3 Most FAQ’s of Chemical Bonding Chapter in Class 10th Physical Science (TS/AP)

8 Marks

LAQ-1 : Explain the formation of the following molecules using valence bond theory.
a) N₂ molecule b) O₂ molecule

Formation of N₂​ and O_2​ Molecules Using Valence Bond Theory

1. N_2​ Molecule Formation

• Electronic Configuration:
  • Nitrogen (N) configuration:
    $$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$
• Overlapping of Orbitals:
  • Formation of σ Bond: Overlapping of 2p_x​ orbitals forms a σ(2p_x​2p_x​) bond along the inter-nuclear axis.
  • Formation of π Bonds: Lateral overlap of 2p_y​ and 2p_z​ orbitals creates two π(2p_y​2p_y​) and π(2p_z​2p_z​) bonds.
• Molecular Formation:
  • Triple Bond: N₂​ molecule consists of one σ bond and two π bonds, forming a triple bond (N≡N).

2. O₂​ Molecule Formation

• Electronic Configuration:
  • Oxygen (O) configuration:
    $$1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$$
• Overlapping of Orbitals:
  • Formation of σ Bond: Overlapping of 2p_y​ orbitals forms a σ(2p_y​2p_y​) bond along the inter-nuclear axis.
  • Formation of π Bond: Lateral overlap of 2p_z​ orbitals creates a π (2p_z​2p_z​) bond.
• Molecular Formation:
  • Double Bond: O₂​ molecule consists of one σ bond and one π bond, forming a double bond (O=O).

Summary

1. N₂​ Molecule: Features a triple bond with one sigma (σ) and two pi (π) bonds.
2. O₂​ Molecule: Comprises a double bond with one sigma (σ) and one pi (π) bond.
3. These formations are explained through the Valence Bond Theory, which emphasizes the overlapping of atomic orbitals to form chemical bonds.

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LAQ-2 : Explain the formation of boron tri-fluoride molecule by hybridization.

Introduction

This guide explains the formation of the boron trifluoride (BF₃​) molecule, an essential concept in chemistry, through the process of hybridization.

Ground State of Boron Atom

1. Initial Configuration: Boron (B) has one unpaired electron.
2. Electronic Configuration:
   $$1s^2 2s^2 2p^1$$

Transition to Excited State

1. Excited State: Boron transitions to an excited state for bonding with fluorine atoms.
2. New Configuration:
   $$1s^2 2s^1 2p^2$$
3. Unpaired Electrons: In this state, boron has three unpaired electrons for bonding.

The Hybridization Process

1. Orbital Merging: Boron’s 2s and 2p orbitals merge and redistribute in the excited state.
2. Formation of Hybrid Orbitals: This results in the formation of three sp₂ hybrid orbitals.

Formation of BF_3​ Molecule

1. Orbital Overlap: Each hybrid orbital of boron overlaps with a fluorine (F) atom’s orbital.
2. Covalent Bonds: This overlap creates three covalent bonds.
3. Combination: Boron and three fluorine atoms combine to form the BF₃​ molecule.

Molecular Structure Details

1. Structure: The BF₃​ molecule has a trigonal planar structure.
2. Bond Angles: The molecule features 120° bond angles.
3. Equal Bond Distribution: This leads to the molecule’s specific trigonal planar structure.

Summary

In summary, the BF₃​ molecule’s formation involves boron’s atomic orbitals undergoing hybridization to form three sp₂ hybrid orbitals. These orbitals then covalently bond with three fluorine atoms, resulting in a molecule with a trigonal planar structure and 120° bond angles. This explanation provides a clear understanding of the molecule’s formation process.

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LAQ-3 : Distinguish between a sigma and a pi bond.

Introduction

Understanding the differences between sigma (�σ) and pi (�π) bonds is essential for studying molecular structures and bond formations.

Sigma (σ) Bond

1. Formation: Formed by head-on or axial overlapping of atomic orbitals.
2. Number of Overlapping Orbitals: Involves one pair of overlapping orbitals.
3. Bond Strength: Generally stronger than a π bond.
4. Electron Density: Concentrated between the two bonding atoms.
5. Rotation: Allows rotation around the bond axis.
6. Examples: Present in all single (s), double (s), and triple (s) bonds.

Pi (π) Bond

1. Formation: Formed by lateral or side-by-side overlap of atomic orbitals.
2. Number of Overlapping Orbitals: Involves two pairs of overlapping orbitals.
3. Bond Strength: Generally weaker than a σ bond.
4. Electron Density: Lies above and below the plane of the atoms.
5. Rotation: Prevents free rotation around the bond axis.
6. Examples: Present only in double (p) and triple (p) bonds.

Summary

In conclusion, sigma and pi bonds differ in their formation, overlapping orbitals, bond strength, electron density distribution, and molecular rotation capabilities. σ bonds are stronger, allow rotation, and have electron density concentrated between the atoms. In contrast, π bonds are weaker, prevent free rotation, and have electron density distributed above and below the atomic plane. These distinctions are key to understanding the structure and properties of various molecules.