JEE Main · Chemistry ↓ Falling

Chemical Bonding and Molecular Structure appeared 42 times across 3 years — 4.9% of Chemistry. This question is from Ionic Bond and Lattice Energy.

Year 2026 2025 2024 Total
Questions 12 14 16 42
Practice all Chemical Bonding and Molecular Structure questions → View full JEE Main weightage & trends →

Which of the following is least ionic?

Solution & Explanation

Core Logic

According to Fajan's rules, covalent character is favored by high charge and small size of the cation, and by cations with a pseudo-noble gas configuration. Ag^+ has a pseudo-noble gas configuration (ns²np⁶nd¹⁰), which results in high polarizing power compared to s-block and typical transition elements. Therefore, AgCl has the maximum covalent character and is the least ionic among the given options.

Ionic character order: AgCl < CoCl₂ < BaCl₂ < KCl

Step 1: Final Selection

Because AgCl is the most covalent, it is the least ionic. Hence, option (2) is correct.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

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

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)