. Sand

  • D. All of the above

• Q2: SiO2 reacts with which acid?

  • A. HCl

  • B. HF

  • C. H2SO4

  • D. HNO3

• Q3: Kieselghur is used in:

  • A. Catalysts

  • B. Filtration

  • C. Electronics

  • D. Medicine

Answer Key

• Q1: D - Quartz, silica, and sand are forms of SiO2.

• Q2: B - SiO2 reacts with HF.

• Q3: B - Kieselghur is used in filtration plants.

Memory Aids

• SiO2 = quartz, sand, silica; attacked only by HF and NaOH.

• Quartz = piezoelectric; silica gel = drying agent.

• Kieselghur = filtration aid.

Silicones

Structure and Preparation

• Organosilicon polymers with repeating unit (R2SiO).

• Starting materials: alkyl/aryl substituted silicon chlorides RnSiCl4-n.

• Prepared by hydrolysis of dialkyldichlorosilanes or diaryldichlorosilanes.

• Preparation steps:

  • Silicon + methyl chloride (CH3Cl) with Cu catalyst at 573 K forms dimethyldichlorosilane.

  • Hydrolysis replaces Cl with OH forming silanols.

  • Polymerization of silanols releases water and forms silicone polymer.

  • Chain length controlled by adding (CH3)3SiCl to block ends.

Properties and Uses

• Water repelling due to non-polar alkyl groups.

• High thermal stability, dielectric strength, and chemical resistance.

• Used as sealants, greases, electrical insulators, and waterproofing agents.

• Biocompatible; used in surgical and cosmetic products.

Solved Examples

• Ex 1: Write the reaction for hydrolysis of dimethyldichlorosilane.

  • (CH3)2SiCl2 + 2H2O \(\rightarrow\) (CH3)2Si(OH)2 + 2HCl

• Ex 2: Polymerization of silanols.

  • 2 (CH3)2Si(OH)2 \(\rightarrow\) [(CH3)2SiO]n + H2O

Practice Set

• Q1: Silicones are polymers with repeating unit:

  • A. (SiO2)

  • B. (R2SiO)

  • C. (SiCl4)

  • D. (SiH4)

• Q2: Which compound is used to control polymer chain length in silicones?

  • A. (CH3)3SiCl

  • B. SiCl4

  • C. H2O

  • D. CH3Cl

• Q3: Silicones are used as:

  • A. Electrical insulators

  • B. Sealants

  • C. Biocompatible materials

  • D. All of the above

Answer Key

• Q1: B - Silicones have repeating unit (R2SiO).

• Q2: A - (CH3)3SiCl blocks polymer chain ends.

• Q3: D - Silicones are used in all listed applications.

Memory Aids

• Silicones = (R2SiO) polymers from hydrolysis and polymerization.

• Chain length controlled by trimethylsilyl chloride.

• Used as sealants, insulators, and biocompatible materials.

Silicates

Structure and Types

• Compounds with SiO44− tetrahedral units as basic units.

• Units may be discrete or linked by sharing 1–4 oxygen atoms forming chains, rings, sheets, or 3D structures.

Zeolites

• Three-dimensional aluminosilicates with some Si replaced by Al3+.

• Negative charge balanced by cations (Na+, K+, Ca2+, etc.).

• Used as ion exchangers for water softening.

• Act as catalysts in petrochemical industries.

• ZSM-5 zeolite converts alcohols directly into gasoline.

Solved Examples

• Ex 1: Describe the basic unit of silicates.

  • SiO44− tetrahedral unit.

• Ex 2: Name a zeolite used in catalysis.

  • ZSM-5

Practice Set

• Q1: The basic structural unit of silicates is:

  • A. SiO2

  • B. SiO44−

  • C. AlO45−

  • D. SiCl4

• Q2: Zeolites are used as:

  • A. Ion exchangers

  • B. Catalysts

  • C. Both A and B

  • D. None

• Q3: ZSM-5 is used to convert:

  • A. Alcohols to gasoline

  • B. Methane to CO

  • C. CO to CO2

  • D. Water to hydrogen

Answer Key

• Q1: B - SiO44− is the basic silicate unit.

• Q2: C - Zeolites act as ion exchangers and catalysts.

• Q3: A - ZSM-5 converts alcohols to gasoline.

Memory Aids

• Silicates = SiO44− units linked in various structures.

• Zeolites = aluminosilicates, ion exchangers, catalysts.

• ZSM-5 = alcohol to gasoline catalyst.