p-Block Elements: JEE Advanced PYQ, 2007–2026

Q1 · Chemistry · p-Block Elements

Sulphur has a much greater tendency for catenation (S–S chains and rings) than oxygen has (O–O chains). What is the main reason?

• AThe O–O single bond is weak because of strong lone pair–lone pair repulsion between the small oxygen atoms, whereas the S–S bond is stronger.
• BOxygen is more electronegative, so O–O bonds are more polar and hence break more readily than S–S bonds.
• CSulphur forms strong pπ–pπ multiple bonds with itself, which link the atoms into chains.
• DOxygen exists as a gas at room temperature, so it cannot form chains.

Q2 · Chemistry · p-Block Elements

In the contact process, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic (ΔH° = –196.6 kJ mol⁻¹). Plants use V₂O₅ at about 720 K and only about 2 bar, although lower temperature and higher pressure would each favour SO₃ at equilibrium. Which statement best explains this choice of conditions?

• ALower temperature would raise the equilibrium conversion but make the rate too slow. The conversion is already high at about 2 bar, so a much higher pressure is not worth its cost.
• BV₂O₅ shifts the equilibrium towards SO₃ only at high temperature, so about 720 K is needed to make the catalyst raise the equilibrium yield.
• CThe forward reaction is endothermic, so high temperature is needed to favour SO₃ and only a moderate pressure is needed to keep SO₂ from decomposing.
• DHigh pressure would favour SO₂ because the reaction produces fewer moles of gas, so a low pressure is used to push the equilibrium forward.

Q3 · Chemistry · p-Block Elements

Which statement about the allotropes of sulphur is correct?

• AMonoclinic sulphur is the stable form at room temperature and changes to rhombic sulphur above 369 K.
• BRhombic and monoclinic sulphur both consist of S₈ molecules, but they differ in crystal packing, and rhombic sulphur is stable below 369 K.
• CRhombic sulphur contains S₈ rings, whereas monoclinic sulphur contains S₆ rings.
• DBoth forms contain planar S₈ rings, and they interconvert at 273 K.

Q4 · Chemistry · p-Block Elements

In the solid state, phosphorus pentachloride exists as an ionic solid made of [PCl₄]⁺ and [PCl₆]⁻ ions, whereas in the gas phase it is a trigonal bipyramidal molecule. Which row correctly gives the shape and the hybridisation of phosphorus in the two solid-state ions?

• A[PCl₄]⁺ is tetrahedral (sp³) and [PCl₆]⁻ is octahedral (sp³d²)
• B[PCl₄]⁺ is see-saw (sp³d) and [PCl₆]⁻ is octahedral (sp³d²)
• C[PCl₄]⁺ is tetrahedral (sp³) and [PCl₆]⁻ is trigonal bipyramidal (sp³d)
• D[PCl₄]⁺ is square planar (dsp²) and [PCl₆]⁻ is octahedral (d²sp³)

Q5 · Chemistry · p-Block Elements

Which of the following correctly states the trend in thermal stability of the hydrides MH₄ of group 14 elements, with the reason?

• ACH₄ > SiH₄ > GeH₄ > SnH₄ > PbH₄, because the M–H bond enthalpy decreases as the size of M increases
• BPbH₄ > SnH₄ > GeH₄ > SiH₄ > CH₄, because the +4 oxidation state becomes more stable down the group
• CAll MH₄ hydrides are equally stable, because all are electron-precise molecules
• DCH₄ is the least stable, because the C–H bond is the shortest and most strained

Q6 · Chemistry · p-Block Elements

One mole of PCl₃ is hydrolysed completely by excess water, giving a phosphorus oxoacid and HCl. The resulting solution is then titrated with NaOH until every ionisable hydrogen of all the acids present has been neutralised. How many moles of NaOH are required?

• A3
• B4
• C5
• D6

Q7 · Chemistry · p-Block Elements

CO₂ is a gas at room temperature, but SiO₂ (quartz) is a hard, high-melting solid. Which explanation is correct?

• ASiO₂ is ionic and made of Si⁴⁺ and O²⁻ ions, whereas CO₂ is covalent.
• BCarbon is sp³ hybridised in CO₂ and silicon is sp hybridised in SiO₂, so SiO₂ is the more strongly bonded.
• CCarbon forms stable pπ–pπ double bonds with oxygen, giving discrete O=C=O molecules, whereas silicon forms strong Si–O single bonds in a three-dimensional network of corner-sharing SiO₄ tetrahedra.
• DSilicon forms pπ–pπ double bonds with oxygen, so SiO₂ molecules link into long chains and form a solid.

Q8 · Chemistry · p-Block Elements

Which of the following statements about interhalogen compounds of the type XX′n (n = 1, 3, 5, 7) is correct?

• AExcept for fluorine itself, interhalogens are generally more reactive than the halogens because the X–X′ bond is weaker than the X–X bond in the parent halogens.
• BAll interhalogen compounds are gases at room temperature because they are covalent molecules of low molar mass.
• CIn XX′n, X is the smaller and more electronegative halogen, and n takes odd values because X′ is always the larger atom.
• DBrF₅ is square planar, with two lone pairs on bromine in an octahedral arrangement of electron pairs.

Q9 · Chemistry · p-Block Elements

Which of the following correctly gives the trend in catenation tendency of group 14 elements and its basis?

• AC > Si > Ge ≈ Sn > Pb, because the M–M bond enthalpy decreases down the group (C–C ≈ 348, Si–Si ≈ 297, Ge–Ge ≈ 260, Sn–Sn ≈ 240 kJ mol⁻¹).
• BSi > C > Ge ≈ Sn > Pb, because silicon has vacant d orbitals which strengthen Si–Si bonds.
• CPb > Sn > Ge > Si > C, because larger atoms form longer chains with greater bond strength.
• DC > Si > Ge ≈ Sn > Pb, because the number of valence electrons decreases down the group.

Q10 · Chemistry · p-Block Elements

Dimethyldichlorosilane, (CH₃)₂SiCl₂, containing a small amount of trimethylchlorosilane, (CH₃)₃SiCl, is hydrolysed and the products are allowed to undergo condensation polymerisation. Which of the following correctly describes the product?

• AA three-dimensional cross-linked silicone, because (CH₃)₃SiCl contributes an extra Si–O bond at each branching point
• BLinear silicone chains whose average length is increased, because (CH₃)₃SiCl acts as a chain-extending unit
• COnly cyclic silicone rings, because the trimethylsilyl groups force the (CH₃)₂Si units to close into rings
• DLinear silicone chains of reduced average length, with (CH₃)₃Si– groups capping the chain ends