Chemical Bonding and Molecular Structure NEET PYQs (2005–2026) with Solutions

10 solved Chemical Bonding and Molecular Structure PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasySigma bonds, pi bonds and lone pairs

   Match List I with List II: Choose the correct answer from the options given below:

   1. AOption A: A–III, B–IV, C–II, D–I
   2. BOption B: A–IV, B–I, C–III, D–II
   3. COption C: A–I, B–II, C–IV, D–III
   4. DOption D: A–II, B–III, C–I, D–IV

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   Correct answer: (B) A–IV, B–I, C–III, D–II

   Explanation

   C₂H₄ has 5σ and 1π bond; C₂H₂ has 3σ and 2π bonds; CH₄ has 4σ bonds; NH₃ has 3σ bonds and one lone pair. Therefore A–IV, B–I, C–III, D–II. Hence option (B) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasyVSEPR and hybridisation geometries

   Match the species in List-I with their geometry in List-II. List-I: A. $\mathrm{PCl_5}$ B. $\mathrm{BrF_5}$ C. $\mathrm{BF_4^-}$ D. $\mathrm{[Ni(CN)_4]^{2-}}$ List-II: I. Tetrahedral II. Square Planar III. Trigonal bipyramidal IV. Square pyramidal Choose the correct answer from the options given below:

   1. AOption A: A-IV, B-III, C-I, D-II
   2. BOption B: A-III, B-IV, C-I, D-II
   3. COption C: A-III, B-I, C-II, D-IV
   4. DOption D: A-III, B-II, C-I, D-IV

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   Correct answer: (B) A-III, B-IV, C-I, D-II

   Explanation

   A. $\mathrm{PCl_5}$: 5 bond pairs, no lone pair, $sp^3d$, so trigonal bipyramidal (III). B. $\mathrm{BrF_5}$: 5 bond pairs + 1 lone pair, $sp^3d^2$, so square pyramidal (IV). C. $\mathrm{BF_4^-}$: 4 bond pairs, $sp^3$, so tetrahedral (I). D. $\mathrm{[Ni(CN)_4]^{2-}}$: $\mathrm{Ni^{2+}}$ is $3d^8$. The strong-field $\mathrm{CN^-}$ ligands pair the electrons, giving $dsp^2$ hybridisation, so square planar (II). Match: A-III, B-IV, C-I, D-II, option (B).

3. Question 3 (NEET 2026)

   NEET 2026EasyVSEPR Theory & Molecular Shapes

   Identify the correct statement about ClF₃ from the following options:

   1. AOption A: It has T-shaped geometry with two lone pairs on Cl atom
   2. BOption B: It has T-shaped geometry with three lone pairs on Cl atom
   3. COption C: It has trigonal pyramidal geometry with two lone pairs on Cl atom
   4. DOption D: It has planar trigonal geometry with two lone pairs on Cl atom

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   Correct answer: (A) It has T-shaped geometry with two lone pairs on Cl atom

   Explanation

   ClF₃ has five electron pairs around Cl (3 bond pairs + 2 lone pairs). Electron geometry is trigonal bipyramidal and molecular geometry is T-shaped. Hence option (A) is correct.

4. Question 4 (NEET 2026)

   NEET 2026MediumIonic & Covalent Bonding

   The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are:

   1. AOption A: −1, 0, +1
   2. BOption B: 0, +1, −1
   3. COption C: 0, 0, 0
   4. DOption D: +1, 0, −1

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   Correct answer: (B) 0, +1, −1

   Explanation

   Formal charge = Valence electrons − (Nonbonding electrons + ½×Bonding electrons). O₂ has 0 charge, O₁ has +1 charge and O₃ has −1 charge. Hence the order (2,1,3) is 0, +1, −1. Therefore option (B) is correct.

5. Question 5 (NEET 2026)

   NEET 2026EasyMolecular orbital configuration of Ne2

   The highest occupied molecular orbital for $\mathrm{Ne_2}$ is

   1. AOption A: $\pi_{2p}$
   2. BOption B: $\sigma_{2p}$
   3. COption C: $\pi^*_{2p}$
   4. DOption D: $\sigma^*_{2p}$

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   Correct answer: (D) $\sigma^*_{2p}$

   Explanation

   $\mathrm{Ne_2}$ would have $2 \times 10 = 20$ electrons. For O, F and Ne, the $\sigma_{2p_z}$ orbital lies below $\pi_{2p}$. Configuration: $$\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ (\pi2p_x^2 = \pi2p_y^2)\ (\pi^*2p_x^2 = \pi^*2p_y^2)\ \sigma^*2p_z^2$$ The last electrons go into $\sigma^*_{2p}$, so it is the highest occupied molecular orbital: option (D). The bond order is$\frac{10 - 10}{2} = 0$, which is why$\mathrm{Ne_2}$ does not exist as a stable molecule.

6. Question 6 (NEET 2025)

   NEET 2025EasyMolecular 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 :

   1. AOption A: Both Statement I and Statement II are true
   2. BOption B: Both Statement I and Statement II are false
   3. COption C: Statement I is true but Statement II is false
   4. DOption D: Statement I is false but Statement II is true

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   Correct answer: (B) Both Statement I and Statement II are false

   Explanation

   A molecule having bond order zero is unstable and generally does not exist independently, so Statement I is false. Bond length decreases as bond order increases because stronger bonding pulls atoms closer together. Hence Statement II is also false. Therefore, both statements are false.

7. Question 7 (NEET 2025)

   NEET 2025MediumValence Bond Theory & Hybridization

   Match List-I with List-II Choose the correct answer from the options given below:

   1. AOption A: A-II, B-I, C-IV, D-III
   2. BOption B: A-II, B-I, C-III, D-IV
   3. COption C: A-IV, B-II, C-III, D-I
   4. DOption D: A-IV, B-II, C-I, D-III

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   Correct answer: (A) A-II, B-I, C-IV, D-III

   Explanation

   XeO3 has sp3 hybridization with pyramidal shape. XeF2 has sp3d hybridization with linear geometry. XeOF4 has sp3d2 hybridization with square pyramidal shape. XeF6 has sp3d3 hybridization with distorted octahedral geometry. Therefore, the correct matching is A-II, B-I, C-IV, D-III, which corresponds to option (A).

8. Question 8 (NEET 2025)

   NEET 2025HardBond Parameters

   Among the given compounds I–III, the correct order of bond dissociation energy of C–H bond marked with * is :

   1. AOption A: II > I > III
   2. BOption B: I > II > III
   3. COption C: III > II > I
   4. DOption D: II > III > I

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   Correct answer: (A) II > I > III

   Explanation

   Bond dissociation energy increases with increasing s-character of the hybrid orbital. In compound II, the marked C–H bond is on an sp carbon, so it has highest bond dissociation energy. In compound I, the carbon is sp² hybridised, while in compound III the bond is on an sp³ carbon. Hence the order is II > I > III.

9. Question 9 (NEET 2025)

   NEET 2025MediumBond Length, Dipole Moment and Molecular Properties

   Identify the correct orders against the property mentioned. A. $H_2O$ > $NH_3$ > $CHCl_3$ – dipole moment B. $XeF_4$ > $XeO_3$ > $XeF_2$ – number of lone pairs on central atom C. O-H > C-H > N-O – bond length D. $N_2$ > $O_2$ > $H_2$ – bond enthalpy Choose the correct answer from the options given below :

   1. AOption A: A, D only
   2. BOption B: B, D only
   3. COption C: A, C only
   4. DOption D: B, C only

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   Correct answer: (A) A, D only

   Explanation

   A is correct because dipole moment order is H2O > NH3 > CHCl3. D is also correct because bond enthalpy order is N2 > O2 > H2. Statement B is incorrect since XeF2 has more lone pairs than XeO3. Statement C is incorrect because actual bond length order differs from the given sequence.

10. Question 10 (NEET 2024)

    NEET 2024MediumDipole Moment and Resonance

    Identify the correct answer.

    1. AOption A: Dipole moment of NF₃ is greater than that of NH₃.
    2. BOption B: Three canonical forms can be drawn for CO₃²⁻ ion.
    3. COption C: Three resonance structures can be drawn for ozone.
    4. DOption D: BF₃ has non-zero dipole moment.

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    Correct answer: (B) Three canonical forms can be drawn for CO₃²⁻ ion.

    Explanation

    For carbonate ion, three equivalent resonance (canonical) structures can be drawn. NF₃ has lower dipole moment than NH₃. Ozone has only two major resonance structures. BF₃ is trigonal planar and has zero dipole moment. Hence, option (B) is correct.