Hybridization — sp, sp², sp³, sp³d, sp³d²
Chemical Bonding
15
JEE Qs
8%
Hard
75
min
Always draw the correct Lewis structure first to accurately count sigma bonds and lone pairs on the central atom, which is crucial for determining hybridization and geometry.
🧮 Key Formulas
✅ Key Points for JEE
- 1Hybridization is a hypothetical concept of mixing atomic orbitals (s, p, d) of comparable energy to form a new set of equivalent hybrid orbitals that minimize electron-pair repulsion and maximize bond strength.
- 2The type of hybridization on the central atom is determined by its steric number, which is the sum of sigma bonds and lone pairs; pi bonds do not contribute to hybridization.
- 3Each type of hybridization (sp, sp², sp³, sp³d, sp³d²) corresponds to a specific electron-pair geometry and ideal bond angles (e.g., sp = linear, 180°; sp³ = tetrahedral, 109.5°).
- 4Lone pair-lone pair, lone pair-bond pair, and bond pair-bond pair repulsions vary, with lone pairs exerting greater repulsion and causing deviations from ideal bond angles in actual molecular geometries.
- 5Electronegativity differences and size of surrounding atoms can subtly influence bond angles even for molecules with the same hybridization and number of lone pairs.
⚠️ Common Mistakes
- ✕Including pi bonds in the calculation of the steric number for hybridization, leading to an incorrect hybridization state.
- ✕Failing to correctly identify the number of lone pairs or sigma bonds on the central atom from the Lewis structure, especially for molecules with multiple bonds or resonance structures.
- ✕Confusing electron-pair geometry (determined by hybridization) with molecular geometry (which is further influenced by the presence and position of lone pairs according to VSEPR theory).
- ✕Incorrectly assuming that all atoms in a molecule, particularly terminal atoms, undergo hybridization or trying to hybridize atoms that don't need to be (e.g., hydrogen).
📝 Practice Questions
See allQ75.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
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
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
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
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
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
NCERT Chapters
- Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure