JEE (Main) · Chemistry, Class 12
p-Block Elements: JEE Main PYQ, 2020–2025
p-Block Elements (Chemistry, Class 12) has been asked 151 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.4 questions a paper, 6% of all Chemistry questions. It is the 6th most-asked of 21 Chemistry chapters. The 2025 shifts had 13 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 16in 14 papers |
| 2021 | 35in 26 papers |
| 2022 | 40in 24 papers |
| 2023 | 20in 12 papers |
| 2024 | 27in 18 papers |
| 2025 | 13in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Group 15 elements54
- Group 13 elements38
- Group 17 and 18 elements25
- Group 16 elements21
- Group 14 elements13
Try it
10 p-Block Elements questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · p-Block Elements
Which of the following is the strongest acid in aqueous solution?
Answer: (C)
Acid strength of the hydrogen halides increases down the group, HF < HCl < HBr < HI, as the H–X bond becomes longer and weaker and the proton is released more easily.
Q2 · Chemistry · p-Block Elements
Going down group 15 from NH₃ to BiH₃, the basic character of the hydrides:
Answer: (A)
All group 15 hydrides have a lone pair on the central atom. Down the group the atom grows larger, the lone pair is spread over a larger volume and is less available for donation, so basic character decreases: NH₃ > PH₃ > AsH₃ > SbH₃ ≥ BiH₃.
Q3 · 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?
Answer: (A)
Apply Le Chatelier's principle and kinetics together. The reaction is exothermic and reduces the number of gas moles from 3 to 2, so low T and high P favour SO₃ at equilibrium. At low T the rate is too slow even with V₂O₅, so about 720 K is a compromise. Because the conversion at about 2 bar is already high, the extra cost and corrosion of high pressure are not justified. A catalyst never changes the position of equilibrium.
Q4 · Chemistry · p-Block Elements
Which of the following hydrides of group 15 is the strongest reducing agent?
Answer: (B)
The E–H bond dissociation enthalpy falls from NH₃ to BiH₃, so the stability of the hydrides decreases down the group. A hydride that loses hydrogen more easily is a better reducing agent, so the reducing character increases down the group and BiH₃ is the strongest.
Q5 · 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?
Answer: (A)
Principle: catenation depends on the strength of the element–element single bond. Oxygen atoms are very small, and the lone pairs on neighbouring O atoms repel strongly, so the O–O single bond is weak. In sulphur the bigger atomic size reduces this repulsion, so the S–S bond is considerably stronger and allows rings such as S₈ and polysulphides. The other options attribute the difference to polarity, to multiple bonding in sulphur, or to physical state, and none of these is the cause.
Q6 · 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?
Answer: (A)
Principle: use the number of bond pairs and lone pairs (VSEPR) to find the geometry, then the hybridisation. In [PCl₄]⁺, P has 5 valence electrons, loses 1 for the positive charge, and forms 4 P–Cl bonds. That gives 4 bond pairs, no lone pair, a tetrahedral shape and sp³ hybridisation. In [PCl₆]⁻, P has 5 electrons, gains 1 for the negative charge, and forms 6 P–Cl bonds. That gives 6 bond pairs, no lone pair, an octahedral shape and sp³d² hybridisation. The trigonal bipyramidal sp³d description applies only to gaseous PCl₅.
Q7 · Chemistry · p-Block Elements
At room temperature water is a liquid while hydrogen sulphide is a gas, although H₂S has the higher molar mass. The reason is that:
Answer: (B)
Oxygen is small and highly electronegative, so the O–H bond is strongly polar and water molecules associate through hydrogen bonding. Sulphur is larger and much less electronegative, so H₂S molecules are held only by weak van der Waals forces and H₂S is a gas.
Q8 · Chemistry · p-Block Elements
Which statement about the allotropes of sulphur is correct?
Answer: (B)
Principle: rhombic (α) and monoclinic (β) sulphur are polymorphs built from the same crown-shaped S₈ ring, and they differ only in crystal packing. The transition temperature is 369 K. Below 369 K rhombic sulphur is stable, and above 369 K monoclinic sulphur is stable. So the correct statement is that both contain S₈ and rhombic is stable below 369 K.
Q9 · 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?
Answer: (C)
Hydrolysis of PCl₃ gives H₃PO₃, with phosphorus staying in the +3 state, and 3 HCl: PCl₃ + 3H₂O → H₃PO₃ + 3HCl. HCl is a strong monobasic acid and takes 3 mol of NaOH. In H₃PO₃ the structure is HP(=O)(OH)₂. Only the two O–H hydrogens ionise, since the P–H hydrogen is not acidic, so H₃PO₃ is dibasic and takes 2 mol of NaOH. Total NaOH = 3 + 2 = 5 mol.
Q10 · Chemistry · p-Block Elements
The boiling points of the noble gases rise steadily from helium to xenon because:
Answer: (A)
Noble gas atoms are held in the liquid only by weak dispersion forces, which is why their boiling points are very low. These forces grow with the number of electrons and the size of the atom, so the boiling point rises from helium, the lowest of any substance, to xenon.
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All 151 real JEE Main questions from p-Block Elements.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About p-Block Elements in JEE Main.
How many questions come from p-Block Elements in JEE Main?
p-Block Elements (Chemistry, Class 12) has been asked 151 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.4 questions a paper, 6% of all Chemistry questions. It is the 6th most-asked of 21 Chemistry chapters. The 2025 shifts had 13 from it in all.
Which topics of p-Block Elements are asked most in JEE Main?
Group 15 elements (54), Group 13 elements (38), Group 17 and 18 elements (25), Group 16 elements (21), counted over the papers from 2020–2025.
Is p-Block Elements important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 6 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve p-Block Elements JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from p-Block Elements with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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