Chemical Bonding and Molecular Structure — NEET Chemistry MCQs with answers

Chemical Bonding and Molecular Structure is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 907 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Lewis structures, octet and formal charge · Ionic bonding, lattice enthalpy and Fajans rules · VSEPR and molecular shapes · Hybridisation · Molecular orbital theory · Bond polarity and dipole moment · Hydrogen bonding and intermolecular forces · Bond parameters and resonance · Valence bond theory and orbital overlap · Bond parameters · Ionic and covalent bonding · Metallic bonding · Hybridization · Chemical Bonding & Molecular Structure · Chemical Bonding · Lattice Energy · VSEPR + Hybridisation + Polarity · MOT bond order and magnetism · VBT resonance and bond-length averaging · Lattice energy and Born-Haber cycle logic · MOT bond order across a homonuclear series · VSEPR + hybridisation + polarity chain · Hybridisation and bond angle chain · Fajans + lattice energy + melting point chain · Hydrogen bonding + boiling point + hybridisation chain · VSEPR applied to an unfamiliar species · MOT applied to an unfamiliar diatomic ion · Fajans rule applied to an unfamiliar pair · Hybridisation and shape for an unfamiliar oxyanion · Polarity prediction for an unfamiliar mixed-substituent molecule · Hydrogen bonding applied to an unfamiliar pair · VSEPR shape and magnetic behaviour for an unfamiliar radical ion · Limiting case: lone pair count to zero · Limiting case: electronegativity difference to zero · Limiting case: lattice enthalpy as ionic radius sum grows very large · Limiting case: MOT bond order as antibonding electrons saturate · Limiting case: bond angle as lone pair count increases within TBP family · Assertion-Reason: VSEPR shape and hybridisation · Assertion-Reason: MOT and paramagnetism · Assertion-Reason: Fajans rule and covalency

8 free Chemical Bonding and Molecular Structure practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Hydrogen bonding and intermolecular forces

   Ice is less dense than liquid water at 0°C, which is unusual for a solid-liquid pair (most solids are denser than their liquids). What is the correct hydrogen-bonding-based explanation?

   • A.In ice, each water molecule is hydrogen-bonded to four neighbours in a rigid, open tetrahedral (hexagonal) lattice with more empty space than in liquid water, where the H-bond network is more disordered and molecules can pack somewhat more closely on average
   • B.Ice is less dense because hydrogen bonds are broken completely upon freezing, allowing molecules to spread apart randomly
   • C.Ice is less dense simply because solids are always less dense than their liquids as a universal rule
   • D.Ice is less dense because water molecules in ice vibrate faster than in the liquid, pushing them further apart
   Show answer and explanation

   Answer: A. In ice, each water molecule is hydrogen-bonded to four neighbours in a rigid, open tetrahedral (hexagonal) lattice with more empty space than in liquid water, where the H-bond network is more disordered and molecules can pack somewhat more closely on average

   In ice, every water molecule engages in four hydrogen bonds (two as donor via its two O–H, two as acceptor via its two lone pairs) arranged in a rigid, open hexagonal lattice, which has considerable empty space, lowering density. In liquid water, hydrogen bonding is more dynamic and less perfectly ordered, allowing molecules to pack somewhat more efficiently on average, giving liquid water a higher density than ice.

   Each water molecule can potentially form 4 hydrogen bonds (2 as H-donor through its two O–H bonds, 2 as H-acceptor through its two lone pairs on O). In ice, this maximal 4-coordinate hydrogen bonding is fully realised in a rigid, ordered hexagonal crystal lattice; this specific tetrahedral arrangement, while maximising hydrogen bond count, is geometrically 'open' and leaves substantial void space between molecules, lowering the overall density. In liquid water, thermal motion continuously breaks and reforms hydrogen bonds; the average number of hydrogen bonds per molecule is somewhat less than 4 and the structure is far less ordered, allowing molecules on average to be packed slightly more closely (denser) than the rigid open lattice of ice. This is why ice floats on water — density (ice) ≈ 0.92 g/cm³ < density (liquid water at 0°C) ≈ 1.00 g/cm³.

   Common mistake: Assuming freezing always breaks intermolecular bonds, or that solids are universally denser than their corresponding liquids.

2. Question 2 · medium · Hydrogen bonding and intermolecular forces

   Why does the boiling point of the noble gas argon (Ar, -186°C) differ from that of neon (Ne, -246°C), even though both are monatomic and nonpolar?

   • A.Ar has more electrons and a larger, more polarisable electron cloud than Ne, giving stronger London dispersion forces between Ar atoms, hence a higher boiling point
   • B.Ar atoms are held together by weak ionic bonds while Ne atoms are not, explaining the higher boiling point of Ar
   • C.Ar has a permanent dipole moment while Ne does not, so Ar experiences dipole-dipole forces in addition to dispersion forces
   • D.Ar's boiling point is higher only because it has a higher atomic number, and atomic number by itself directly determines boiling point regardless of electron count or polarisability
   Show answer and explanation

   Answer: A. Ar has more electrons and a larger, more polarisable electron cloud than Ne, giving stronger London dispersion forces between Ar atoms, hence a higher boiling point

   Both Ar and Ne are monatomic and nonpolar, interacting only via London dispersion forces. Ar has 18 electrons (vs Ne's 10) and a larger, more diffuse, more easily polarisable electron cloud, giving stronger instantaneous-dipole-induced-dipole attractions between Ar atoms than between Ne atoms, hence Ar's higher boiling point.

   Since Ar and Ne are both monatomic noble gases with complete, spherically symmetric electron configurations, they cannot form any chemical bonds (ionic, covalent) with atoms of the same element, nor do they possess any permanent dipole moment. The only intermolecular (inter-atomic) attractive force present is London dispersion, which arises from instantaneous fluctuations in electron distribution inducing temporary dipoles. The strength of this force increases with the number of electrons and the size/diffuseness (polarisability) of the electron cloud: Ar (atomic number 18, 18 electrons, larger atomic radius) has a substantially larger, more polarisable electron cloud than Ne (atomic number 10, 10 electrons, smaller atomic radius). This gives Ar atoms stronger mutual dispersion attraction, requiring more thermal energy to separate them into the gas phase, hence Ar's higher boiling point (-186°C) compared to Ne (-246°C).

   Common mistake: Attributing the boiling point difference to bonding, dipole moment, or atomic number directly rather than to electron count/polarisability's effect on dispersion force strength.

3. Question 3 · medium · Ionic bonding, lattice enthalpy and Fajans rules

   According to Fajans' rules, covalent character in a predominantly ionic compound is favoured by which combination of factors?

   • A.Small, highly charged cation and large, highly charged anion
   • B.Large cation and small anion, both singly charged
   • C.Large cation and large anion, both singly charged
   • D.Small cation and small anion, both singly charged
   Show answer and explanation

   Answer: A. Small, highly charged cation and large, highly charged anion

   Covalent character is favoured when the cation strongly polarises the anion, i.e. a small, highly charged cation (high polarising power) paired with a large, highly charged anion (high polarisability).

   Fajans' rules state that covalent character increases when: (i) the cation is small and highly charged (high charge density, so high polarising power), (ii) the anion is large and highly charged (loosely held outer electrons, so high polarisability), and (iii) the cation does not have a noble-gas electronic configuration (e.g. pseudo-inert-gas d10 cations polarise more than expected). Option A combines both the cation and anion conditions correctly. The other options either invert the size requirement for the cation or choose a poorly polarisable small anion.

   Common mistake: Swapping which ion should be small/large, forgetting the cation must be small and highly charged while the anion must be large and highly charged.

4. Question 4 · medium · Ionic bonding, lattice enthalpy and Fajans rules

   Consider the following statements based on Fajans' rules. I. A cation with a pseudo-noble-gas (18-electron) configuration polarises an anion more than a same-sized cation with a noble-gas (8-electron) configuration. II. Increasing the anion's negative charge, keeping its size fixed, decreases the compound's covalent character. III. Decreasing the cation's size, keeping its charge fixed, increases the compound's covalent character. IV. Covalent character in an ionic compound generally correlates with a lower melting point compared to a purely ionic compound of similar lattice type. Which of the statements are correct?

   • A.I, II and III only
   • B.II, III and IV only
   • C.I, III and IV only
   • D.I, II, III and IV
   Show answer and explanation

   Answer: C. I, III and IV only

   Statements I, III and IV correctly apply Fajans' rules; statement II is false because a higher anion charge increases (not decreases) polarisability and hence covalent character.

   I is correct: pseudo-noble-gas (d10, 18-electron) cations like Ag+ or Zn2+ shield the nucleus poorly and polarise anions more than noble-gas cations of similar size, e.g. Na+. II is false: increasing the anion's charge (for fixed size) makes its electron cloud more loosely held by its own nucleus relative to charge, increasing polarisability and therefore increasing, not decreasing, covalent character. III is correct: a smaller cation of fixed charge has higher charge density, hence higher polarising power and greater covalent character (e.g. LiCl more covalent than KCl). IV is correct: greater covalent character typically means weaker long-range ionic lattice forces and more directional bonding, generally correlating with lower melting points (e.g. AgI melts and decomposes far more easily in terms of lattice-breaking tendency than AgF, and BeCl2 melts far below CaCl2).

   Common mistake: Believing that increasing an anion's charge decreases covalent character, when in fact it increases polarisability and covalent character.

5. Question 5 · medium · Ionic bonding, lattice enthalpy and Fajans rules

   Consider the following statements defining polarising power and polarisability. I. Polarising power is a property mainly associated with the cation, describing its ability to distort a neighbouring anion's electron cloud. II. Polarisability is a property mainly associated with the anion, describing how easily its electron cloud can be distorted. III. A cation's polarising power increases with increasing size at constant charge. IV. An anion's polarisability increases with increasing negative charge at constant size. Which statements are correct?

   • A.I and II only
   • B.I, II and IV
   • C.I, III and IV
   • D.II, III and IV
   Show answer and explanation

   Answer: B. I, II and IV

   Polarising power belongs to the cation and polarisability to the anion (I, II correct); polarising power decreases, not increases, with cation size at constant charge (III false); but polarisability does increase with anionic charge at constant size (IV correct).

   I is correct by definition: polarising power describes how strongly a cation can distort a nearby anion. II is correct by definition: polarisability describes how easily an anion's electron cloud is distorted. III is false: a larger cation at the same charge has lower charge density, so its polarising power decreases with increasing size, not increases. IV is correct: for anions of the same size, increasing the negative charge means more electrons are held by the same effective nuclear attraction, loosening the outer electron cloud and increasing polarisability (this is part of Fajans' rules, e.g. O2- is more polarisable than F- of similar size). Correct statements: I, II, IV.

   Common mistake: Believing polarising power increases with cation size, confusing it with the opposite (correct) trend of decrease with size.

6. Question 6 · medium · Ionic bonding, lattice enthalpy and Fajans rules

   Consider the following statements about the consequences of increased covalent character in a predominantly ionic compound (per Fajans' rules). I. Melting and boiling points tend to be lower than for a comparable, more purely ionic compound. II. Solubility in polar solvents like water tends to increase, while solubility in non-polar organic solvents tends to decrease. III. The compound is more likely to show colour due to easier promotion/distortion of electron density (as seen in going from AgF to AgI). IV. Electrical conductivity in the molten state tends to be lower than that of a purely ionic melt, since fewer free ions are available. Which statements are correct?

   • A.I and II only
   • B.I, III and IV only
   • C.II, III and IV only
   • D.I, II, III and IV
   Show answer and explanation

   Answer: B. I, III and IV only

   Increased covalent character generally lowers melting/boiling points (I), can produce visible colour from easier electron cloud distortion (III), and reduces the free-ion population and hence conductivity in the melt (IV); however, it decreases water solubility and increases organic-solvent solubility, so statement II (which claims the opposite pattern) is false.

   I is a standard Fajans consequence: greater covalent character weakens the 3-D ionic lattice, generally lowering melting and boiling points relative to a more purely ionic analogue (e.g. AgI vs AgF, BeCl2 vs CaCl2). II is false as stated: covalent compounds are typically less soluble in polar solvents like water and more soluble in non-polar organic solvents ('like dissolves like'), the reverse of what the statement claims. III is correct: the classic AgF (colourless) to AgI (yellow) progression reflects increasing covalent character, since easier distortion of the anion's electron cloud lowers the energy gap for charge-transfer-type electronic transitions, producing colour. IV is correct: in the molten state, a more covalent compound has fewer fully dissociated free ions available to carry current, so its electrical conductivity tends to be lower than that of a purely ionic melt. Correct statements: I, III, IV.

   Common mistake: Assuming increased covalent character increases water solubility, when it actually favours solubility in non-polar solvents instead.

7. Question 7 · medium · Ionic bonding, lattice enthalpy and Fajans rules

   A student states: "A smaller anion is closer to the cation, so the cation should be able to pull its electron cloud more strongly. Therefore smaller anions should be more polarisable than larger ones." Evaluate this claim.

   • A.The claim is correct: proximity to the cation is the only factor that determines polarisability
   • B.The claim is incorrect: what matters for polarisability is how loosely the anion's own nucleus holds its outer electrons; larger anions have more diffuse, more shielded electron clouds that are intrinsically easier to distort, so larger anions are more polarisable despite being farther from the cation
   • C.The claim is correct only for anions with a single negative charge, and incorrect for multiply charged anions
   • D.The claim is incorrect, but only because closer proximity actually makes an anion less polarisable due to repulsion between electron clouds
   Show answer and explanation

   Answer: B. The claim is incorrect: what matters for polarisability is how loosely the anion's own nucleus holds its outer electrons; larger anions have more diffuse, more shielded electron clouds that are intrinsically easier to distort, so larger anions are more polarisable despite being farther from the cation

   Polarisability is primarily an intrinsic property of how loosely an anion holds its own outer electrons, which increases with anion size (more shells, more shielding, more diffuse electron cloud) — larger anions are more polarisable even though they sit farther from the cation.

   The student's reasoning conflates cation-anion distance with the anion's intrinsic electronic 'looseness.' Polarisability is fundamentally about how easily the anion's own electron cloud can be distorted, which is governed mainly by how strongly the anion's own nucleus holds onto its valence electrons. Larger anions have their valence electrons in higher shells, farther from their own nucleus and more shielded by inner electrons, so these electrons are inherently more loosely held and easier to distort — regardless of the cation's proximity. This is why polarisability increases down a group of anions (F- < Cl- < Br- < I-) even though smaller anions do sit closer to a given cation. The claimed mechanism (closeness to cation increasing polarisability) is not how polarisability is defined or determined.

   Common mistake: Believing that a smaller anion, being closer to the cation, must be more polarisable, instead of recognising polarisability as an intrinsic property tied to anion size.

8. Question 8 · easy · Lewis structures, octet and formal charge

   How many lone pairs of electrons are present on the central nitrogen atom in the Lewis structure of NH3?

   • A.2
   • B.1
   • C.0
   • D.3
   Show answer and explanation

   Answer: B. 1

   Nitrogen has 5 valence electrons; 3 are used to form 3 N-H bonds, leaving 2 electrons (1 lone pair) on N.

   N contributes 5 valence electrons. Each N-H single bond uses 1 electron from N (the other from H). Three N-H bonds use 3 of N's electrons, leaving 5 - 3 = 2 electrons, i.e., exactly one lone pair, on the central nitrogen atom. This lone pair is responsible for NH3's pyramidal shape and its Lewis base character.

   Common mistake: Miscounting bonding electrons as lone pairs or vice versa

Questions about Chemical Bonding and Molecular Structure for NEET

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