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ChemistryMediumClass 11

Resonance ā€” Structures, stability

Chemical Bonding

15

JEE Qs

8%

Hard

75

min

Master the rules for drawing valid resonance structures and meticulously apply the stability criteria hierarchy to correctly compare their contributions and the overall stability of the resonance hybrid.

šŸ§® Key Formulas

Formal Charge (FC) = V - NB - (1/2)B
Where V = valence electrons of the atom, NB = number of non-bonding electrons (lone pair electrons), B = number of bonding electrons (electrons in shared pairs)

āœ… Key Points for JEE

  • 1Resonance structures (canonical forms) are hypothetical Lewis structures; the true structure is a resonance hybrid, which is a weighted average of all contributing canonical forms.
  • 2Only pi (Ļ€) electrons and lone pair electrons can delocalize; the sigma (Ļƒ) bond framework and atom connectivity must remain unchanged in all resonance structures.
  • 3All valid resonance structures for a given molecule must have the same total number of valence electrons and obey the octet rule for second-period elements where possible.
  • 4Stability of resonance structures follows this hierarchy: 1. Structures with complete octets for all atoms (especially 2nd period) > 2. Structures with less charge separation > 3. Structures with negative charge on more electronegative atoms and positive charge on less electronegative atoms > 4. Structures with a greater number of covalent bonds.
  • 5The most stable resonance structures contribute most to the resonance hybrid, and greater delocalization of electrons (leading to more stable significant contributors) results in greater resonance energy and enhanced stability of the molecule.

āš ļø Common Mistakes

  • āœ•Breaking sigma bonds or changing the atomic connectivity while drawing resonance structures.
  • āœ•Violating the octet rule for second-period elements (e.g., carbon, nitrogen, oxygen, fluorine) or drawing structures with like charges on adjacent atoms.
  • āœ•Incorrectly applying the hierarchy of stability criteria, especially misjudging the importance of complete octets versus minimal charge separation.

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