Chemical Bonding and Molecular Structure - NCERT Solutions

Chapter Overview

Chemical Bonding explains how atoms combine to form molecules and compounds. This chapter covers various types of chemical bonds, molecular geometry, and theories explaining bond formation.

Key Concepts:

• Chemical Bond: Force that holds atoms together
• Ionic Bond: Transfer of electrons between atoms
• Covalent Bond: Sharing of electron pairs
• VSEPR Theory: Prediction of molecular geometry
• Hybridization: Mixing of atomic orbitals

NCERT Exercise Solutions

Exercise 4.1

Question: Explain the formation of a chemical bond.

Solution: A chemical bond is formed when:

1. Energy Consideration: The system achieves a state of minimum energy
2. Octet Rule: Atoms tend to attain the noble gas configuration (8 electrons in valence shell)
3. Electrostatic Forces: Attractive and repulsive forces balance at optimum distance

Mechanism:

• Ionic Bond Formation: One atom transfers electrons to another
• Covalent Bond Formation: Atoms share electrons to complete octets
• Coordinate Bond: One atom donates both electrons for sharing

Exercise 4.2

Question: Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Solution: Lewis dot symbols represent valence electrons as dots around the element symbol:

• Mg (Magnesium): Group 2, 2 valence electrons

  Mg••
  

• Na (Sodium): Group 1, 1 valence electron

  Na•
  

• B (Boron): Group 13, 3 valence electrons

   •
  B•
   •
  

• O (Oxygen): Group 16, 6 valence electrons

  ••
  

O•• ••

- **N (Nitrogen)**: Group 15, 5 valence electrons

• N•• •

- **Br (Bromine)**: Group 17, 7 valence electrons

•• Br•• •• •

### Exercise 4.3

**Question:** Draw Lewis structures for the following molecules: (i) NF₃, (ii) CO₃²⁻, (iii) HCN, (iv) SO₃

**Solution:**

**(i) NF₃ (Nitrogen trifluoride):**
- Total valence electrons = 5 (N) + 3 × 7 (F) = 26 electrons
- Central atom: N (least electronegative)
- Structure: N at center with 3 F atoms
- Lone pairs: 1 on N, 3 on each F

••

F••• N•• F••• ••

**(ii) CO₃²⁻ (Carbonate ion):**
- Total valence electrons = 4 (C) + 3 × 6 (O) + 2 (charge) = 24 electrons
- Central atom: C
- Structure: Resonance structures with double bonds
- Each O has 1 double bond and 2 single bonds in resonance

**(iii) HCN (Hydrogen cyanide):**
- Total valence electrons = 1 (H) + 4 (C) + 5 (N) = 10 electrons
- Structure: H-C≡N (triple bond between C and N)
- Linear molecule

**(iv) SO₃ (Sulfur trioxide):**
- Total valence electrons = 6 (S) + 3 × 6 (O) = 24 electrons
- Central atom: S
- Structure: Double bonds with all O atoms (resonance)
- Trigonal planar geometry

### Exercise 4.4

**Question:** Define the octet rule. Write its significance and limitations.

**Solution:**

**Definition:** The octet rule states that atoms tend to gain, lose, or share electrons to attain a stable electronic configuration with 8 electrons in their valence shell, similar to noble gases.

**Significance:**
1. **Stability**: Explains why atoms form bonds
2. **Compound Formation**: Helps predict formulae of compounds
3. **Bond Type**: Determines type of bond (ionic or covalent)
4. **Molecular Structure**: Aids in drawing Lewis structures

**Limitations:**
1. **Incomplete Octet**: Be, B, and Al form stable compounds with incomplete octets
2. **Expanded Octet**: Elements in period 3 and beyond can expand octet (P, S, Cl)
3. **Odd Electron Species**: NO, NO₂ have odd number of electrons
4. **Electron Deficient Compounds**: BCl₃, AlCl₃ are stable with incomplete octets
5. **Noble Gas Compounds**: Some noble gases form compounds violating octet rule

## NCERT Highlights for JEE/NEET

### 🎯 Important Concepts for JEE:

1. **Dipole Moment**: $\vec{\mu} = q \times d$ (in Debye units)
2. **VSEPR Theory**: AXₙEₘ notation for molecular geometry
3. **Molecular Orbital Theory**: Energy level diagrams and bond order
4. **Hybridization**: sp, sp², sp³, sp³d, sp³d²

### 🧬 Key Points for NEET:

1. **Bond Parameters**: Bond length, bond angle, bond enthalpy, bond order
2. **Resonance**: Delocalization of π-electrons
3. **Hydrogen Bonding**: Intermolecular and intramolecular H-bonds
4. **Polarity of Molecules**: Based on molecular geometry and electronegativity

## JEE/NEET Practice Questions

### Question 1 (JEE Level)

**Statement:** The correct order of bond angles in the following species is:
(A) NH₃ > H₂O > BF₃ > BeCl₂
(B) BeCl₂ > BF₃ > NH₃ > H₂O
(C) BF₃ > BeCl₂ > NH₃ > H₂O
(D) BeCl₂ > NH₃ > BF₃ > H₂O

**Solution:**
Analyzing each molecule using VSEPR theory:

- **BeCl₂**: Linear (AX₂), bond angle = 180°
- **BF₃**: Trigonal planar (AX₃), bond angle = 120°
- **NH₃**: Trigonal pyramidal (AX₃E), bond angle = 107° (reduced from 109.5° due to lone pair)
- **H₂O**: Bent (AX₂E₂), bond angle = 104.5° (further reduced due to two lone pairs)

Order: BeCl₂ (180°) > BF₃ (120°) > NH₃ (107°) > H₂O (104.5°)

**Answer:** Option (B) BeCl₂ > BF₃ > NH₃ > H₂O

### Question 2 (NEET Level)

**Statement:** Which of the following molecules has zero dipole moment?
(A) CO₂
(B) H₂O
(C) NH₃
(D) CHCl₃

**Solution:**
Analyzing molecular geometry and symmetry:

- **CO₂**: Linear molecule with O=C=O structure
  - Two C=O bonds of equal magnitude but opposite directions
  - Dipole moments cancel out
  - Net dipole moment = 0

- **H₂O**: Bent molecule, O-H bonds don't cancel
  - Net dipole moment exists

- **NH₃**: Trigonal pyramidal, N-H bonds don't cancel
  - Net dipole moment exists

- **CHCl₃**: Tetrahedral geometry, asymmetrical distribution
  - Net dipole moment exists

**Answer:** Option (A) CO₂ has zero dipole moment

## VSEPR Theory and Molecular Geometry

### Geometry Predictions Using VSEPR:

| Steric Number | Geometry | Example | Bond Angle |
|---------------|----------|---------|------------|
| 2 (AX₂) | Linear | BeCl₂, CO₂ | 180° |
| 3 (AX₃) | Trigonal planar | BF₃, AlCl₃ | 120° |
| 3 (AX₂E) | Bent | SO₂ | 119° |
| 4 (AX₄) | Tetrahedral | CH₄, NH₄⁺ | 109.5° |
| 4 (AX₃E) | Trigonal pyramidal | NH₃, PCl₃ | 107° |
| 4 (AX₂E₂) | Bent | H₂O, H₂S | 104.5° |
| 5 (AX₅) | Trigonal bipyramidal | PCl₅ | 90°, 120° |
| 6 (AX₆) | Octahedral | SF₆ | 90° |

## Hybridization Types

### sp Hybridization
- **Geometry**: Linear
- **Angle**: 180°
- **Example**: BeCl₂, C₂H₂ (acetylene)
- **Orbitals**: 1 s + 1 p = 2 sp orbitals

### sp² Hybridization
- **Geometry**: Trigonal planar
- **Angle**: 120°
- **Example**: BF₃, C₂H₄ (ethylene)
- **Orbitals**: 1 s + 2 p = 3 sp² orbitals

### sp³ Hybridization
- **Geometry**: Tetrahedral
- **Angle**: 109.5°
- **Example**: CH₄, NH₃, H₂O
- **Orbitals**: 1 s + 3 p = 4 sp³ orbitals

## Important Formulas

| Concept | Formula | Description |
|---------|---------|-------------|
| Dipole moment | $\vec{\mu} = q \times d$ | Product of charge and distance |
| Bond order | $\frac{\text{Bonding electrons} - \text{Antibonding electrons}}{2}$ | Single, double, triple bonds |
| Formal charge | FC = V - (L + B/2) | V=valence, L=lone pair, B=bonding |
| Electronegativity difference | | Determines bond type |

## Success Tips

1. **Master VSEPR Theory**: Essential for predicting molecular geometry
2. **Practice Lewis Structures**: Draw them systematically
3. **Understand Hybridization**: Connect to molecular geometry
4. **Memorize Exceptions**: Octet rule violations and their reasons

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## Master Chemical Bonding

Chemical bonding explains why and how atoms combine. Focus on concepts and practice structure determination!

> **Remember:** Chemical bonding is the language of chemistry. Master it to understand molecular behavior and reactivity!

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*For additional practice, explore our [Chemistry Practice Problems](/additional-problems-chapterwise/chemistry/) section.*