NEET Chemical Bonding and Molecular Structure PYQs & Practice Questions

10 free NEET Chemistry questions on Chemical Bonding and Molecular Structure, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: VSEPR Theory · Bond Order and Molecular Orbital Theory · Dipole Moment · Hydrogen Bonding and Boiling Point · Hydrogen Bonding Strength · Hybridisation and Bond Angle · Electronegativity · Ionisation Enthalpy and Half-Filled Stability · Dipole Moment, Lone Pairs, Bond Length, Bond Enthalpy Comparisons · Lewis Acids and Bases · Sigma Bonds, Pi Bonds and Lone Pairs · Molecular Orbital Theory

Q1NEET 2026VSEPR Theory - Molecular Geometry
Match the species in List I with their geometry in List II: List I: A) PCl5, B) BrF5, C) BF4⁻, D) [Ni(CN)4]2⁻ List II: I) Tetrahedral, II) Square Planar, III) Trigonal bipyramidal, IV) Square pyramidal Choose the correct answer from the options given below:
- A. A-III, B-IV, C-I, D-II
- B. A-III, B-I, C-II, D-IV
- C. A-III, B-II, C-I, D-IV
- D. A-IV, B-III, C-I, D-II
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Answer: —
PCl5 (sp3d hybridization, no lone pairs on P) is trigonal bipyramidal (III). BrF5 (sp3d2, one lone pair on Br) is square pyramidal (IV). BF4⁻ (sp3, no lone pairs on B) is tetrahedral (I). [Ni(CN)4]2⁻: CN⁻ is a strong field ligand, causing Ni2+ (d8) to adopt dsp2 hybridization, giving a square planar (II) diamagnetic complex. So A-III, B-IV, C-I, D-II.
Q2NEET 2026Molecular Orbital Theory - HOMO
The highest occupied molecular orbital (HOMO) for Ne2 is:
- A. σ2p
- B. π*2p
- C. σ*2p
- D. π2p
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Answer: B — π*2p
Ne2 (hypothetical) would have 20 electrons total (10 from each Ne atom), completely filling all 10 molecular orbitals formed from 1s and 2s/2p atomic orbitals: σ1s, σ*1s, σ2s, σ*2s, σ2pz, π2px, π2py, π*2px, π*2py, σ*2pz. With all orbitals completely filled (including the highest, σ*2p), the HOMO is σ*2p, and the bond order is zero (which is why Ne2 does not exist as a stable molecule).
Q3NEET 2026Formal charges on oxygen atoms in ozone
The structure of ozone (O3) with oxygen atoms numbered 1, 2, 3 is shown. The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are:
- A. 0, 0, 0
- B. -1, 0, +1
- C. +1, 0, -1
- D. 0, +1, -1
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Answer: C — +1, 0, -1
In ozone's resonance structure, the central O (2) has formal charge 0... per official key, formal charges on atoms 2,1,3 are +1, 0, -1.
Q4NEET 2026Matching bonds in simple molecules
Match List I with List II: List I — A. C2H4, B. C2H2, C. CH4, D. NH3. List II — I. 3σ bonds, 2π bonds, II. 3σ bonds, one lone pair, III. 4σ bonds, IV. 5σ bonds, 1π bond.
- A. A-IV, B-I, C-III, D-II
- B. A-III, B-IV, C-II, D-I
- C. A-II, B-III, C-I, D-IV
- D. A-I, B-II, C-IV, D-III
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Answer: —
C2H4 (ethylene) has 5σ+1π bonds (IV); C2H2 (acetylene) has 3σ+2π bonds (I); CH4 has 4σ bonds (III); NH3 has 3σ bonds + 1 lone pair (II). A-IV, B-I, C-III, D-II.

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Q5NEET 2025Bond Order and Molecular Orbital Theory
Given below are two statements: Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Both Statement I and Statement II are true
- B. Both Statement I and Statement II are false
- C. Statement I is true but Statement II is false
- D. Statement I is false but Statement II is true
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Answer: A — Both Statement I and Statement II are true
Statement I is false: a bond order of zero means the molecule does not exist/is not stable (e.g., He2). Statement II is false: as bond order increases, bond length actually decreases (bonds become stronger and shorter), not increases. Both statements are false.
Q6NEET 2025Dipole Moment, Lone Pairs, Bond Length, Bond Enthalpy Comparisons
Identify the correct orders against the property mentioned: A. H2O > NH3 > CHCl3 – dipole moment. B. XeF4 > XeO3 > XeF2 – number of lone pairs on central atom. C. O–H > C–H > N–O – bond length. D. N2 > O2 > H2 – bond enthalpy. Choose the correct answer from the options given below:
- A. A, D only
- B. B, D only
- C. A, C only
- D. B, C only
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Answer: —
A is correct: dipole moments H2O(1.85D) > NH3(1.47D) > CHCl3(1.04D). B is incorrect: lone pairs are XeF4(2) = XeF2(3)... actual order differs (XeF2 has 3 lone pairs, XeF4 has 2, XeO3 has 1), so stated order is wrong. C is incorrect bond length order stated. D is correct: bond enthalpy N2(941) > O2(498) > H2(436) kJ/mol (triple bond strongest). So A, D only.
Q7NEET 2023Lewis Acids and Bases
Amongst the given options, which of the following molecules/ion acts as a Lewis acid?
- A. H2O
- B. BF3
- C. OH-
- D. NH3
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Answer: A — H2O
BF3 has boron with an incomplete octet (only 6 electrons around it), making it electron-deficient and able to accept an electron pair, so it acts as a Lewis acid. H2O, OH-, and NH3 all have lone pairs available to donate and act as Lewis bases.
Q8NEET 2023Sigma Bonds, Pi Bonds and Lone Pairs
The number of sigma bonds, pi bonds and lone pairs of electrons in pyridine, respectively, are:
- A. 12, 3, 0
- B. 11, 3, 1
- C. 12, 2, 1
- D. 11, 2, 0
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Answer: A — 12, 3, 0
Pyridine (C5H5N) has 11 sigma bonds (5 C-C/C-N sigma bonds in the ring plus 5 C-H bonds plus... counted as 11 total sigma bonds in the ring system and substituents), 3 pi bonds (from the three alternating double bonds in the aromatic ring), and 1 lone pair of electrons on the nitrogen atom (which lies in the plane of the ring and is not part of the aromatic pi system).

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Q9NEET 2023Molecular Orbital Theory
The correct order of energies of molecular orbitals of N2 molecule is:
- A. sigma1s < sigma*1s < sigma2s < sigma*2s < sigma2pz < (pi2px = pi2py) < (pi*2px = pi*2py) < sigma*2pz
- B. sigma1s < sigma*1s < sigma2s < sigma*2s < sigma2pz < sigma*2pz < (pi2px = pi2py) < (pi*2px = pi*2py)
- C. sigma1s < sigma*1s < sigma2s < sigma*2s < (pi2px = pi2py) < (pi*2px = pi*2py) < sigma2pz < sigma*2pz
- D. sigma1s < sigma*1s < sigma2s < sigma*2s < (pi2px = pi2py) < sigma2pz < (pi*2px = pi*2py) < sigma*2pz
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Answer: C — sigma1s < sigma*1s < sigma2s < sigma*2s < (pi2px = pi2py) < (pi*2px = pi*2py) < sigma2pz < sigma*2pz
For N2 (and other molecules with atomic number up to 7, where there is significant s-p mixing), the correct molecular orbital energy order is: sigma1s < sigma*1s < sigma2s < sigma*2s < (pi2px = pi2py) < sigma2pz < (pi*2px = pi*2py) < sigma*2pz. This is because the pi2p orbitals are lower in energy than the sigma2pz orbital for these lighter diatomic molecules.
Q10NEET 2023Octet Rule Exceptions
Amongst the following, the total number of species NOT having eight electrons around the central atom in its outermost shell, is: NH3, AlCl3, BeCl2, CCl4, PCl5.
- A. 2
- B. 4
- C. 1
- D. 3
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Answer: C — 1
NH3 has 8 electrons around N (3 bond pairs + 1 lone pair) and CCl4 has 8 electrons around C (4 bond pairs) — both satisfy the octet. AlCl3 has only 6 electrons around Al, BeCl2 has only 4 electrons around Be, and PCl5 has 10 electrons around P — none of these three satisfy the octet rule. So 3 species (AlCl3, BeCl2, PCl5) do NOT have eight electrons around the central atom.

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