75 free MCQs on Chemical Bonding and Molecular Structure with worked answers and explanations. Understand how atoms join to form molecules. Covers ionic and covalent bonding, VSEPR theory, hybridization (sp, sp², sp³), molecular orbital theory, bond polarity, and intermolecular forces like hydrogen bonding.
Take the timed Chemical Bonding and Molecular Structure chapterwise test →Below are 75 practice questions on Chemical Bonding and Molecular Structure, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Chemical Bonding and Molecular Structure notes.
VSEPR theory predicts molecular shape from the number of bonding and lone electron pairs around the central atom.
Q1.
Which of the following substances is expected to have the highest melting point?
Answer: A. NaCl
Why: NaCl is an ionic solid held by strong electrostatic forces, so it has a much higher melting point than the molecular solids CO<sub>2</sub>, I<sub>2</sub> and CH<sub>4</sub>.
Q2.
Which of the following molecules is non-polar despite containing polar bonds?
Answer: B. BF<sub>3</sub>
Why: BF<sub>3</sub> is trigonal planar and symmetric, so the three bond dipoles cancel, giving a net dipole moment of zero.
Q3.
The number of lone pairs of electrons present on the central oxygen atom in a water molecule is:
Answer: C. 2
Why: Oxygen in H<sub>2</sub>O forms two bond pairs and retains two lone pairs, giving a bent shape.
Q4.
A molecule in which the central atom is sp<sup>2</sup> hybridised with no lone pairs has a bond angle of:
Answer: D. 120 degrees
Why: Three sp<sup>2</sup> hybrid orbitals point to the corners of an equilateral triangle, giving a trigonal planar geometry with 120 degree bond angles.
Q5.
Among carbon-carbon single, double and triple bonds, the shortest and strongest is the:
Answer: A. triple bond
Why: A triple bond has the greatest number of shared electron pairs, so it is the shortest and strongest, while the single bond is the longest and weakest.
Q6.
A covalent bond is formed by:
Answer: A. Sharing of electrons between atoms
Why: Covalent bonds form when two atoms share one or more pairs of electrons.
Q7.
An ionic bond is formed by:
Answer: A. Transfer of electrons from metal to non-metal
Why: Ionic bonds result from the electrostatic attraction between oppositely charged ions formed by electron transfer.
Q8.
Lewis dot structure represents:
Answer: A. Valence electrons as dots around the symbol
Why: Lewis structures use dots to represent valence electrons around the chemical symbol.
Q9.
How many bonding pairs are in a double bond?
Answer: A. Two pairs (4 electrons)
Why: A double bond consists of two shared electron pairs (4 electrons total).
Q10.
The VSEPR theory is used to predict:
Answer: A. Shape of molecules
Why: VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry based on electron pair repulsion.
Q11.
The shape of a water molecule is:
Answer: A. Bent (V-shape)
Why: Water has 2 bonding pairs and 2 lone pairs on oxygen; repulsion from lone pairs bends the molecule to ~104.5°.
Q12.
The shape of ammonia (NH<sub>3</sub>) is:
Answer: A. Trigonal pyramidal
Why: NH<sub>3</sub> has 3 bonding pairs and 1 lone pair; this gives a trigonal pyramidal shape.
Q13.
Which molecule has a linear shape?
Answer: A. CO<sub>2</sub>
Why: CO<sub>2</sub> has two double bonds and no lone pairs on carbon, giving a linear geometry.
Q14.
The shape of methane (CH<sub>4</sub>) is:
Answer: A. Tetrahedral
Why: CH<sub>4</sub> has 4 bonding pairs and no lone pairs; this gives a regular tetrahedral shape with 109.5° bond angles.
Q15.
A sigma bond is formed by:
Answer: A. Head-on (axial) overlap of orbitals
Why: Sigma bonds form by head-on overlap of atomic orbitals along the internuclear axis.
Q16.
A pi bond is formed by:
Answer: A. Lateral (sideways) overlap of p orbitals
Why: Pi bonds are formed by lateral overlap of parallel p orbitals, above and below the internuclear axis.
Q17.
Which type of bond is strongest?
Answer: A. Sigma bond
Why: Sigma bonds have maximum orbital overlap (head-on) and are the strongest covalent bonds.
Q18.
Electronegativity difference greater than 1.7 generally indicates:
Answer: A. Ionic bond
Why: A large electronegativity difference (>1.7) leads to significant charge transfer, producing an ionic bond.
Q19.
In a coordinate (dative) bond, both electrons are donated by:
Answer: A. One atom (the donor/Lewis base)
Why: In a coordinate/dative bond, one atom (Lewis base/donor) donates both electrons to a Lewis acid acceptor.
Q20.
Which of the following is a Lewis acid?
Answer: A. BF<sub>3</sub>
Why: BF<sub>3</sub> (boron trifluoride) is electron-deficient and accepts electron pairs, making it a Lewis acid.
Q21.
The bond angle in methane is:
Answer: A. 109.5°
Why: Methane has perfect tetrahedral geometry with bond angles of 109.5°.
Q22.
Which gas has a triple bond?
Answer: A. N<sub>2</sub>
Why: N<sub>2</sub> (nitrogen gas) has a triple bond (one sigma + two pi bonds) making it very stable.
Q23.
Hydrogen bonding is possible only when hydrogen is bonded to:
Answer: A. Highly electronegative atom with lone pair (F, O, N)
Why: Hydrogen bonding requires H to be attached to a highly electronegative atom (F, O, or N) that has lone pairs.
Q24.
Which compound has the strongest hydrogen bonds?
Answer: A. HF
Why: HF has the strongest hydrogen bonds due to fluorine's extreme electronegativity, though HF forms only one H-bond per molecule.
Q25.
The octet rule states that atoms tend to:
Answer: A. Have 8 electrons in their valence shell
Why: The octet rule says atoms prefer 8 valence electrons (like noble gases) by forming bonds.
Q26.
The hybridisation of the sulfur atom in the SF<sub>4</sub> molecule is:
Answer: B. sp<sup>3</sup>d
Why: SF<sub>4</sub> has 4 bond pairs and 1 lone pair (5 electron domains), requiring sp<sup>3</sup>d hybridisation and giving a see-saw shape.
Q27.
According to VSEPR theory, the molecular shape of ClF<sub>3</sub> is:
Answer: C. T-shaped
Why: ClF<sub>3</sub> has 3 bond pairs and 2 lone pairs in a trigonal bipyramidal arrangement; the lone pairs occupy equatorial positions, giving a T-shaped molecule.
Q28.
Using molecular orbital theory, the bond order of the O<sub>2</sub><sup>+</sup> ion is:
Answer: D. 2.5
Why: O<sub>2</sub><sup>+</sup> has 15 electrons (10 bonding, 5 antibonding), so bond order = (10 - 5)/2 = 2.5.
Q29.
A molecule of ethene, C<sub>2</sub>H<sub>4</sub>, contains how many sigma and pi bonds respectively?
Answer: A. 5 sigma and 1 pi
Why: Ethene has four C-H sigma bonds plus one C-C sigma bond (5 sigma) and one C-C pi bond, so 5 sigma and 1 pi.
Q30.
The correct decreasing order of bond angle among CH<sub>4</sub>, NH<sub>3</sub> and H<sub>2</sub>O is:
Answer: B. CH<sub>4</sub> > NH<sub>3</sub> > H<sub>2</sub>O
Why: Increasing lone pairs on the central atom compress the bond angle: CH<sub>4</sub> (109.5) > NH<sub>3</sub> (107) > H<sub>2</sub>O (104.5).
Q31.
The hybridisation of carbon in ethyne (acetylene) is:
Answer: A. sp
Why: In ethyne (C≡C), each carbon forms one sigma bond with H and one sigma bond with C, using sp hybridisation (linear, 180°).
Q32.
What is the hybridisation of nitrogen in ammonia?
Answer: A. sp<sup>3</sup>
Why: Nitrogen in NH<sub>3</sub> is sp<sup>3</sup> hybridised: three bonds and one lone pair in sp<sup>3</sup> orbitals.
Q33.
The hybridisation of BF<sub>3</sub> is:
Answer: A. sp<sup>2</sup>
Why: Boron in BF<sub>3</sub> has 3 bonding pairs and no lone pairs; sp<sup>2</sup> hybridisation gives trigonal planar geometry.
Q34.
PCl<sub>5</sub> has which hybridisation?
Answer: A. sp<sup>3</sup>d
Why: Phosphorus in PCl<sub>5</sub> uses 5 bonding pairs in sp<sup>3</sup>d hybridisation, giving trigonal bipyramidal geometry.
Q35.
SF<sub>6</sub> has which hybridisation?
Answer: A. sp<sup>3</sup>d<sup>2</sup>
Why: Sulfur in SF<sub>6</sub> has 6 bonding pairs and uses sp<sup>3</sup>d<sup>2</sup> hybridisation, giving octahedral geometry.
Q36.
According to VSEPR, lone pairs take up more space than bonding pairs because:
Answer: A. Lone pairs are only on one nucleus and spread more (more repulsion)
Why: Lone pairs are attracted to only one nucleus and occupy more angular space than bonding pairs shared between two nuclei.
Q37.
Bond order = (bonding electrons - antibonding electrons) / 2. The bond order of O<sub>2</sub> is:
Answer: A. 2
Why: O<sub>2</sub> has 8 bonding MO electrons and 4 antibonding MO electrons: bond order = (8-4)/2 = 2.
Q38.
The bond order of NO is:
Answer: A. 2.5
Why: NO has 8 bonding and 3 antibonding electrons: BO = (8-3)/2 = 2.5.
Q39.
Which molecule is paramagnetic according to MO theory?
Answer: A. O<sub>2</sub>
Why: MO theory predicts O<sub>2</sub> has two unpaired electrons in degenerate pi* orbitals, making it paramagnetic.
Q40.
Bent's rule states that:
Answer: A. Atoms preferentially direct more p character toward electronegative substituents
Why: Bent's rule: an atom uses more p character in orbitals directed toward more electronegative substituents, and more s character toward less electronegative groups.
Q41.
Why does water have a higher boiling point than H<sub>2</sub>S?
Answer: A. Water has strong hydrogen bonds; H<sub>2</sub>S only has weak van der Waals forces
Why: Water molecules form extensive hydrogen bonds; H<sub>2</sub>S cannot form hydrogen bonds because sulfur's electronegativity is too low.
Q42.
Which of the following is an exception to the octet rule by being electron-deficient?
Answer: A. BF<sub>3</sub>
Why: BF<sub>3</sub> has only 6 electrons around boron (three bonding pairs); it is electron deficient (incomplete octet).
Q43.
Resonance structures differ only in:
Answer: A. Position of electrons (bonds/lone pairs), not nuclei
Why: Resonance structures are drawn with the same atomic framework; only electron (bond) positions differ.
Q44.
The bond length order in a triple bond vs double bond vs single bond is:
Answer: A. Triple < double < single
Why: More shared electron pairs pull nuclei closer: triple bonds are shortest, single bonds are longest.
Q45.
The formal charge of an atom in a Lewis structure is:
Answer: A. Valence electrons - non-bonding electrons - (bonding electrons/2)
Why: Formal charge = (valence electrons) - (lone pair electrons) - (½ × bonding electrons).
Q46.
Fajan's rules predict that a compound has more covalent character when:
Answer: A. The cation is small and highly charged, and the anion is large and highly charged
Why: Fajan's rules: high charge density of the cation polarises the large anion, increasing covalent character.
Q47.
Which bond is more polar: H-F or H-Cl?
Answer: A. H-F (larger electronegativity difference)
Why: Fluorine is much more electronegative than chlorine, so H-F has a larger dipole moment and is more polar.
Q48.
The dipole moment of CO<sub>2</sub> is zero because:
Answer: A. Two equal C=O dipoles point in opposite directions and cancel
Why: CO<sub>2</sub> is linear; the two C=O bond dipoles are equal in magnitude but opposite in direction, giving zero net dipole.
Q49.
Which type of intermolecular force is present in all molecules?
Answer: A. London dispersion (van der Waals) forces
Why: London dispersion forces arise from instantaneous dipoles and are present in all molecules, polar or non-polar.
Q50.
The lattice energy of an ionic compound depends on:
Answer: A. Charges and sizes of ions (higher charge and smaller size = higher lattice energy)
Why: Lattice energy is proportional to (charge product)/(ion distance); small, highly charged ions give the highest lattice energies.
Q51.
The molecular shape of the triiodide ion, I<sub>3</sub><sup>-</sup>, as predicted by VSEPR theory is:
Answer: C. Linear
Why: The central iodine in I<sub>3</sub><sup>-</sup> has 2 bond pairs and 3 lone pairs; the lone pairs occupy the equatorial positions of a trigonal bipyramid, leaving a linear ion.
Q52.
Among the following species, which has the highest bond order?
Answer: D. O<sub>2</sub><sup>+</sup>
Why: Bond orders are O<sub>2</sub> = 2, O<sub>2</sub><sup>-</sup> = 1.5, O<sub>2</sub><sup>2-</sup> = 1 and O<sub>2</sub><sup>+</sup> = 2.5; removing an antibonding electron raises the bond order, so O<sub>2</sub><sup>+</sup> is highest.
Q53.
According to molecular orbital theory, the number of unpaired electrons present in a molecule of O<sub>2</sub> is:
Answer: A. 2
Why: In O<sub>2</sub>, the two highest electrons occupy the degenerate pi* antibonding orbitals singly, giving 2 unpaired electrons and explaining its paramagnetism.
Q54.
The formal charge on the central nitrogen atom in the ammonium ion, NH<sub>4</sub><sup>+</sup>, is:
Answer: C. +1
Why: Formal charge = 5 - 0 - (8/2) = 5 - 4 = +1; nitrogen forms four bonds with no lone pair, carrying the positive charge of the ion.
Q55.
The measured dipole moment of HCl is 1.03 D, while the value calculated for 100% ionic character is 6.12 D. The percent ionic character of the H-Cl bond is about:
Answer: B. 17%
Why: Percent ionic character = (observed/ionic) x 100 = (1.03/6.12) x 100, which is approximately 17%.
Q56.
In MO theory, bonding MOs are formed by:
Answer: A. Constructive interference of atomic wavefunctions
Why: Constructive interference of atomic orbitals lowers energy to give bonding MOs; destructive interference raises energy to give antibonding MOs.
Q57.
The HOMO of H<sub>2</sub> is the:
Answer: A. Sigma 1s bonding MO (filled)
Why: In H<sub>2</sub>, both electrons occupy the lowest sigma 1s bonding MO; that is both the HOMO and the only occupied MO.
Q58.
Why is the bond angle in water (104.5°) less than in NH<sub>3</sub> (107°)?
Answer: A. Water has two lone pairs vs one in NH<sub>3</sub>; greater lone pair repulsion compresses the angle more in water
Why: Two lone pairs in H<sub>2</sub>O cause greater repulsion than one lone pair in NH<sub>3</sub>, reducing the bond angle further from tetrahedral.
Q59.
Which of the following has the highest bond dissociation energy?
Answer: A. N≡N
Why: N≡N has a triple bond (945 kJ/mol), the highest among common diatomic molecules. F-F is surprisingly weak due to lone pair repulsion.
Q60.
Why is F-F bond energy (159 kJ/mol) unexpectedly low compared to Cl-Cl (243 kJ/mol)?
Answer: A. Lone pair-lone pair repulsion between the very close F atoms weakens the bond
Why: Fluorine atoms are tiny; adjacent lone pairs on the two F atoms experience intense repulsion, weakening the F-F bond.
Q61.
In the valence bond (VB) theory, a covalent bond forms when:
Answer: A. Atomic orbitals overlap so that electrons of opposite spin are shared between two atoms
Why: VB theory: a bond forms by overlap of two half-filled atomic orbitals with antiparallel (opposite) electron spins.
Q62.
Hyperconjugation involves delocalisation of electrons from:
Answer: A. C-H sigma bond into an adjacent empty or pi orbital
Why: Hyperconjugation (no-bond resonance) occurs when electrons in a C-H sigma bond are delocalised into an adjacent vacant or pi orbital.
Q63.
The Born-Haber cycle is used to calculate:
Answer: A. Lattice enthalpy of ionic compounds indirectly
Why: The Born-Haber cycle applies Hess's law to ionic formation, combining measurable enthalpies to calculate the lattice energy.
Q64.
According to VSEPR, in PCl<sub>5</sub> (trigonal bipyramidal), lone pairs prefer the equatorial position because:
Answer: A. Equatorial positions have only 2 nearby 90° interactions, vs 3 for axial positions
Why: Axial positions have 3 adjacent equatorial positions at 90° (more repulsion); equatorial positions have only 2 axial neighbours at 90°.
Q65.
The resonance energy of benzene is approximately 150 kJ/mol. This means:
Answer: A. Benzene is more stable than a hypothetical cyclohexatriene by 150 kJ/mol
Why: Resonance energy is the extra stabilisation relative to a single resonance structure; benzene's delocalised pi system stabilises it by ~150 kJ/mol.
Q66.
The MO electron configuration of N<sub>2</sub> is (sigma1s)2(sigma*1s)2(sigma2s)2(sigma*2s)2(pi2p)4(sigma2p)2. Its bond order is:
Answer: A. 3
Why: Bond order = (10 bonding - 4 antibonding)/2 = 6/2 = 3. N<sub>2</sub> has a triple bond.
Q67.
Walsh diagrams show how molecular orbital energies change with:
Answer: A. Bond angle (geometric changes)
Why: Walsh diagrams plot MO energies as a function of bond angle, predicting the preferred geometry of a molecule.
Q68.
In which molecule does back-bonding (pi donation from F to empty p of central atom) significantly occur?
Answer: A. BF<sub>3</sub>
Why: In BF<sub>3</sub>, fluorine's lone pairs donate into boron's empty 2p orbital (p-pi back-bonding); this is strongest with F due to good orbital size match.
Q69.
The dipole moment of NH<sub>3</sub> is greater than that of NF<sub>3</sub> because:
Answer: A. In NF<sub>3</sub>, the N-F bond dipoles partially cancel the lone pair direction, while in NH<sub>3</sub> the bond dipoles and lone pair add together
Why: In NH<sub>3</sub>, lone pair and bond dipoles point in the same direction (additive); in NF<sub>3</sub>, the electronegative F atoms reverse the bond dipoles, opposing the lone pair direction.
Q70.
Drago's rule (hard-soft concept) states that hard bases prefer:
Answer: A. Hard acids (ionic, high charge density, low polarisability)
Why: In HSAB theory, hard acid-hard base and soft acid-soft base combinations are preferred. Hard species are small, highly charged, and low in polarisability.
Q71.
The concept of equivalent and non-equivalent resonance structures relates to:
Answer: A. Whether all resonance forms have the same energy (equivalent) or different energies
Why: Equivalent resonance structures have identical energy (e.g., benzene); non-equivalent ones have different energies and contribute unequally to the hybrid.
Q72.
Isoelectronic species have the same number of electrons AND identical bond properties. Which pair is isoelectronic?
Answer: A. CO and N<sub>2</sub> (both have 14 electrons and triple bonds)
Why: CO has 14 electrons (6+8) and N<sub>2</sub> also has 14 electrons (7+7); both have triple bonds and are isoelectronic.
Q73.
In a double bond, one bond is sigma and one is pi. The pi bond is WEAKER than the sigma bond because:
Answer: A. Lateral p orbital overlap is less effective than head-on axial overlap
Why: Pi bonds form by sideways (lateral) overlap of p orbitals, which is less efficient than the direct head-on overlap of sigma bonds.
Q74.
The bond dissociation energy of Cl<sub>2</sub> > F<sub>2</sub> > Br<sub>2</sub> > I<sub>2</sub> (halogen series) because:
Answer: A. Cl<sub>2</sub> has optimal overlap of 3p-3p; F-F has lone pair repulsion; Br and I have weaker, more diffuse overlap
Why: F-F is weakened by lone pair repulsion in tiny F atoms; Br and I have larger, more diffuse orbitals with poorer overlap; Cl<sub>2</sub> hits the sweet spot.
Q75.
The van der Waals radius is larger than the covalent radius for the same atom. The ionic radius of a cation compared to the neutral atom is:
Answer: A. Smaller (loss of electrons reduces shielding, pulls remaining electrons in)
Why: Cations form by losing electrons; fewer electrons but same nuclear charge means greater attraction and smaller radius.