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ChemistryMediumClass 11

VSEPR — Bond angles in molecules

Chemical Bonding

15

JEE Qs

8%

Hard

75

min

Master determining the steric number and lone pairs to establish the basic geometry, then systematically apply the repulsion order and electronegativity effects to precisely compare bond angles.

🧮 Key Formulas

Repulsion Order: Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bond Pair (LP-BP) > Bond Pair-Bond Pair (BP-BP)

✅ Key Points for JEE

  • 1VSEPR theory predicts molecular geometry and bond angles by minimizing electron pair repulsions (both bonding and non-bonding) around the central atom.
  • 2Lone pairs occupy more space than bonding pairs, leading to greater repulsion and compression of adjacent bond angles compared to ideal geometries.
  • 3The electronegativity of the central atom and surrounding atoms significantly influences bond angles: higher electronegativity of surrounding atoms reduces bond angles, while higher electronegativity of the central atom increases them (in specific cases, like comparing NH₃ and NF₃).
  • 4Multiple bonds are treated as a single 'super pair' for determining the basic geometry, but they exert greater repulsion than single bonds, affecting specific bond angles.
  • 5Compare bond angles by first determining hybridization and ideal geometry, then account for lone pair repulsions, and finally consider electronegativity effects and steric factors if necessary.

⚠️ Common Mistakes

  • Incorrectly determining the number of lone pairs on the central atom or miscalculating the steric number, leading to an incorrect predicted geometry.
  • Forgetting to apply the repulsion hierarchy (LP-LP > LP-BP > BP-BP) when comparing molecules with different numbers of lone pairs.
  • Not considering the effect of electronegativity of the central atom or surrounding atoms on the fine-tuning of bond angles.
  • Assuming ideal bond angles (e.g., 109.5° for sp³-hybridized atoms) even when lone pairs are present or surrounding atoms vary.

📝 Practice Questions

See all

Q75.Total number of non bonded electrons present in NO2− ion based on Lewis theory is 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025·NumericalEasy

Q71.The number of molecules/ions that show linear geometry among the following is ________ SO2, BeCl2, CO2, N−3 , NO2, F2O, XeF2, NO+2 , I−3 , O3

2025·NumericalMedium

Q69.Arrange the following compounds in increasing order of their dipole moment : HBr, H2 S, NF3 and CHCl3 (1) H2 S < HBr < NF3 < CHCl3 (2) NF3 < HBr < H2 S < CHCl3 (3) HBr < H2 S < NF3 < CHCl3 (4) CHCl3 < NF3 < HBr < H2 S

2025·MCQMedium

Q64.Which of the following statement is true with respect to H2O, NH3 and CH4 ? A. The central atoms of all the molecules are sp3 hybridized. B. The H −O −H, H −N −H and H −C −H angles in the above molecules are 104.5∘, 107.5∘ and 109.5∘ , respectively. C. The increasing order of dipole moment is CH4 < NH3 < H2O. D. Both H2O and NH3 are Lewis acids and CH4 is a Lewis base. E. A solution of NH3 in H2O is basic. In this solution NH3 and H2O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below: (1) A, B and C Only (2) A, D and E Only (3) C, D and E Only (4) A, B, C and E Only

2025·MCQMedium

Q70.Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z -direction] ? A. 2pz and 2px B. 2 s and 2px C. 3 dxy 2025 (24 Jan Shift 1) JEE Main Previous Year Paper and 3 dx2−y2 D. 2 s and 2pz E. 2pz and 3dx2 −y2 Choose the correct answer from the options given below: (1) A and B Only (2) D Only (3) E Only (4) C and D Only

2025·MCQMedium

Q53.Given below are two statements : Statement (I): Experimentally determined oxygen-oxygen bond lengths in the O3 are found to be same and the bond length is greater than that of a O = O (double bond) but less than 2025 (24 Jan Shift 2) JEE Main Previous Year Paper that of a single (O −O) bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O = O) but more than that of a single bond (O −O). In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true

2025·Assertion ReasoningMedium

NCERT Chapters

  • Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure