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Chemical Bonding and Molecular Structure appeared 42 times across 3 years — 4.9% of Chemistry. This question is from Dipole Moment and Formal Charge.

Year 2026 2025 2024 Total
Questions 12 14 16 42

Given below are statements about some molecules/ions. Identify the CORRECT statements. A. The dipole moment value of NF₃ is higher than that of NH₃. B. The dipole moment value of BeH₂ is zero. C. The bond order of O₂²⁻ and F₂ is same. D. The formal charge on the central oxygen atom of ozone is -1. E. In NO₂, all the three atoms satisfy the octet rule, hence it is very stable. Choose the correct answer from the options given below :

Solution & Explanation

Core Logic

(A) Dipole moment of NF₃ is lower than that of NH₃ because in NF₃, the orbital dipole due to the lone pair and the bond dipoles (N-F) are in opposite directions, whereas in NH₃ they reinforce each other. (Statement A is false).

(B) BeH₂ is sp hybridized, a linear molecule. The two Be-H bond dipoles cancel out, resulting in a zero net dipole moment. (Statement B is true).

(C) Bond order of O₂²⁻ (peroxide ion, 18 e^-) = 1. Bond order of F₂ (18 e^-) = 1. They are isoelectronic and have the same bond order. (Statement C is true).

(D) The formal charge on the central oxygen atom in ozone (O₃) is +1, not -1. (Statement D is false).

(E) In NO₂, nitrogen is an odd-electron species (7+16 = 23 e^-), thus it does not follow the octet rule. (Statement E is false).

Step 1: Final Conclusion

Only statements B and C are correct.

Pattern Recognition

Ammonia vs Nitrogen trifluoride is a classic NCERT dipole exception. Central atoms with odd electrons (like N in NO, NO₂, ClO₂) inherently violate the octet rule.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 9

Q77 jee_main_2024_31_jan_morning Molecular Orbital Theory
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:
  • A. A, B, C only
  • B. A and C only
  • C. B, C, D only
  • D. B and D only

Solution

Core Logic

Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:

  • The combining atomic orbitals must have the same or nearly the same energy.
  • The combining atomic orbitals must have the same symmetry about the molecular axis.
  • The combining atomic orbitals must overlap to the maximum extent (not minimum).
Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp³ hybrid orbitals in its bonding is NH₃, SO₂, SiO₂, BeCl₂, CO₂, H₂O, CH₄, BF₃
Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Analyzing the hybridization of the central atom in each species:

  • NH₃: 3 bp + 1 lp = 4 electron domains arrow sp³
  • SO₂: 2 bp + 1 lp = 3 electron domains arrow sp²
  • SiO₂: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally arrow sp³
  • BeCl₂: 2 bp + 0 lp = 2 electron domains arrow sp
  • CO₂: 2 bp + 0 lp = 2 electron domains arrow sp
  • H₂O: 2 bp + 2 lp = 4 electron domains arrow sp³
  • CH₄: 4 bp + 0 lp = 4 electron domains arrow sp³
  • BF₃: 3 bp + 0 lp = 3 electron domains arrow sp²
  • Total species with sp³ hybridization: NH₃, SiO₂, H₂O, CH₄. Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Questions — jee_main_2026_24_january_morning

Practice all Chemical Bonding and Molecular Structure previous-year questions →

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