Chemistry · Volume 1 · Chapter 2

Samacheer Class 12 Chemistry - p-Block Elements-I

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Q.1 An aqueous solution of borax is a) neutral b) acidic c) basic d) amphoteric
Answer: c

Borax (sodium tetraborate) hydrolyses in water to give borate ions which react with water producing OH−; the solution is basic.

Q.2 Boric acid is an acid because its molecule (NEET) a) contains replaceable H+ ion b) gives up a proton c) combines with proton to form water molecule d) accepts OH- from water ,releasing proton.
Answer: d

Boric acid, B(OH)3, behaves as a Lewis acid: it accepts OH− from water to form B(OH)4−. This shifts equilibrium and releases H+ (i.e. it acts acidic by accepting OH−).

Q.3 Which among the following is not a borane? a)BH3 b)B2H6 c)B3H6 d) none of these
Answer: d

All listed species correspond to boron–hydrogen compounds (boranes or borane-related). BH3 is not isolable as a free monomer but is still a borane species (often observed as complexed). Therefore, among the given choices, none is strictly 'not a borane'.

Q.4 Which of the following metals has the largest abundance in the earth's crust? a) Aluminium b) Calcium c) Magnesium d) Sodium
Answer: a

Aluminium is the most abundant metal in the Earth's crust (about 8% by weight), more abundant than Ca, Mg or Na.

Q.5 In diborane, the number of electrons that accounts for banana bonds is a) six b) two c) four d) three
Answer: c

Diborane (B2H6) has two three-centre two-electron (3c–2e) bonds (the B–H–B bridge bonds). Each bridge uses 2 electrons, so total electrons accounting for the banana (bridge) bonds = 2 × 2 = 4.

Q.6 The element that does not show catenation among the following p-block elements is a) Carbon b) silicon c) Lead d) germanium
Answer: c

Catenation decreases down the group as element–element bond strength decreases. Lead (Pb) shows negligible catenation compared to C, Si and Ge.

Q.7 Carbon atoms in fullerene with formula C60 have a) sp3 hybridised b) sp hybridised c) sp2 hybridised d) partially sp2 and partially sp3 hybridised
Answer: c

In C60 each carbon is bonded to three neighbours in a trigonal planar arrangement with one p orbital contributing to the delocalised π system — characteristic of sp2 hybridisation.

Q.8 Oxidation state of carbon in its hydrides a) +4 b) -4 c) +3 d) +2
Answer: b

In the simplest hydride CH4, hydrogen is assigned +1, so carbon has oxidation state −4. This is the lowest oxidation state of carbon in its hydrides.

Q.9 The basic structural unit of silicates is (NEET) a) SiO3( )- b) SiO4( )- c) SiO( )- d) SiO4( )-
Answer: b

Silicates are built from the tetrahedral unit SiO4^4− (silicate tetrahedron), which is the basic structural unit.

Q.10 The repeating unit in silicone is a) SiO2 b) Si R O R c) Si R O O R d) Si R O O R
Answer: b

Silicones (polysiloxanes) have a repeating –Si–O– backbone with organic R groups on silicon: general repeating unit is –[R2Si–O]–. Option (b) corresponds to the siloxane repeating unit.

Q.11 Which of these is not a monomer for a high molecular mass silicone polymer? a) Me3SiCl b) PhSiCl3 c) MeSiCl3 d) Me2SiCl2
Answer: a

Me3SiCl is mono-functional (trimethylsilyl chloride) and acts as a chain terminator rather than a monomer for forming high molecular mass silicone polymers. Multi-functional chlorosilanes like SiCl4, RSiCl3, R2SiCl2 serve as monomers.

Q.12 Which of the following is not sp2 hybridised? a) Graphite b) graphene c) Fullerene d) dry ice
Answer: d

Dry ice (solid CO2) has linear CO2 molecules in which carbon is sp-hybridised. Graphite, graphene and fullerene carbons are sp2-hybridised.

Q.13 The geometry at which carbon atom in diamond are bonded to each other is a) Tetrahedral b) hexagonal c) Octahedral d) none of these
Answer: a

In diamond each carbon is covalently bonded to four others in a tetrahedral arrangement (sp3 hybridisation).

Q.14 Which of the following statements is not correct? a) Beryl is a cyclic silicate b) Mg2SiO4 is an orthosilicate c) SiO4 4- is the basic structural unit of silicates d) Feldspar is not aluminosilicate
Answer: d

Feldspars are aluminosilicates (framework silicates containing Al3+ substituting for Si4+). Thus statement (d) is incorrect; the others are correct (beryl = cyclic silicate, Mg2SiO4 is olivine orthosilicate, SiO4^4− is basic unit).

Q.15Match items in column - I with the items of column - II and assign the correct code. Column-I: A Borazole B(OH)3 B Boric acid B3N3H6 C Quartz Na2[B4O5(OH)4]·8H2O D Borax SiO2v
Solution

Borazole = B3N3H6 (analogue of benzene); boric acid = B(OH)3; quartz = SiO2; borax = sodium tetraborate Na2[B4O5(OH)4]·8H2O.

Answer:

A–B3N3H6, B–B(OH)3, C–SiO2, D–Na2[B4O5(OH)4]·8H2O

Q.16 Duralumin is an alloy of a) Cu,Mn b) Cu,Al,Mg c) Al,Mn d) Al,Cu,Mn,Mg
Answer: d

Duralumin is a high-strength aluminium alloy containing Al as base with Cu, Mn and Mg as major alloying elements (commonly Al–Cu–Mn–Mg).

Q.17 The compound that is used in nuclear reactors as protective shields and control rods is a) Metal borides b) metal oxides c) Metal carbonates d) metal carbide
Answer: a

Boron-containing compounds (e.g., boron carbide B4C, metal borides) are used in control rods and shielding because boron has a high neutron absorption cross-section.

Q.18 The stability of +1 oxidation state increases in the sequence a) Al < Ga < In < Tl b) Tl < In < Ga < Al c) In < Tl < Ga < Al d) Ga< In < Al < Tl
Answer: a

Due to the inert pair effect, the stability of the +1 oxidation state increases down the group 13: Al < Ga < In < Tl.

Answer the following questionsAnswer the following questions18 questions
Q.1Write a short note on anomalous properties of the first element of p-block.v
Solution

Key anomalous features of boron (compared with other group 13 elements): - Small atomic and ionic size and high ionisation energy. - Higher electronegativity and greater tendency to form covalent rather than metallic compounds. - Existence of electron-deficient compounds (e.g., boranes) and multicentre bonding (3c–2e bonds). - Does not show typical metallic properties shown by heavier group members. - Forms stable complexes and network solids (e.g., borax, borates) and polymeric structures like B12 icosahedra. - Exhibits diagonal relationship with silicon (similarities in properties).

Answer:

Anomalous properties of boron (first p-block element)

Q.2Describe briefly allotropism in p- block elements with specific reference to carbon.v
Solution

Allotropy: existence of two or more different structural forms (allotropes) of the same element in the same physical state. Carbon allotropes (brief): - Diamond: each carbon is sp3-hybridised, tetrahedrally bonded to four others, 3D covalent network; very hard, high thermal conductivity, electrical insulator. - Graphite: layers of hexagonal rings; each carbon is sp2-hybridised with delocalised π electrons between layers; soft, lubricating, good electrical conductor along planes. - Graphene: single layer of sp2-bonded carbon atoms in a hexagonal lattice; exceptional strength, high electrical and thermal conductivity. - Fullerenes (e.g., C60): spherical/cage molecules of sp2 carbons with pentagons and hexagons; molecular solids with distinct properties. - Carbon nanotubes: rolled graphene sheets (single- or multi-walled) with remarkable mechanical and electronic properties. - Amorphous carbon: charcoal, soot — lack long-range order. These allotropes illustrate how differences in bonding and structure among p-block elements lead to varied physical and chemical properties.

Answer:

Allotropy in p-block elements — example of carbon

Q.3Give the uses of Borax.v

Principal uses of borax include:

  • As a flux in metallurgy (removes oxides during welding and soldering).
  • Manufacture of borosilicate glass and enamels.
  • In laundry and household cleaners (buffering and emulsifying agent).
  • Water softening and pH buffering in various processes.
  • Used as a fire retardant and a preservative in limited applications.
  • Precursor for boric acid (H3BO3) and other boron compounds; laboratory reagent.
Q.4What is catenation ? describe briefly the catenation property of carbon.v
Solution

Definition: Catenation = formation of stable bonds between atoms of the same element to give chains/rings. Carbon's exceptional catenation arises from: (i) high C–C bond strength, (ii) tetravalency enabling four bonds, (iii) small atomic size giving good overlap, (iv) multiple hybridisations (sp3, sp2, sp) allowing variety of structures. Consequences: enormous diversity of organic compounds, long-chain hydrocarbons, polymers, cyclic and polycyclic systems and allotropes (diamond, graphite, fullerenes). Other elements (Si, S) show catenation but to a much lesser extent.

Answer:

Catenation is the ability of an element to form stable chains of its own atoms. Carbon shows very strong catenation due to strong C–C bonds, small size and tetravalency, forming long chains, rings and diverse allotropes (alkanes, polymers, aromatic rings, graphene, diamond).

Q.5Write a note on Fisher tropsch synthesis.v
Solution

Overview: CO and H2 (syngas) are passed over catalysts (Fe or Co, supported) at 200–350 °C and moderate pressure. Surface reaction polymerises CO and H to form long-chain hydrocarbons and water: nCO + (2n+1)H2 → CnH2n+2 + nH2O. Products include alkanes, alkenes, and oxygenates; chain length/distribution depends on catalyst and conditions. Industrial use: manufacture of synthetic diesel and other fuels where petroleum is scarce.

Answer:

Fischer–Tropsch synthesis converts syngas (CO + H2) over transition metal catalysts (Fe, Co) to produce hydrocarbons (paraffins/olefins) and water; used to make synthetic fuels from coal, natural gas or biomass.

Q.6Give the structure of CO and CO2.v
Solution

CO: Lewis representation can be written as :C≡O: (or :C←O: to show lone-pair donation). Formal charges: C carries slight negative charge and O slight positive in some descriptions; overall neutral. Bond order ≈ 3; molecule is linear. CO2: O=C=O, linear (180°) with two resonance forms equivalent, each C=O double bond; nonpolar overall. Both structures explain bonding, polarity and reactivity (CO is a good σ-donor and π-acceptor ligand).

Answer:

CO: a molecule with a formal triple bond between C and O (one bond is dative from O→C), bond order 3; CO2: linear O=C=O with two equivalent double bonds, bond order 2 and molecular geometry linear.

Q.7Give the uses of silicones.v
Solution

Applications derive from thermal stability, hydrophobicity, low surface tension and chemical inertness: (1) Sealants and caulks, (2) Lubricants and greases, (3) Heat- and weather-resistant coatings, (4) Electrical insulation and potting compounds, (5) Silicone rubber for moulds/gaskets and medical implants (biocompatible), (6) Personal-care products (emollients, anti-foaming agents), (7) Release agents and anti-foaming agents in industry.

Answer:

Silicones (polysiloxanes, e.g., PDMS) are used as sealants, lubricants, heat-resistant coatings, electrical insulators, moulding/adhesive materials, medical implants and devices, water-repellent treatments and in cosmetics.

Q.8Describe the structure of diborane.v
Solution

Structure details: Each boron is bonded to two terminal H (normal 2-center–2-electron bonds) and shares two bridging H atoms with the other B via 3-center–2-electron bonds. There are four terminal B–H bonds and two bridging B–H–B bonds. The bridging bonds are longer and weaker than terminal B–H. The molecule is electron-deficient (each B has only 6 valence electrons) and is stabilized by multicenter bonding.

Answer:

Diborane B2H6 has a bridged structure with two terminal B–H bonds on each B and two bridging hydrogen atoms forming two three-center two-electron (3c–2e) B–H–B bonds; electron-deficient and C2h symmetry.

Q.9Write a short note on hydroboration.v
Solution

Reaction: R2BH + alkene → organoborane (syn addition). Regiochemistry: boron attaches to the less substituted carbon (anti-Markovnikov). Oxidation: organoborane + H2O2/NaOH → alcohol (retains stereochemistry; syn addition gives syn product). Hydroboration–oxidation is a valuable method to prepare primary/secondary alcohols from alkenes without rearrangement.

Answer:

Hydroboration is the addition of BH3 (or R2BH) across C=C giving organoboranes; it proceeds syn and anti-Markovnikov (B attaches to the less substituted carbon); subsequent oxidation (H2O2/NaOH) yields alcohols (anti-Markovnikov hydration).

Q.10Give one example for each of the following (i) icosagens (ii) tetragens (iii) pnictogens (iv) chalcogensv
Solution

Group names and examples: (i) Icosagens = Group 13 (example B). (ii) Tetragens = Group 14 (example C). (iii) Pnictogens = Group 15 (example N or P). (iv) Chalcogens = Group 16 (example O or S).

Answer:

(i) Icosagen: Boron (B) (ii) Tetragen: Carbon (C) (iii) Pnictogen: Nitrogen (N) (iv) Chalcogen: Oxygen (O)

Q.11Write a note on metallic nature of p-block elements.v
Solution

Explanation: Metallic character depends on ease of losing electrons (low ionization energy) and increases with increasing atomic size and decreasing ionization energy down a group. Across a period, electronegativity increases and ionization energy rises, so metallic character decreases. Many p-block elements show multiple oxidation states, and some exhibit amphoteric behaviour (e.g., Al2O3), while heavier p-block elements may be metallic (e.g., Tl).

Answer:

Metallic character in the p-block increases down a group and decreases across a period from left to right. Elements at left of p-block (e.g., Al, Ga) show metallic properties; those at right (e.g., O, F) are non‑metallic; intermediate elements show metalloids behavior.

Q.12Complete the following reactions a b c . . . B OH +NH Na B O + H SO H O B H + 2NaOH ( ) →  + →  + 2H O B H + CH OH BF + 9 H O HCOOH + H SO →  →  →  d e f . . . SiCl + NH SiCl + C H OH B + NaOH →  →  →  →  g h i . . . j. H B O Red hot  → v
Solution

Notes on each equation: (a) Borax + sulfuric acid → boric acid (hydration shown). (b) Boric acid + base → tetrahydroxyborate ion; on heating gives metaborate. (c) BF3 hydrolyses to boric acid and HF. (d) Boric acid reacts with alcohols to give borate esters. (e) Formic acid is dehydrated by conc. H2SO4 to give CO and H2O. (f) Diborane hydrolyses to boric acid and H2. (g) SiCl4 reacts with NH3 at high temperature to give silicon nitride + HCl. (h) SiCl4 reacts with alcohols to give tetraalkoxysilanes used in sol–gel chemistry. (i) Boron oxides react with alkali to give borates. (j) On strong heating boric acid loses water to form metaboric acid and ultimately boron trioxide.

Answer:

Reconstructed common transformations (a–j) related to boron and silicon chemistry: (a) Na2B4O7·10H2O + H2SO4 + 5 H2O → 4 H3BO3 + Na2SO4 (b) H3BO3 + NaOH → Na[B(OH)4] (or on heating → NaBO2 + H2O) (c) BF3 + 3 H2O → B(OH)3 + 3 HF (d) H3BO3 + CH3OH ⇌ B(OCH3)3 + 3 H2O (esterification) (e) HCOOH —(conc. H2SO4)→ CO + H2O (dehydration) (f) B2H6 + 6 H2O → 2 B(OH)3 + 6 H2 (hydrolysis) (g) SiCl4 + 4 NH3 → Si3N4 + 12 HCl (formation of silicon nitride under forcing conditions) (h) SiCl4 + 4 C2H5OH → Si(OC2H5)4 + 4 HCl (formation of tetraalkoxysilane) (i) B + NaOH → (depends on conditions) e.g. B2O3 + 2 NaOH → 2 NaBO2 + H2O (j) 2 H3BO3 —(red hot)→ B2O3 + 3 H2O

Q.13How will you identify borate radical?v
Solution

Procedure: Acidify a borate-containing sample, add excess methanol and a little concentrated H2SO4 to form B(OCH3)3 (trimethyl borate). Ignite a little of this; the flame shows a characteristic green colour due to boron — a positive test for borates.

Answer:

Convert the borate to trimethyl borate by adding methanol and conc. H2SO4; the ester on ignition burns with a characteristic green flame (boron test).

Q.14Write a note on zeolites.v
Solution

Structure: a 3D network of SiO4 and AlO4 tetrahedra linked by shared oxygens producing cavities and channels. Properties: high surface area, uniform pore size, cation-exchange capacity (to balance AlO4− units), reversible dehydration. Uses: molecular sieves (separation by size), catalysts in petrochemical industry (zeolite Y, ZSM-5), water softening and gas adsorption.

Answer:

Zeolites are crystalline aluminosilicate frameworks with open porous structures and exchangeable cations and water; they act as molecular sieves, ion exchangers and heterogeneous catalysts (important in petroleum cracking and adsorption).

Q.15How will you convert boric acid to boron nitride?v
Solution

Stepwise: (1) 2 H3BO3 —(heat)→ B2O3 + 3 H2O (dehydration). (2) B2O3 + 2 NH3 —(900–1200 °C)→ 2 BN + 3 H2O. The product BN (hexagonal boron nitride) is analogous to graphite.

Answer:

Heat boric acid to form B2O3, then react B2O3 with ammonia at high temperature: B2O3 + 2 NH3 → 2 BN + 3 H2O.

Q.16A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). identify A, B and C.v
Solution

Lithium hydride reacts with boron halides to give lithium borohydride, a strong reducing agent. LiBH4 is an important hydride reducing agent (reduces esters, carbonyls etc.). The balanced equation commonly written: 4 LiH + BCl3 → LiBH4 + 3 LiCl.

Answer:

A = LiH, B = BCl3 (or BCl3 source), C = LiBH4 (lithium borohydride). Example reaction: 4 LiH + BCl3 → LiBH4 + 3 LiCl.

Q.17A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify A and B.v
Solution

Potash alum (KAl(SO4)2·12H2O) contains K+ (fourth period alkali metal). On heating (decomposition) it gives potassium sulfate (K2SO4) among other decomposition products. Aqueous K2SO4 (contains SO42−) gives a white precipitate with BaCl2 due to BaSO4. Alizarin (or alizarin indicator) gives characteristic coloration with potassium salts (commonly used as a test in qualitative analysis), hence the red-colour observation corresponds to potassium presence.

Answer:

A = K (potassium); the double salt is potash alum KAl(SO4)2·12H2O. On heating it yields potassium sulfate (B = K2SO4).

Q.18CO is a reducing agent. Justify with an example.v
Solution

In these reactions carbon monoxide is oxidised to CO2 while the metal oxide is reduced to the metal. Example: CuO(s) + CO(g) → Cu(s) + CO2(g). This demonstrates CO acting as a reducing agent (electron donor) in metallurgical and laboratory reductions.

Answer:

CO reduces metal oxides to metals; e.g., Fe2O3 + 3 CO → 2 Fe + 3 CO2 (in blast furnace) or CuO + CO → Cu + CO2.