Answer key (B) & worked solution
Correct answer: Option B
Chemical Bonding and Molecular Structure carries 9.2 of 100 Chemistry marks in NEET by past-paper share: 65 of the 707 tagged NEET Chemistry PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 1 of 26 Chemistry chapters.
How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Chemistry PYQs in the bank (65 of 707), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.
16 facts the paper keeps asking, each with how it is asked. Copy it onto one page of your notebook: that page is this chapter's handbook.
Atoms bond by transferring or sharing valence electrons to reach eight in the valence shell; is stable with a duplet.
Asked as: Which statement about the Kössel–Lewis electronic theory of bonding is correct?
Incomplete octet (, , , , ), odd-electron (, ), expanded octet (, , ); obeys the octet.
Asked as: Which molecule does NOT obey the octet rule / which has an expanded octet?
In the central O is and the singly bonded end O is .
Asked as: Formal charge on the central O of ; the most stable Lewis structure has the smallest formal charges.
Stability of an ionic compound is measured by its lattice enthalpy, not by octet formation; ionisation is always endothermic, electron gain can be either.
Asked as: Why does form even though is positive?
Covalent radius = half the distance between two bonded like atoms; van der Waals radius = half the distance between like atoms of separate molecules (: vs ).
Asked as: Compare the covalent and van der Waals radius of , or which one is larger.
, , ; for polyatomics such as , use the average bond enthalpy.
Asked as: Arrange the bonds by strength; why the two bonds of need different energies to break.
Isoelectronic species have equal bond order: and ; , , . Higher bond order means higher bond enthalpy and shorter bond length.
Asked as: Pick the pair with the same bond order / arrange by bond length.
The canonical forms have no real existence, and the hybrid is lower in energy than any single form; , and have equal, averaged bonds.
Asked as: Which statement about resonance is incorrect (e.g. "the molecule oscillates between forms")?
, , , and have zero dipole; . because the lone-pair dipole opposes the bond dipoles.
Asked as: Which has zero dipole moment, and why is ?
Covalent character rises with a smaller cation, a larger anion and a higher cation charge; for cations of the same size and charge, an (transition-metal) cation polarises more than one with a noble-gas configuration.
Asked as: Arrange halides/salts by increasing covalent character.
A lone pair sits equatorial in (see-saw), and ( lp) is T-shaped.
Asked as: Why is the bond angle less than (both from ); shape of or .
A bond (axial overlap) is stronger than a bond (sidewise overlap); a bond forms only in addition to .
Asked as: Count the and bonds in or ; which overlap cannot form a bond.
( s, , ), (, ), ( s, , ). The number of hybrids equals the number of orbitals mixed; promotion is not essential, and filled orbitals can take part.
Asked as: Hybridisation of each C in ; which statement about hybridisation is false.
is (trigonal bipyramidal), and its axial bonds are longer and weaker than its equatorial bonds. is (octahedral), is (square pyramidal), and and are (square planar).
Asked as: Why axial bonds are longer; match each species to its hybridisation and shape.
and have bond order and do not exist; has bond order and is diamagnetic; 's double bond is both ; has bond order and is paramagnetic with unpaired .
Asked as: Bond order and magnetic nature of , , , or ; which species does not exist.
Hydrogen bonds form only with , or . They are intermolecular (, water, alcohol) or intramolecular (o-nitrophenol), and are strongest in the solid state and weakest in the gas.
Asked as: Which compound shows intramolecular H-bonding / why is associated.
The MO energy order switches between N₂ and O₂. For B₂, C₂ and N₂, π2p lies below σ2pz. For O₂ and F₂, σ2pz lies below π2p. Using the wrong order gives the wrong magnetic nature (e.g. B₂, C₂); the bond order comes out the same either way.
8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.
Correct answer: Option B
Correct answer: Option B
Correct answer: Option B
Correct answer: Option D
Correct answer: Option D
Correct answer: Option C
Correct answer: Option B
Correct answer: Option C
The 12 busiest of 20 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.
| Topic | PYQs | Practice questions |
|---|---|---|
| VSEPR Theory | 18 | 119 |
| Molecular Orbital Theory | 10 | 103 |
| Hybridisation | 6 | 115 |
| Dipole Moment of a Bond | 4 | 56 |
| Valence Bond Theory | 4 | 29 |
| Electrovalent or Ionic Bond | 4 | 25 |
| Kossel and Lewis Approach | 3 | 18 |
| Covalent Bond | 3 | 17 |
| Metallic Bond | 3 | 3 |
| Bond Parameters | 2 | 25 |
| Intermolecular Forces | 2 | 7 |
| Hydrogen Bonding | 1 | 52 |
Unknown is not zero. Until you answer at least 3 graded questions in Chemical Bonding and Molecular Structure, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.
Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.
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