25 questions in 25 minutes on the NTA computer-based test interface, all from Chemical Bonding and Molecular Structure. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Chemical Bonding and Molecular Structure, of which 15 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
Revising first? Read the Chemical Bonding and Molecular Structure chapter notes, then come back and take this test to check it stuck.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Chemical Bonding and Molecular Structure mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 15 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Chemical Bonding and Molecular Structure?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Chemistry
- Q1In which of the following is the central atom sp hybridised?
- CO₂
- NH₃
- H₂O
- BF₃
Answer: (A) CO₂
In CO₂ the carbon has 2 σ bonds and no lone pairs (the two double bonds count as 2 domains), giving sp hybridisation and a linear shape. BF₃ is sp², H₂O and NH₃ are sp³.
- Q2The correct order of decreasing bond angle is:
- H₂O > NH₃ > CH₄
- NH₃ > CH₄ > H₂O
- CH₄ > NH₃ > H₂O
- CH₄ > H₂O > NH₃
Answer: (C) CH₄ > NH₃ > H₂O
As lone pairs increase, repulsion compresses bond angles: CH₄ (0 lp, 109.5°) > NH₃ (1 lp, 107°) > H₂O (2 lp, 104.5°).
- Q3Which of the following is a polar molecule?
- XeF₄
- BF₃
- SF₄
- SiF₄
Answer: (C) SF₄
SF₄ has 4 bond pairs + 1 lone pair (see-saw, sp³d); the asymmetry leaves a net dipole, so it is polar. BF₃ (trigonal planar), SiF₄ (tetrahedral) and XeF₄ (square planar) are all symmetric with μ = 0.
- Q4The formal charge on the central nitrogen atom in the nitrate ion NO₃⁻ (one N=O double bond, two N-O single bonds) is:
- 0
- +2
- +1
- -1
Answer: (C) +1
For N: V = 5, lone pairs L = 0, bonding electrons B = 8 (one double + two single = 4 bonds). FC = 5 − 0 − 8/2 = 5 − 4 = +1.
- Q5Which of the following molecules has the maximum dipole moment?
- CO₂
- CH₄
- NH₃
- NF₃
Answer: (C) NH₃
CO₂ (linear) and CH₄ (tetrahedral) are symmetric with μ = 0. Between NH₃ and NF₃, in NH₃ the N-H bond dipoles point toward N and add to the lone-pair dipole (μ ≈ 1.47 D), while in NF₃ the N-F bond dipoles point away from N and oppose the lone-pair dipole (μ ≈ 0.24 D). Hence NH₃ has the maximum dipole moment.
- Q6In which of the following pairs are the two species isostructural?
- BF₃ and NF₃
- BrO₃⁻ and XeO₃
- SF₄ and XeF₄
- SO₃²⁻ and NO₃⁻
Answer: (B) BrO₃⁻ and XeO₃
BrO₃⁻ (3 bp + 1 lp) and XeO₃ (3 bp + 1 lp) are both pyramidal (sp³) → isostructural. SF₄ (see-saw) vs XeF₄ (square planar), SO₃²⁻ (pyramidal) vs NO₃⁻ (planar), and BF₃ (planar) vs NF₃ (pyramidal) are each mismatched.
- Q7Which of the following species has the minimum bond length?
- O₂⁺
- O₂⁻
- O₂
- O₂²⁻
Answer: (A) O₂⁺
Bond length is shortest for the highest bond order. Bond orders: O₂⁺ = 2.5 (highest), O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. Hence O₂⁺ has the minimum bond length.
- Q8The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is:
- 4
- 3
- 2
- 1
Answer: (C) 2
Element 16 is sulphur; its diatomic molecule S₂ has the same valence MO pattern as O₂. The last two electrons occupy the two degenerate π* orbitals singly, giving 2 unpaired electrons (paramagnetic).
- Q9The dipole moment of NH₃ (1.47 D) is greater than that of NF₃ (0.24 D). The main reason is:
- NF₃ is planar
- in NH₃ the lone-pair dipole and N-H bond dipoles add up, while in NF₃ they oppose each other
- N-F bonds are non-polar
- fluorine is less electronegative than hydrogen
Answer: (B) in NH₃ the lone-pair dipole and N-H bond dipoles add up, while in NF₃ they oppose each other
In NH₃ the bond dipoles point from H toward N, in the same direction as the lone-pair dipole, so they add. In NF₃ the bond dipoles point from N toward F, opposing the lone-pair dipole, giving a small net μ.
- Q10Which one of the following does NOT apply to a metallic bond?
- overlapping valence orbitals
- highly directed bonds
- delocalised electrons
- mobile valence electrons
Answer: (B) highly directed bonds
A metallic bond is the attraction between a ‘sea’ of delocalised, mobile valence electrons and the positive metal kernels – it is non-directional. So ‘highly directed bonds’ does not apply.
- Q11The angle between an overlapping s-orbital and one p-orbital (axial overlap) is:
- 120°60′
- 109°28′
- 120°
- 180°
Answer: (D) 180°
For effective σ-overlap the s-orbital meets the p-orbital head-on along its axis, so the two lobes of the p-orbital and the s-orbital lie in a straight line – the angle is 180°.
- Q12The hybridisation of the central atom in XeF₄ is:
- sp³d
- sp³
- sp²
- sp³d²
Answer: (D) sp³d²
Xe in XeF₄ has 4 bond pairs + 2 lone pairs = 6 electron domains, requiring sp³d² hybridisation (octahedral electron geometry, square planar shape).
- Q13Among N₂, O₂, O₂⁺ and O₂⁻, which has the highest bond order?
- O₂⁻
- O₂
- N₂
- O₂⁺
Answer: (C) N₂
Bond orders: N₂ = 3, O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5. N₂ has the highest bond order (triple bond) and hence the shortest, strongest bond.
- Q14Which one of the following species does NOT exist under normal conditions?
- Li₂
- Be₂⁺
- B₂
- Be₂
Answer: (D) Be₂
Be₂ (8 e⁻) has N_b = 4, Nₐ = 4, bond order = 0, so it does not exist. Be₂⁺ (B.O. 0.5), B₂ (B.O. 1) and Li₂ (B.O. 1) all have non-zero bond order.
- Q15Which of the following does NOT have a linear arrangement of atoms?
- C₂H₂
- BeH₂
- H₂S
- CO₂
Answer: (C) H₂S
H₂S has two lone pairs on S (sp³) → bent (≈92°), not linear. C₂H₂ (sp carbons), BeH₂ (sp, no lone pair) and CO₂ (sp) are all linear.
- Q16The ion that is isoelectronic with CO is:
- O₂⁻
- O₂⁺
- N₂⁺
- CN⁻
Answer: (D) CN⁻
CO has 6 + 8 = 14 electrons. CN⁻ has 6 + 7 + 1 = 14 electrons – isoelectronic. N₂⁺ (13), O₂⁻ (17) and O₂⁺ (15) differ.
- Q17Which one of the following hydrides has the highest dipole moment?
- PH₃
- NH₃
- AsH₃
- SbH₃
Answer: (B) NH₃
All are pyramidal. Nitrogen is the most electronegative and smallest central atom of the group, so N-H bonds are most polar and NH₃ has the largest dipole moment: NH₃ > PH₃ > AsH₃ > SbH₃.
- Q18Which of the following molecules has a net dipole moment of zero?
- H₂O
- NH₃
- SO₂
- CO₂
Answer: (D) CO₂
CO₂ is linear and symmetric, so its two equal C=O bond dipoles point in opposite directions and cancel (μ = 0). H₂O and SO₂ are bent and NH₃ is pyramidal, so all have non-zero dipole moments.
- Q19The correct sequence of the bond enthalpy of the C-X bond is:
- CH₃-Cl > CH₃-I > CH₃-F > CH₃-Br
- CH₃-F > CH₃-Cl > CH₃-Br > CH₃-I
- CH₃-F < CH₃-Cl > CH₃-Br > CH₃-I
- CH₃-F < CH₃-Cl < CH₃-Br < CH₃-I
Answer: (B) CH₃-F > CH₃-Cl > CH₃-Br > CH₃-I
Down the halogen group the atomic size increases (F < Cl < Br < I), so the C-X bond length increases and the bond becomes weaker. Hence bond enthalpy decreases F → I: CH₃-F > CH₃-Cl > CH₃-Br > CH₃-I.
- Q20Mark the INCORRECT statement among the following:
- The bond order in O₂⁺ and O₂⁻ decreases as O₂⁺ > O₂⁻
- Electrons in an antibonding MO contribute to repulsion between the two atoms
- With increase in bond order, bond length decreases and bond strength increases
- The bond energy of a diatomic molecule always increases when an electron is lost
Answer: (D) The bond energy of a diatomic molecule always increases when an electron is lost
Losing an electron can either raise or lower the bond order (and hence bond energy), depending on whether the electron came from a bonding or an antibonding orbital – so it does NOT always increase. The other statements are correct.
- Q21Ortho-nitrophenol has a lower boiling point than para-nitrophenol mainly because:
- ortho isomer has stronger intermolecular hydrogen bonding
- ortho isomer is ionic
- para isomer has a lower molecular mass
- ortho isomer forms intramolecular hydrogen bonds, reducing intermolecular association
Answer: (D) ortho isomer forms intramolecular hydrogen bonds, reducing intermolecular association
In o-nitrophenol the -OH and -NO₂ groups are close enough to form an intramolecular H-bond (chelation), reducing intermolecular H-bonding. The para isomer forms intermolecular H-bonds, raising its boiling point.
- Q22The pair of species with the same bond order is:
- O₂²⁻, B₂
- O₂⁺, NO⁺
- N₂, O₂
- NO, CO
Answer: (A) O₂²⁻, B₂
O₂²⁻ (18 e⁻) has B.O. = (10−8)/2 = 1 and B₂ (10 e⁻) has B.O. = (6−4)/2 = 1 – same bond order. The other pairs differ (O₂⁺ 2.5 vs NO⁺ 3; NO 2.5 vs CO 3; N₂ 3 vs O₂ 2).
- Q23Assertion (A): The bond dissociation energy of N₂ is very high. Reason (R): N₂ has a bond order of 3 with a triple bond between the nitrogen atoms.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer: (A) Both A and R are true and R is the correct explanation of A
N₂ has bond order 3 (one σ and two π bonds), giving an exceptionally strong, short bond and a very high dissociation energy (~945 kJ/mol). R correctly explains A.
- Q24Linus Pauling received the Nobel Prize for his work on:
- thermodynamics
- atomic structure
- chemical bonds
- photosynthesis
Answer: (C) chemical bonds
Linus Pauling won the 1954 Nobel Prize in Chemistry for his research on the nature of the chemical bond and its application to the structure of complex substances.
- Q25Which of the following molecules has an incomplete octet around the central atom?
- CCl₄
- SF₆
- PCl₅
- BCl₃
Answer: (D) BCl₃
In BCl₃, boron has only 6 electrons around it (three B-Cl bonds), an incomplete octet – boron is electron-deficient. PCl₅ and SF₆ have expanded octets; CCl₄ has a complete octet.