Chemical Bonding PYQs with solutions

In Chemistry, chemical bonding refers to the force that holds atoms or ions together in a molecule or compound. These bonds form when atoms share or transfer electrons to achieve stable electron configurations, usually resembling that of noble gases.

There are three main types of chemical bonds: Ionic Bond, Covalent Bond and Metallic Bond.

Chemical bonding is one of the most fundamental and high-weightage topics in both JEE and NEET chemistry syllabi. It lays the foundation for understanding molecular structure, physical and chemical properties, and reactivity patterns of elements and compounds. Mastery of chemical bonding concepts is crucial for scoring well in Physical and Inorganic Chemistry sections of competitive exams.

From the basics of ionic and covalent bonds to advanced topics like hybridization, molecular orbital theory, and VSEPR theory, this chapter integrates conceptual clarity with logical reasoning—skills essential for both JEE Main/Advanced and NEET aspirants.

1.0Key Concepts Covered in Chemical Bonding for IIT JEE

Chemical Bond: Force holding atoms together in a molecule/compound.

Types of Bond: Ionic, Covalent, Coordinate, Metallic, Hydrogen bonding.

Octet Rule (Lewis Concept)

• Atoms combine to complete 8 electrons in their valence shell.
• Duet rule for H, He, Li, Be.

Limitations:

• Doesn’t explain stability of molecules like NO, NO₂.
• Fails in molecules with odd/even number of electrons (like BF₃, SF₆).
• Doesn’t explain shapes or energies.

Ionic Bond

• Formed by transfer of electrons from a metal to a non-metal.
• Electrovalent bond: Bond between oppositely charged ions.

Lattice Energy (U): Energy released when gaseous ions form one mole of ionic crystal.

• Greater lattice energy = stronger bond.

Factors affecting Lattice Energy:

• Charge on ions ↑ → U ↑
• Size of ions ↓ → U ↑

Properties:

• High melting/boiling points.
• Soluble in polar solvents.
• Conduct electricity in molten/aqueous state.

Covalent Bond (Lewis-Langmuir Theory)

• Formed by mutual sharing of electrons between non-metals.
• Bond order = No. of shared electron pairs.

Types:

• Single bond – 1 pair (e.g., H₂)
• Double bond – 2 pairs (e.g., O₂)
• Triple bond – 3 pairs (e.g., N₂)

Bond parameters:

• Bond length: Distance between nuclei of bonded atoms.
• Bond energy: Energy required to break the bond.
• Bond angle: Angle between two bonds from same atom.

Coordinate (Dative) Bond

• One atom donates both electrons for bond formation.
• Example: NH₄⁺, H₃O⁺.

2.0Valence Shell Electron Pair Repulsion (VSEPR) Theory

Electron pairs (bonded or lone) repel each other → arrange to minimize repulsion.

AXₙEₘ notation:

• A = central atom
• X = bonding pair
• E = lone pair

3.0Valence Bond Theory (VBT)

• Covalent bond forms by overlapping half-filled atomic orbitals.
• Greater overlap = stronger bond.

Types of Overlap: s-s, s-p, p-p, sp-sp², etc.

Types of bonds:

• Sigma (σ) bond: End-to-end overlap.
• Pi (π) bond: Side-by-side overlap.

Hybridization:

4.0Molecular Orbital Theory (MOT)

• Atomic orbitals combine to form molecular orbitals.
• Bond Order (BO) = ½ (Nb - Na)
• • Nb = No. of bonding electrons
  • Na = No. of antibonding electrons

Stability: BO > 0 → stable; BO = 0 → unstable

Order of orbitals for O₂, F₂: σ(1s) < σ*(1s) < σ(2s) < σ*(2s) < σ(2pz) < π(2px)=π(2py) < π*(2px)=π*(2py) < σ*(2pz)

Magnetic nature:

• Unpaired electrons → paramagnetic (e.g., O₂)
• Paired → diamagnetic (e.g., N₂)

5.0Bonding in Some Molecules

Hydrogen bonding:

• • Intermolecular (e.g., H₂O)
  • Intramolecular (e.g., o-nitrophenol)

Effects: Raises boiling points, solubility in water, etc.

Resonance: Delocalization of electrons → hybrid structure

Example: O₃, CO₃²⁻, NO₂⁻

Dipole Moment

μ = q × d [Unit: Debye (D)]

Indicates:

• Polarity of bond/molecule.
• Vector quantity (can cancel or add).

Examples:

• H₂O – μ ≠ 0 (polar)
• CO₂ – μ = 0 (non-polar)
  

6.0Important Previous Year Questions 

Q.1 Arrange the following bonds according to their average bond energies in descending order :  C–Cl, C–Br, C–F, C–I

(1) C–I > C–Br > C–Cl > C–F

(2) C–Br > C–I > C–Cl > C–F

(3) C–F > C–Cl > C–Br > C–I

(4) C–Cl > C–Br > C–I > C–F

Ans.(3)

Bond length order in carbon halogen bonds are in the order of

C−F < C−Cl <  C−Br < C−I Hence, Bond energy order

C−F > C−Cl > C−Br > C−I

Q.2 The bond order and the magnetic characteristics

of CN– are :

(1) 3, diamagnetic

(2) 2 $\frac{1}{2}$, paramagnetic

(3) 3, paramagnetic

(4) 2 $\frac{1}{2}$, diamagnetic

Ans. (3)

Solution CN⁻ is a 14 electrons system similar to that of nitrogen molecules. As per the MOT concept, Nitrogen molecule is diamagnetic as all the electrons are paired and bond order is 3.

B.O.=12​[Nb​−Na]=12​[8−2] = 3

Q.3 Given below are two statements :

Statement I : o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Statement II : o-Nitrophenol has high melting due to hydrogen bonding.

In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Statement I is false but Statement II is true

(2) Both statement I and statement II are true

(3) Both statement I and statement II are false

(4) Statement I is true but statement II is false

Ans. (4)

Solution: 

thus it is more volatile due to intramolecular H-bonding.

Melting point depends on packing efficiency not on H-bonding thus statement II is false

Q.3 AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is ______. (Round off to the Nearest Integer).

Ans. AX is a covalent diatomic molecule.

The molecule is NO.

Total no. of electrons is 15.

Q.3 Match List-I with List-II

(1) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

(2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Ans.  (3).

Solution 

In the given list I A represent the formula for overlapping of anti-bonding molecular orbital and D represent the formula for overlapping of bonding molecular orbital. Formula given in B used to determine dipole moment and formula given in C used to determine bond order.

Q. 4 The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol–1 in 200 mL of water at 27°C is _______ Pa. [integer value] (use R = 0.083 L bar mol–1 K –1 ) 

Ans. (415) 

Sol.  = iCRT 

$=\frac{1\times 2}{60000\times 0.2}\times 0.083\times 300$

= 0.00415 bar(∵ 1 bar =10⁵)

So, 0.00415 x 10⁵= 415 Pa

Q.5 Identify the correct order for the given property for the following compounds

Choose the correct answer from the option given below :-

(1) (B), (C) and (D) only

(2) (A), (C) and (E) only

(3) (A), (C) and (D) only

(4) (A), (B) and (E) only

Ans. (2)

Sol. Boiling point of alkyl halide increases with increase in size, mass of halogen atom and size of alkyl group 

Boiling point of isomeric alkyl halide decreases with increase in branching

Density increases with increase in atomic mass of halogen atom

Q.6 The correct order of melting point of dichlorobenzenes is

(1)

(2)

(3)

 (4)

Ans. by NTA (4)

Solution:   

Q.7 Resonance in carbonate ion $(C{O}_3^{2-})$ is

Which of the following is true?

(1) It is possible to identify each structure individually by some physical or chemical method.

(2) All these structures are in dynamic equilibrium with each other.

(3) Each structure exists for equal amount of time.

(4) $C{O}_3^{2-}$ has a single structure i.e., resonance hybrid of the above three structures

Ans. Ans (4)

Sol.   Resonating structures are hypothetical and resonance hybrid is real structure which is weighted average of all the resonating structures.

Q.8 Amongst the following, the number of species having the linear shape is ________.

$Xe{F}_2,I_3^{+},C_3{O}_2,I_3^{-},C{O}_2,S{O}_2,BeCl$ and $BC{l}_2^0$

Ans. (5)

Sol.

Q. 9 Given below are two statements :

Statement (I) : A p bonding MO has lower electron density above and below the inter-nuclear axis.

Statement (II) : The p* antibonding MO has a node between the nuclei.

In the light of the above statements, choose the most appropriate answer from the options given below:

(1) Both Statement I and Statement II are false

(2) Both Statement I and Statement II are true

(3) Statement I is false but Statement II is true

(4) Statement I is true but Statement II is false

Ans.   (3)

Sol.  A pi bonding molecular orbital has higher electron density above and below inter nuclear axis

Q.10 The correct statement/s about Hydrogen bonding is/are :

A.  Hydrogen bonding exists when H is covalently bonded to the highly electro negative atom.

B.  Intermolecular H bonding is present in o-nitro phenol

C.   Intramolecular H bonding is present in HF.

D.  The magnitude of H bonding depends on the physical state of the compound.

E.   H-bonding has a powerful effect on the structure and properties of compounds.

Choose the correct answer from the options given below :

(1) A only                         (2) A, D, E only

(3) A, B, D only               (4) A, B, C only

Ans.   (2)

Solution. 

(A) Generally hydrogen bonding exists when H is covalently bonded to the highly electronegative atom like F, O, N.

(B) Intramolecular H bonding is present in  

(C) Intermolecular Hydrogen bonding is present in HF.

(D) The magnitude of Hydrogen bonding in solid state is greater than liquid state.

(E) Hydrogen bonding has a powerful effect on the structure & properties of compounds like melting point, boiling point, density etc.

Q.11 Number of molecules/ions from the following in which the central atom is involved in sp³ hybridization is ________.

NO₃^–, BCl₃, ClO₂^–, ClO₃

(1) 2                            (2) 4                              (3) 3                            (4) 1

Ans. (1)

Solution : ClO₂^–, ClO₃   are the molecules/ions where the central atom is involved in sp³ hybridization