JEE (Main) · Chemistry, Class 11
Chemical Bonding and Molecular Structure: JEE Main PYQ, 2020–2025
Chemical Bonding and Molecular Structure (Chemistry, Class 11) has been asked 158 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 4th most-asked of 21 Chemistry chapters. The 2025 shifts had 18 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 15in 14 papers |
| 2021 | 36in 26 papers |
| 2022 | 32in 24 papers |
| 2023 | 15in 12 papers |
| 2024 | 42in 18 papers |
| 2025 | 18in 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.
- VSEPR theory55
- Molecular orbital theory36
- Hybridisation25
- Dipole moment and Fajans' rule23
- Hydrogen bonding14
- Ionic bonding and lattice enthalpy5
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10 Chemical Bonding and Molecular Structure 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 · Chemical Bonding and Molecular Structure
What is the molecular shape of the sulphur trioxide molecule, SO₃, according to VSEPR theory?
Answer: (A)
Step 1: Sulphur has 6 valence electrons. Each of its three S–O bonds uses one of them, leaving no lone pair on S in the resonance structures. Step 2: Electron domains around S = 3, all bonding. Step 3: Three bonding domains with no lone pairs give a trigonal planar molecular shape. Lone pairs on the terminal oxygen atoms do not count for the central atom.
Q2 · Chemistry · Chemical Bonding and Molecular Structure
The chlorides LiCl, NaCl, KCl and CsCl are all predominantly ionic. Based on the polarising power of the cation, which of them has the greatest covalent character?
Answer: (D)
By Fajans' rules, covalent character in an ionic bond increases when the cation is small and highly charged, and when the anion is large. All four cations here carry a charge of +1, so the size decides. Li⁺ is the smallest, giving it the highest charge density and the greatest polarising power over the Cl⁻ cloud, so the electron density is pulled partly between the nuclei. The order of covalent character is LiCl > NaCl > KCl > CsCl. Choosing CsCl confuses a large ion being easily polarised with it being a strong polariser, and assuming equal charge means equal character ignores the effect of size.
Q3 · Chemistry · Chemical Bonding and Molecular Structure
What are the hybridisation states of the central atom in XeO₃, SF₄ and XeOF₄, respectively? (Count both σ-bonds and lone pairs on the central atom; π-bonds do not take part in hybridisation.)
Answer: (D)
The hybridisation of the central atom depends on its steric number, which is the number of σ-bonds plus the number of lone pairs. Double bonds such as Xe=O contribute only one σ-bond to this count. XeO₃: 3σ + 1 lone pair = 4, so sp³ (pyramidal). SF₄: 4σ + 1 lone pair = 5, so sp³d (see-saw). XeOF₄: 4 Xe–F σ + 1 Xe–O σ + 1 lone pair = 6, so sp³d² (square pyramidal). The answer is sp³, sp³d, sp³d². The usual errors are forgetting the lone pair on the central atom or counting the π-bonds.
Q4 · Chemistry · Chemical Bonding and Molecular Structure
Which of the following compounds can form hydrogen bonds with other molecules of the same compound in the liquid state?
Answer: (A)
Principle: a hydrogen bond needs an H atom covalently bonded to a small, highly electronegative atom (F, O or N) acting as donor, and a lone pair on F, O or N acting as acceptor. Apply to each option: methanol has O–H (donor) and O lone pairs (acceptor), so it qualifies. Methane has only C–H bonds, which are not polar enough. Chloromethane has no H on Cl, and C–H bonds do not donate. Trimethylamine has lone pairs but no N–H, so it can accept but not donate, and therefore cannot form H-bonds with its own kind.
Q5 · Chemistry · Chemical Bonding and Molecular Structure
In the Lewis structure of carbon monoxide, carbon and oxygen are joined by a triple bond and each atom carries one lone pair. What is the formal charge on the carbon atom?
Answer: (D)
Carbon has 4 valence electrons, 2 electrons in its lone pair and 6 shared electrons in the triple bond. Formal charge = 4 − 2 − ½(6) = −1. Oxygen, by the same count, is 6 − 2 − 3 = +1, and the two add up to zero for the neutral molecule.
Q6 · Chemistry · Chemical Bonding and Molecular Structure
Which of the following ground-state diatomic species has bond order 2.5 and its single unpaired electron in a bonding σ molecular orbital? (Use the MO energy order σ2s < σ*2s < π2p_x = π2p_y < σ2p_z < π*2p_x = π*2p_y < σ*2p_z for species with up to 14 total electrons, and the order with σ2p_z below π2p for species with more than 14 total electrons.)
Answer: (A)
For each species, write the configuration, count the bond order and locate the unpaired electron. C₂⁻ has 9 valence electrons: σ2s², σ*2s², π2p⁴, σ2p_z¹. Nb = 7 and Na = 2, so bond order = 2.5, and the odd electron is in the bonding σ2p_z. O₂⁺ also has bond order 2.5, but its odd electron is in π*2p, so it fails the orbital condition. F₂⁺ has bond order 1.5 with an odd electron in π*2p. Be₂⁺ has bond order 0.5 with an odd electron in σ*2s. Only C₂⁻ meets both conditions.
Q7 · Chemistry · Chemical Bonding and Molecular Structure
Why is a Na⁺ ion smaller than a Na atom, while a Cl⁻ ion is larger than a Cl atom?
Answer: (A)
Cation formation removes electrons. Na⁺ has 10 electrons and 11 protons, so the remaining electrons feel a stronger effective nuclear pull and the ion is smaller. Anion formation adds electrons to the same shell. Cl⁻ has 18 electrons and 17 protons, so repulsion increases and the ion is larger. The ionic radius trends are the basis for the size effects seen in ionic solids.
Q8 · Chemistry · Chemical Bonding and Molecular Structure
The methyl cation (CH₃⁺) and the methyl anion (CH₃⁻) are reactive intermediates. Which row correctly gives the hybridisation of carbon and the shape of each ion?
Answer: (D)
Count the electron pairs around carbon. In CH₃⁺ there are three bond pairs and no lone pair. Carbon is sp² hybridised, the shape is trigonal planar, and an empty unhybridised p orbital lies perpendicular to the plane. In CH₃⁻ there are three bond pairs and one lone pair, four pairs in all. Carbon is sp³ hybridised, and the lone pair occupies one hybrid orbital, so the shape is trigonal pyramidal.
Q9 · Chemistry · Chemical Bonding and Molecular Structure
In SF₄ the central sulphur atom has five electron pairs (four bond pairs and one lone pair) arranged in a trigonal bipyramid. Which of the following correctly gives the shape of the molecule, the position of the lone pair and the reason for that position?
Answer: (A)
Step 1: Count electron pairs on S. Six valence electrons plus four from F gives 10 electrons, i.e. 5 pairs: 4 bond pairs and 1 lone pair. The electron geometry is trigonal bipyramidal. Step 2: Choose the lone pair site. Repulsions at 90° matter most. Axial site: the lone pair has 3 bond pairs at 90°. Equatorial site: the lone pair has 2 bond pairs at 90° (the 120° interactions are weaker). The equatorial site is therefore preferred. Step 3: Name the shape. With the lone pair equatorial, the four F atoms form a see-saw (distorted tetrahedral) arrangement. Step 4: Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, so the axial S–F bonds are pushed back from the lone pair and the axial F–S–F angle is about 173°, below 180°. Only the first option has the correct shape, reason and angle.
Q10 · Chemistry · Chemical Bonding and Molecular Structure
In gaseous beryllium chloride, BeCl₂, what are the hybridisation of the central beryllium atom and the Cl–Be–Cl bond angle?
Answer: (B)
Principle: the steric number (number of bond pairs plus lone pairs on the central atom) determines the hybridisation, and the steric number gives the geometry. For BeCl₂, Be has 2 bond pairs and 0 lone pairs, so the steric number is 2. Two orbitals mean sp hybridisation. Two sp orbitals are 180° apart, so the molecule is linear with a Cl–Be–Cl angle of 180°.
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All 158 real JEE Main questions from Chemical Bonding and Molecular Structure.
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About Chemical Bonding and Molecular Structure in JEE Main.
How many questions come from Chemical Bonding and Molecular Structure in JEE Main?
Chemical Bonding and Molecular Structure (Chemistry, Class 11) has been asked 158 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 4th most-asked of 21 Chemistry chapters. The 2025 shifts had 18 from it in all.
Which topics of Chemical Bonding and Molecular Structure are asked most in JEE Main?
VSEPR theory (55), Molecular orbital theory (36), Hybridisation (25), Dipole moment and Fajans' rule (23), counted over the papers from 2020–2025.
Is Chemical Bonding and Molecular Structure important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 4 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Chemical Bonding and Molecular Structure JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Chemical Bonding and Molecular Structure 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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