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ChemistryMediumClass 11

Hybridization + VSEPR + Shapes

Chemical Bonding

72

JEE Qs

8%

Hard

75

min

Master the step-by-step process from Lewis structure to hybridization, electron geometry, and then molecular geometry by accounting for lone pair repulsions.

🧮 Key Formulas

Steric Number (SN) = (Number of sigma bonds around central atom) + (Number of lone pairs on central atom)
Hybridization based on Steric Number:
SN=2: sp (linear electron geometry)
SN=3: sp^2 (trigonal planar electron geometry)
SN=4: sp^3 (tetrahedral electron geometry)
SN=5: sp^3d (trigonal bipyramidal electron geometry)
SN=6: sp^3d^2 (octahedral electron geometry)
SN=7: sp^3d^3 (pentagonal bipyramidal electron geometry)

āœ… Key Points for JEE

  • 1Hybridization determines the electron domain geometry (arrangement of all electron pairs — bonding and lone pairs) around the central atom, establishing the fundamental spatial arrangement.
  • 2VSEPR theory refines the molecular geometry (arrangement of only atoms) by considering the differential repulsive forces between electron pairs, leading to distortions from ideal bond angles.
  • 3The order of repulsive interactions is Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bond Pair (LP-BP) > Bond Pair-Bond Pair (BP-BP); this dictates the specific molecular shape and bond angle compressions.
  • 4A systematic approach is crucial: 1. Draw Lewis structure. 2. Calculate Steric Number to find hybridization and electron geometry. 3. Count lone pairs on the central atom to determine the actual molecular geometry and predict bond angle distortions.

āš ļø Common Mistakes

  • āœ•Confusing electron geometry (based on hybridization/steric number) with molecular geometry (the arrangement of only atoms, considering lone pair repulsions).
  • āœ•Incorrectly calculating the number of lone pairs or steric number, especially for polyatomic ions or molecules with resonance.
  • āœ•Ignoring the effect of lone pairs on bond angles, assuming ideal angles even when lone pairs are present.

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 104.5∘, 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

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