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ChemistryMediumClass 11

VSEPR Theory ā€” Shapes of molecules

Chemical Bonding

15

JEE Qs

8%

Hard

75

min

Master the accurate calculation of steric number and the repulsion hierarchy to swiftly predict molecular shapes and approximate bond angles for various compounds and polyatomic ions.

šŸ§® Key Formulas

Steric Number (SN) = Number of (sigma bonds + lone pairs) around the central atom
SN = (1/2) * [V + M - C + A]
V = number of valence electrons of the central atom
M = number of monovalent atoms attached to the central atom (e.g., H, F, Cl, Br, I)
C = charge of the cation (if any)
A = charge of the anion (if any)

āœ… Key Points for JEE

  • 1The steric number (SN) around the central atom dictates the electron domain geometry (or electron pair geometry), which is the arrangement of all electron pairs (bonding and non-bonding).
  • 2Molecular geometry (the actual shape of the molecule) is determined by the electron domain geometry and the specific distribution of bond pairs and lone pairs. Lone pairs are 'invisible' in the molecular shape.
  • 3The order of repulsion strength is Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bond Pair (LP-BP) > Bond Pair-Bond Pair (BP-BP); this repulsion minimizes energy by influencing bond angles and distorting ideal geometries.
  • 4Multiple bonds (double or triple bonds) are treated as a single 'super' electron domain for the purpose of calculating steric number, but they are bulkier than single bonds and can slightly affect bond angles.
  • 5Lone pairs occupy more space than bond pairs; in geometries like trigonal bipyramidal, lone pairs preferentially occupy equatorial positions to minimize 90Ā° repulsions.

āš ļø Common Mistakes

  • āœ•Incorrectly calculating the number of lone pairs or the steric number, often by miscounting valence electrons or misidentifying the central atom.
  • āœ•Confusing electron domain geometry with molecular geometry, especially when lone pairs are present.
  • āœ•Failing to apply the repulsion hierarchy (LP-LP > LP-BP > BP-BP) to correctly predict distorted bond angles from ideal geometries (e.g., for H2O, NH3).
  • āœ•Ignoring the contribution of charge on polyatomic ions when calculating the steric number.

šŸ“ Practice Questions

See all

Q75.Total number of non bonded electrons present in NO2āˆ’ ion based on Lewis theory is 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025Ā·NumericalEasy

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

2025Ā·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

2025Ā·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 āˆ’O āˆ’H, H āˆ’N āˆ’H and H āˆ’C āˆ’H angles in the above molecules are 104.5āˆ˜, 107.5āˆ˜ and 109.5āˆ˜ , 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

2025Ā·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 dx2āˆ’y2 D. 2 s and 2pz E. 2pz and 3dx2 āˆ’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

2025Ā·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 āˆ’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 āˆ’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

2025Ā·Assertion ReasoningMedium

NCERT Chapters

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