Chemical Bonding and Molecular Structure

Covers NCERT Class 11 Chemistry, Chapter "Chemical Bonding and Molecular Structure" — the conceptual backbone for organic mechanism, coordination chemistry and molecular shapes across the entire NEET syllabus.

Foundation Advanced High-Yield Revision
Advanced NTA favourite High weightage

⭐ NTA-favourite subtopics in this chapter

VSEPR shapes and bond angles (especially exceptions like NH₃ vs BF₃, SF₄, ClF₃), determining hybridisation from a Lewis structure, Molecular Orbital Theory bond order and magnetic behaviour (O₂, N₂, and their ions), types and relative strength of hydrogen bonding, and dipole moment comparisons across similar molecules.

Exam-oriented notes

NCERT-tagged theory, simplified

NCERT §4.1

Kössel–Lewis approach: atoms bond to achieve a stable noble-gas-like octet, either by complete transfer of electrons (ionic) or by sharing (covalent).

NCERT §4.2

Ionic bond formation is favoured by low ionisation enthalpy of the metal, high (more negative) electron gain enthalpy of the non-metal, and high lattice enthalpy of the resulting solid.

NCERT §4.3

Bond parameters: bond length decreases and bond enthalpy increases as bond order increases (single < double < triple). Bond angle is a direct clue to hybridisation and lone-pair count.

NCERT §4.4

VSEPR: electron pairs (bonding + lone) around a central atom arrange to minimise repulsion, in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. This single ordering explains almost every "why is the bond angle less than ideal" question.

NCERT §4.6

Hybridisation shortcut: count σ-bonds + lone pairs on the central atom. 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d².

NCERT §4.7–4.8

Molecular Orbital Theory: bond order = ½(N_bonding − N_antibonding). A molecule/ion is paramagnetic if it has unpaired electrons in its MO configuration (e.g. O₂), diamagnetic if all electrons are paired (e.g. N₂).

NCERT §4.9

Hydrogen bonding needs H attached to a small, highly electronegative atom (F, O, N). Intermolecular H-bonding raises boiling point (HF, H₂O anomalies); intramolecular H-bonding (e.g. o-nitrophenol) instead lowers it compared to the para isomer.

Applied

Dipole moment (μ = q×d, in Debye) is a vector — symmetric molecules like BF₃ and CO₂ have μ = 0 despite polar bonds, because bond dipoles cancel; NH₃ retains a net dipole because its lone pair breaks that symmetry.

Printable

Formula & shape sheet

Chemical Bonding — key relations

Bond order (MOT)
B.O. = ½(N_b − N_a)
Dipole moment
μ = q × d (Debye, D)
Formal charge
FC = V − N − B/2
% ionic character (Fajans, qualitative)
↑ with charge, ↓ with cation size
HybridisationShapeIdeal bond angleExample
spLinear180°BeCl₂, CO₂
sp²Trigonal planar120°BF₃, SO₃
sp³Tetrahedral109.5°CH₄
sp³ (1 lone pair)Pyramidal~107°NH₃
sp³ (2 lone pairs)Bent / V-shaped~104.5°H₂O
sp³dTrigonal bipyramidal90° & 120°PCl₅
sp³d²Octahedral90°SF₆
PYQ-style practice

Practice set, NEET pattern

Original questions modelled on recurring NEET question types — not verbatim reproductions of any official paper.

Q1. The bond angle in NH₃ is less than the ideal tetrahedral angle because:
  • N is less electronegative than expected
  • Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion
  • NH₃ has sp² hybridisation
  • Hydrogen bonding distorts the molecule
Show solution
The single lone pair on N compresses the H–N–H angle from 109.5° to ~107° via stronger lone pair–bond pair repulsion. Answer: B
Q2. Which of the following is paramagnetic?
  • N₂
  • O₂
  • CO
  • F₂ (all paired)
Show solution
O₂'s MO configuration has two unpaired electrons in π* orbitals, making it paramagnetic — a landmark success of MOT over simple VBT. Answer: B
Q3. BF₃ has zero net dipole moment even though B–F bonds are highly polar, because:
  • Boron has no lone pair to break symmetry, and the three B–F dipoles cancel by symmetry
  • Fluorine is not electronegative enough
  • BF₃ is a linear molecule
  • B–F bonds are actually non-polar
Show solution
Trigonal planar (sp²) symmetry with no lone pair on B means the three equal bond dipoles vectorially cancel. Answer: A
Q4. The hybridisation of the central atom in SF₄ is:
  • sp³
  • sp³d
  • sp³d²
  • sp²
Show solution
S has 4 bond pairs + 1 lone pair = 5 electron domains ⇒ sp³d (see-saw shape). Answer: B
Q5. Ortho-nitrophenol has a lower boiling point than para-nitrophenol mainly because:
  • Ortho isomer has intramolecular H-bonding, reducing intermolecular association
  • Ortho isomer is non-polar
  • Para isomer has a smaller molecular mass
  • Ortho isomer cannot form H-bonds at all
Show solution
Intramolecular ("chelate-like") H-bonding in the ortho isomer uses up the H-bonding capacity internally, so fewer intermolecular H-bonds form, lowering the boiling point relative to the para isomer, which H-bonds intermolecularly. Answer: A
Q6. Bond order of the O₂⁻ (superoxide) ion is:
  • 1
  • 1.5
  • 2
  • 2.5
Show solution
O₂ has bond order 2; adding one electron to an antibonding π* orbital reduces it by 0.5 ⇒ 1.5. Answer: B
Daily Practice Problems

DPP — Chemical Bonding (15 questions)

Scaled from direct NCERT application (Q1–Q6) to mixed NEET-level difficulty (Q7–Q15).

1.State the octet rule and give one exception.
2.Write the Lewis structure of CO₂ and identify its hybridisation.
3.Define bond enthalpy and bond length; state how they trend with bond order.
4.Predict the shape of PCl₅ using VSEPR.
5.Give one example each of intermolecular and intramolecular hydrogen bonding.
6.Calculate the formal charge on each atom in the nitrate ion, NO₃⁻.
7.Explain why the H–O–H angle in water (104.5°) is smaller than the H–N–H angle in ammonia (107°).
8.Determine the hybridisation and shape of XeF₄.
9.Using MOT, find the bond order and magnetic nature of the C₂ molecule.
10.Arrange HF, HCl, HBr, HI in decreasing order of boiling point and explain.
11.Why does CO₂ have zero dipole moment while SO₂ does not, despite both being triatomic?
12.Which has a greater bond angle: ClF₃ or NH₃? Justify with lone pair count.
13.Rank NO, NO⁺, NO⁻ in order of increasing bond length.
14.Why is the ionic character of a bond expected to increase with increasing electronegativity difference?
15.Assertion: SF₆ exists but SH₆ does not. Reason: Sulfur cannot expand its octet using 3d orbitals with a small, highly electronegative atom like F but can with H. Judge the pair.
Show answer key
1. Atoms are stable with 8 valence e⁻; exception e.g. BF₃ (6 e⁻ around B)  |  2. O=C=O, sp  |  3. Both increase together with bond order (except length decreases, enthalpy increases — restate: length↓, enthalpy↑ as order↑)  |  4. Trigonal bipyramidal, sp³d  |  5. Intermolecular: HF···HF; intramolecular: o-nitrophenol  |  6. N: +1, two O: 0 each, one O: −1 (resonance-averaged)  |  7. Water has 2 lone pairs vs ammonia's 1, so greater lp–lp/lp–bp repulsion compresses the angle further  |  8. sp³d², square planar  |  9. Bond order 2, diamagnetic  |  10. HF > HI > HBr > HCl (HF anomalously high due to strong H-bonding)  |  11. CO₂ is linear (dipoles cancel); SO₂ is bent due to a lone pair on S (dipoles don't cancel)  |  12. NH₃ (1 lone pair) > ClF₃ (2 lone pairs compress it further)  |  13. NO⁺ < NO < NO⁻ (bond order 3 > 2.5 > 2)  |  14. Greater electronegativity difference pulls the shared pair almost fully to one atom, approaching full electron transfer  |  15. Assertion true; reason's explanation of "why" is the accepted one — correct pairing, answer A-type
Story mode

A memory-only mythology narrative for bonding types

Purely a recall device — not a reinterpretation of the source stories.

📖 The Ardhanarishvara Union — covalent bonding

Ardhanarishvara, the composite form of Shiva and Parvati, is one whole made of two equal halves. Picture a non-polar covalent bond (like H–H or Cl–Cl) the same way: two identical atoms merging into one shared electron cloud, contributing and holding the pair with perfectly equal claim. Where the two halves differ in "strength" — one atom pulling the shared pair closer — you get a polar covalent bond instead: still shared, but no longer symmetric.

📖 Indra's court — ionic bonding as tribute, not theft

Think of a metal atom as a minor chieftain who finds one lone electron a burden to guard, and a non-metal atom as a well-fortified court that has room for exactly one more attendant to complete its guard of eight. The chieftain hands the electron over willingly — it's a transfer both sides are better off for, not a struggle. That willing, complete transfer, followed by the two now-charged ions holding together purely by electrostatic attraction, is the ionic bond: NCERT's own language of "complete transfer of an electron" mapped onto something easier to hold in memory.

🧠 Quick VSEPR repulsion order

"Loners push hardest" — lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Every "smaller than ideal angle" question in this chapter reduces to counting lone pairs and applying this one line.