Covalent Bonding

Master covalent bond formation, Lewis structures, electronegativity, bond parameters, and polar vs nonpolar bonds for JEE Main & Advanced.

17 min read

Prerequisites

Before studying covalent bonding, review:

The Hook: Diamond vs Graphite Paradox

Connect: The Carbon Mystery

Same element. Completely different properties.

Diamond: Hardest natural substance, insulator, transparent, priceless jewelry Graphite: Soft enough for pencils, conductor, black, costs pennies

How can the same atom (carbon) create materials so different? The answer isn’t what atoms are bonded—it’s how they’re bonded!

In Oppenheimer (2023), the physicists used graphite as a moderator in nuclear reactors, but diamond wouldn’t work. Why? Because covalent bonding creates different structures with wildly different properties. Master this chapter, and you’ll understand everything from DNA to semiconductors to the fuel in your car!

JEE Reality Check: This topic is a goldmine—expect 4-5 questions worth 16-20 marks! Lewis structures, formal charge, and bond parameters are JEE favorites.

The Core Concept

What is a Covalent Bond?

A covalent bond is formed by mutual sharing of electrons between two atoms to achieve stable electronic configuration (usually octet).

graph LR
    A[H atom
1 electron] -->|shares| C[H-H bond
2 electrons shared]
    B[H atom
1 electron] -->|shares| C
    C --> D[H₂ molecule
Both achieve He config]

    style A fill:#ffe5e5
    style B fill:#ffe5e5
    style C fill:#c8e6c9
    style D fill:#a5d6a7

In simple terms: Instead of one atom stealing electrons (ionic), two atoms share electrons like roommates sharing rent. Both benefit, both achieve stability!

Formation of Covalent Bonds

The Octet Rule

Atoms tend to gain, lose, or share electrons to achieve 8 electrons in their valence shell (noble gas configuration).

Exceptions:

  • H, He: Need only 2 electrons (duet rule)
  • Be, B: Can have less than 8 (electron deficient: BeCl₂, BF₃)
  • P, S, Cl, Br, I: Can have more than 8 (expanded octet: PCl₅, SF₆, IF₇)
Why Octet?

Noble gases (He, Ne, Ar) are extremely stable because of filled valence shells. Other atoms “want” this stability—so they share electrons to simulate a noble gas configuration!

ns²np⁶ = the magic configuration (8 electrons for most elements)

Lewis Dot Structures

How to Draw Lewis Structures

Step 1: Count total valence electrons Step 2: Identify central atom (usually least electronegative, never H) Step 3: Connect atoms with single bonds Step 4: Complete octets of outer atoms (except H) Step 5: Place remaining electrons on central atom Step 6: If central atom doesn’t have octet, form multiple bonds

Example: CO₂

Step 1: Total electrons = 4 (C) + 2×6 (O) = 16

Step 2: Central atom = C (less electronegative than O)

Step 3-4: Initial structure

O - C - O  (uses 4 electrons, 12 remaining)

Step 5: Complete octets

:Ö: - C - :Ö:  (each O has 6 more electrons as lone pairs)

Step 6: C only has 4 electrons! Form double bonds:

O=C=O  (each O has 2 lone pairs, C has 8 electrons)

Final Lewis structure:

$$\boxed{\text{:}\ddot{O}=C=\ddot{O}\text{:}}$$

Interactive Demo: Visualize Molecular Bonding

See how atoms share electrons to form covalent bonds. Explore different molecular geometries and electron sharing patterns.

Formal Charge: The Stability Predictor

Formula

$$\boxed{\text{Formal Charge} = V - L - \frac{B}{2}}$$

Where:

  • V = Valence electrons in free atom
  • L = Lone pair electrons
  • B = Bonding electrons (shared electrons)

Alternative formula:

$$\text{FC} = V - (\text{Lone pairs} + \text{Number of bonds})$$

Rules for Stability

Formal Charge Guidelines

Most stable structure:

  1. Lowest total formal charges (ideally all zeros)
  2. Negative FC on more electronegative atom
  3. Adjacent atoms should NOT have same sign charges
  4. Large negative/positive charges are unstable

Memory trick: “Low FC, Negative on Electronegative, No Like charges = Stable WINNER”

Example: CO₃²⁻ (Carbonate Ion)

Structure 1: One C=O double bond, two C-O single bonds

For C: FC = 4 - 0 - 8/2 = 0
For O (double bond): FC = 6 - 4 - 4/2 = 0
For O (single bond): FC = 6 - 6 - 2/2 = -1 (two of these)

Total charge = 0 + 0 + (-1) + (-1) = -2 ✓ (matches ion charge!)

Resonance: The double bond can be with any of the three oxygens, so we have 3 resonance structures.

Quick Check

Question: Why is CO₂ linear and stable, but CO₃²⁻ needs a negative charge?

Answer: CO₂ has 16 electrons (matches two C=O bonds perfectly). CO₃²⁻ has 24 electrons—too many for three C=O double bonds, so some must be single bonds with lone pairs (creating negative charges on O atoms).

Bond Parameters

1. Bond Length

Definition: Equilibrium distance between nuclei of two bonded atoms.

Factors affecting bond length:

FactorEffectExample
Bond orderHigher order → Shorter bondC-C (154 pm) > C=C (134 pm) > C≡C (120 pm)
Atomic sizeLarger atoms → Longer bondC-F (138 pm) < C-Cl (177 pm) < C-Br (194 pm) < C-I (214 pm)
HybridizationMore s-character → Shortersp (50% s) < sp² (33% s) < sp³ (25% s)

Memory trick: “Triple bonds are tight, single bonds are spacious!”

2. Bond Angle

Definition: Angle between two adjacent bonds at an atom.

MoleculeHybridizationBond Angle
BeCl₂sp180°
BF₃sp²120°
CH₄sp³109.5°
NH₃sp³107° (lone pair repulsion)
H₂Osp³104.5° (two lone pairs)

3. Bond Order

Definition: Number of electron pairs shared between two atoms.

$$\boxed{\text{Bond Order} = \frac{\text{Number of bonds}}{2\text{ (for bond pairs)}}}$$

Simple counting method:

  • Single bond: Bond order = 1
  • Double bond: Bond order = 2
  • Triple bond: Bond order = 3

For resonance structures:

$$\text{Bond Order} = \frac{\text{Total bonds between two atoms}}{\text{Number of resonance structures}}$$

Example: In CO₃²⁻, each C-O bond has:

  • Total bonds: 1 double + 2 singles = 4 bonds for 3 C-O bonds
  • Bond order = 4/3 = 1.33

4. Bond Energy (Bond Enthalpy)

Definition: Energy required to break one mole of bonds in gaseous state.

$$\text{H-H}(g) \rightarrow 2\text{H}(g) \quad \Delta H = +436 \text{ kJ/mol}$$

Trend:

  • Higher bond order → Higher bond energy
  • Shorter bond → Higher bond energy
BondBond Energy (kJ/mol)
C-C348
C=C614
C≡C839
C-H413
O-H463
N≡N946 (strongest!)
JEE Shortcut

Stronger bond = Shorter bond = Higher bond energy = Higher bond order

All four properties correlate! If one increases, all increase (generally).

Electronegativity: The Electron Puller

Definition

Electronegativity (EN): The ability of an atom in a molecule to attract shared electrons toward itself.

Pauling Scale (most common):

$$\boxed{\text{F} > \text{O} > \text{N} \approx \text{Cl} > \text{Br} > \text{C} \approx \text{S} \approx \text{I} > \text{H} > \text{P}}$$

Actual values:

ElementElectronegativity
F4.0
O3.5
N, Cl3.0
Br2.8
C2.5
H2.1
P2.1

Memory trick: “FONCl Brings Charges” → F, O, N, Cl have highest EN

Across a period (left → right): EN increases

  • Reason: Nuclear charge increases, atomic size decreases

Down a group (top → bottom): EN decreases

  • Reason: Atomic size increases, shielding increases
graph LR
    A[Li: 1.0] --> B[Be: 1.5] --> C[B: 2.0] --> D[C: 2.5] --> E[N: 3.0] --> F[O: 3.5] --> G[F: 4.0]

    style A fill:#ffebee
    style G fill:#b71c1c

Polar vs Nonpolar Bonds

Nonpolar Covalent Bond

ΔEN ≈ 0 (electronegativity difference near zero)

Electrons shared equally between atoms.

Examples:

  • H-H (ΔEN = 0)
  • C-C (ΔEN = 0)
  • Cl-Cl (ΔEN = 0)

Polar Covalent Bond

0 < ΔEN < 1.7

Electrons shared unequally—partial charges develop.

Example: H-Cl

  • EN(H) = 2.1, EN(Cl) = 3.0
  • ΔEN = 0.9 → polar covalent
  • Chlorine pulls electrons more → δ⁻ on Cl, δ⁺ on H
$$\boxed{H^{\delta+} - Cl^{\delta-}}$$

Ionic Bond

ΔEN > 1.7

Complete electron transfer (covered in previous chapter).

Quick Decision Chart

ΔEN = 0: Nonpolar covalent (H₂, Cl₂, O₂) ΔEN = 0.1 - 0.4: Weakly polar (C-H) ΔEN = 0.4 - 1.7: Polar covalent (H-Cl, C-O, C-N) ΔEN > 1.7: Ionic (NaCl, MgO)

JEE tip: If they ask “which bond is most polar?” → find largest ΔEN!

Dipole Moment: Measuring Polarity

Definition

Dipole moment (μ): Product of charge (q) and distance of separation (d).

$$\boxed{\mu = q \times d}$$

Units: Debye (D), where 1 D = 3.34 × 10⁻³⁰ C·m

Vector Nature

Dipole moment is a vector — direction matters!

Convention: Arrow points from positive to negative (or sometimes δ⁺ → δ⁻)

Molecular Dipole Moment

For polyatomic molecules, vector sum of all bond dipoles determines net dipole.

Examples:

CO₂: Linear, symmetrical

O=C=O
← →  (dipoles cancel)
Net μ = 0 (nonpolar molecule despite polar C=O bonds!)

H₂O: Bent, asymmetrical

   H
    \
     O
    /
   H
Net μ = 1.85 D (polar molecule!)

CCl₄: Tetrahedral, symmetrical

All four C-Cl dipoles cancel due to symmetry
Net μ = 0 (nonpolar!)

CHCl₃: Tetrahedral, asymmetrical

Three Cl atoms pull more than one H
Net μ = 1.04 D (polar!)
Common Mistake: Polar Bonds ≠ Polar Molecule

Mistake: “CO₂ has polar C=O bonds, so it must be polar!”

Correct thinking:

  • Individual bonds CAN be polar
  • But if molecule is symmetrical, dipoles cancel
  • Result: Nonpolar molecule

Examples of nonpolar molecules with polar bonds:

  • CO₂ (linear)
  • BF₃ (trigonal planar)
  • CCl₄ (tetrahedral)
  • PCl₅ (trigonal bipyramidal)
  • SF₆ (octahedral)

Fajan’s Rules (Revisited for Covalent Character)

When we studied ionic bonding, we saw that “ionic” compounds can have covalent character. Now we understand why!

Polarization

Polarization: Distortion of electron cloud of an ion by the electric field of a nearby ion.

graph LR
    A[Spherical anion
unpolarized] -->|Small cation
approaches| B[Distorted anion
polarized]
    B --> C[Electron density
shared between ions
= Covalent character]

    style A fill:#b3e5fc
    style B fill:#fff59d
    style C fill:#c8e6c9

Fajan’s Rules Summary

Covalent character INCREASES when:

  1. Cation is small (high charge density → strong polarizing power)

    • Li⁺ > Na⁺ > K⁺

  2. Cation has high charge (more polarizing)

    • Al³⁺ > Mg²⁺ > Na⁺

  3. Anion is large (easily polarizable)

    • I⁻ > Br⁻ > Cl⁻ > F⁻

  4. Cation has pseudo noble gas configuration (18e⁻ or 18+2e⁻)

    • Ag⁺ (4d¹⁰) > Na⁺ (noble gas)
    • Cu⁺ (3d¹⁰) has high polarizing power
Application Example

Question: Why is AgCl insoluble in water but NaCl is highly soluble?

Answer:

  • Ag⁺ has 18e⁻ configuration (pseudo noble gas) → highly polarizing
  • AgCl has significant covalent character (polarization)
  • Covalent compounds are less soluble in polar solvents like water
  • NaCl is truly ionic → dissolves easily in water

Proof: AgCl is white (covalent compounds often colored/white), has low melting point compared to typical ionic compounds, and is used in photographic films (covalent character helps sensitivity to light).

Resonance: When One Structure Isn’t Enough

Definition

Resonance: When a molecule can be represented by two or more Lewis structures that differ only in electron positions, not atom positions.

The actual structure is a resonance hybrid—a blend of all resonance forms.

Rules for Resonance

  1. Same atomic positions (only electrons move)
  2. Same number of paired/unpaired electrons
  3. All structures must be valid Lewis structures
  4. Lower energy (more stable) structures contribute more to hybrid

Example: Benzene (C₆H₆)

Two equivalent resonance structures:

    H         H
     \       /
      C=C-C
     /  |  \
   H-C  C-C-H
      \ | /
       C=C
      /   \
     H     H

↔ (double bonds alternate positions)

Reality: All C-C bonds are equal (139 pm), intermediate between C-C (154 pm) and C=C (134 pm)

Bond order = 1.5 for each C-C bond

Example: Ozone (O₃)

Two resonance structures:

$$O=\ddot{O}-\ddot{O} \leftrightarrow \ddot{O}-\ddot{O}=O$$

Result: Both O-O bonds are equal with bond order = 1.5

Resonance Stability

More resonance structures = More stable molecule

Why? Energy is distributed over multiple bonds, lowering overall energy.

Examples:

  • Benzene: 2 equivalent structures → very stable (aromatic stability)
  • Carbonate (CO₃²⁻): 3 equivalent structures → stable
  • Nitrate (NO₃⁻): 3 equivalent structures → stable

JEE tip: If a question asks “which is more stable?” and options include resonance, choose the one with more resonance structures!

Memory Tricks & Patterns

Bond Parameter Correlations

“Triple bonds are Short, Strong, and Stingy with length!”

PropertySingle < Double < Triple
Bond Order1 < 2 < 3
Bond EnergyWeak < Medium < Strong
Bond LengthLong < Medium < Short

“FONCl Browns Crackers In Homes”

  • F, O, N, Cl, Br, C, I, H (descending order)

Formal Charge Quick Calculation

For common atoms:

  • C with 4 bonds, no lone pairs → FC = 0 ✓
  • N with 3 bonds, 1 lone pair → FC = 0 ✓
  • O with 2 bonds, 2 lone pairs → FC = 0 ✓
  • O with 1 bond, 3 lone pairs → FC = -1 (anion)
  • N with 4 bonds, no lone pairs → FC = +1 (cation)

When to Use Covalent Bonding Concepts

Decision Tree

Use Lewis structures when:

  • Predicting molecular formulas and shapes
  • Calculating formal charges for stability
  • Drawing resonance structures

Use electronegativity when:

  • Predicting bond polarity (polar vs nonpolar)
  • Comparing bond strengths (more EN difference → more polar → stronger)
  • Determining dipole moments

Use bond parameters when:

  • Comparing stability (higher BO → more stable)
  • Estimating enthalpy changes (ΔH = Σbonds broken - Σbonds formed)
  • Understanding reactivity (weaker bonds break first)

Use Fajan’s rules when:

  • Predicting ionic vs covalent character
  • Explaining solubility (AgCl, BeCl₂)
  • Understanding color/properties of “ionic” compounds

Common Mistakes to Avoid

Trap #1: Octet Violation

Mistake: Assuming ALL atoms need exactly 8 electrons

Exceptions you MUST remember:

  • H, He: 2 electrons (duet)
  • Be, B: Less than 8 (BeCl₂ has 4, BF₃ has 6 on central atom)
  • Elements with d-orbitals (period 3+): Can exceed 8
    • PCl₅: P has 10 electrons
    • SF₆: S has 12 electrons
    • IF₇: I has 14 electrons

Why? Period 3+ elements have vacant d-orbitals for electron accommodation!

Trap #2: Dipole Moment Confusion

Mistake: “Polar bonds → Polar molecule”

Correct approach:

  1. Identify polar bonds (ΔEN > 0)
  2. Draw molecular geometry
  3. Check symmetry — if symmetrical, dipoles cancel → nonpolar

Classic JEE traps:

  • CO₂: Polar bonds, but linear → μ = 0 (nonpolar)
  • BCl₃: Polar bonds, but trigonal planar → μ = 0
  • CCl₄: Polar bonds, but tetrahedral symmetry → μ = 0

Polar despite bonds: H₂O, NH₃, CHCl₃ (asymmetrical!)

Trap #3: Formal Charge ≠ Oxidation State

Mistake: Using formal charge and oxidation state interchangeably

Formal charge: Assumes equal sharing of bonding electrons Oxidation state: Assumes complete transfer to more electronegative atom

Example: CO molecule

  • Formal charges: C (0), O (0) — structure is :C≡O:
  • Oxidation states: C (+2), O (-2)

For JEE: Use formal charge for structure stability, oxidation state for redox reactions!

Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1

Question: Draw the Lewis structure of H₂O and calculate formal charges.

Solution:

Step 1: Total electrons = 2(1) + 6 = 8

Step 2: Central atom = O

Step 3: Connect H-O-H (uses 4 electrons)

Step 4: Complete octet of O (4 more electrons as 2 lone pairs)

Final structure:

   H
    \
     O: (2 lone pairs)
    /
   H

Formal charges:

  • O: FC = 6 - 4 - 4/2 = 0
  • H: FC = 1 - 0 - 2/2 = 0

Answer: All formal charges are zero → stable structure ✓

Problem 2

Question: Arrange the following in order of increasing bond length: C-C, C=C, C≡C

Solution: Higher bond order → shorter bond

  • C≡C: Triple bond, bond order = 3
  • C=C: Double bond, bond order = 2
  • C-C: Single bond, bond order = 1

Answer: C≡C < C=C < C-C

Actual values: 120 pm < 134 pm < 154 pm

Level 2: JEE Main Type

Problem 3

Question: Which of the following has zero dipole moment? (A) H₂O (B) NH₃ (C) BF₃ (D) CHCl₃

Solution:

(A) H₂O: Bent shape (104.5°) → asymmetrical → μ ≠ 0

(B) NH₃: Trigonal pyramidal → asymmetrical → μ ≠ 0

(C) BF₃: Trigonal planar (120°) → symmetrical → μ = 0 ✓

(D) CHCl₃: Tetrahedral but three Cl and one H → asymmetrical → μ ≠ 0

Answer: (C) BF₃

Explanation: BF₃ is planar and symmetrical. The three B-F bond dipoles cancel perfectly due to 120° angles.

Problem 4

Question: Calculate the bond order of O₂⁻ (superoxide ion).

Using Lewis structure:

  • O₂ has 12 valence electrons
  • O₂⁻ has 13 valence electrons

Lewis structure: :Ö-Ö̈: with one extra electron

  • One O-O single bond (2 electrons)
  • One O=O double bond contribution
  • Total bonds ≈ 1.5

Bond order = 1.5

Alternative: Will be confirmed by MOT (coming in molecular orbital theory!)

Answer: Bond order = 1.5

Level 3: JEE Advanced Type

Problem 5: Resonance Challenge

Question: Draw all resonance structures of the carbonate ion (CO₃²⁻) and calculate: (a) C-O bond order (b) Formal charge on each atom in one structure

Solution:

(a) Resonance structures:

Structure 1:        Structure 2:        Structure 3:
   O⁻                  O                   O
   |                   ||                  |
   C                   C                   C
  / \\                / \                 // \
 O   O⁻              O⁻  O⁻              O⁻   O⁻

Each structure has:

  • One C=O double bond
  • Two C-O single bonds

Total bonds for 3 C-O bonds: 1(double) + 2(single) = 4 bonds distributed among 3 positions

Bond order = 4/3 ≈ 1.33

(b) Formal charges (Structure 1):

  • C: FC = 4 - 0 - 8/2 = 0
  • O (double bonded): FC = 6 - 4 - 4/2 = 0
  • O (single bonded): FC = 6 - 6 - 2/2 = -1 (×2)

Total charge = 0 + 0 + (-1) + (-1) = -2 ✓ (matches ion charge!)

Answer: (a) Bond order = 1.33 (b) FC: C = 0, O(double) = 0, O(single) = -1 each

Problem 6: Multi-Concept

Question: Explain why BeCl₂ is linear and covalent despite Be being a metal and Cl being a non-metal.

Solution:

Part 1: Why linear?

  • Be has 2 valence electrons
  • Forms 2 bonds with 2 Cl atoms
  • No lone pairs on Be
  • VSEPR: 2 bond pairs, 0 lone pairs → linear (180°)
  • Hybridization: sp (will study in hybridization chapter)

Part 2: Why covalent?

Applying Fajan’s rules:

  1. Be²⁺ is very small (31 pm) → high charge density
  2. Be²⁺ has +2 charge → high polarizing power
  3. Cl⁻ is relatively large (181 pm) → polarizable
  4. Result: Be²⁺ distorts Cl⁻ electron cloud → covalent character

Evidence of covalent character:

  • BeCl₂ is molecular (exists as discrete molecules)
  • Low melting point (405°C vs typical ionic >800°C)
  • Soluble in organic solvents (like ether, benzene)
  • Poor conductor even in molten state

Answer: BeCl₂ is linear due to sp hybridization (2 bond pairs, no lone pairs) and is covalent due to high polarizing power of small Be²⁺ ion (Fajan’s rules).

Quick Revision Box

SituationFormula/Approach
Drawing Lewis structureCount e⁻ → Central atom → Connect → Octets → Multiple bonds if needed
Formal chargeFC = V - L - B/2 (want zeros!)
Bond orderSingle = 1, Double = 2, Triple = 3; or (bonds/positions) for resonance
Polar vs nonpolar bondΔEN < 0.4: nonpolar; 0.4-1.7: polar; >1.7: ionic
Polar molecule?Check symmetry! Symmetrical → μ = 0
Covalent characterSmall cation + Large anion + High charge

Must-Remember Values

BondLength (pm)Energy (kJ/mol)
C-C154348
C=C134614
C≡C120839
C-H109413
O=O121498
N≡N110946

Electronegativity Scale

F (4.0) > O (3.5) > N = Cl (3.0) > Br (2.8) > C (2.5) > H = P (2.1)

Real-World Applications

Where Covalent Bonding Rules

1. Organic Chemistry = Covalent Chemistry Every organic compound (from DNA to plastics) is held together by covalent bonds!

2. Semiconductors: Silicon (Si-Si covalent bonds) in computer chips. Understanding band gaps = understanding covalent bonding!

3. Polymers: Nylon, polyester, PVC — all long chains of covalent bonds. The strength of materials depends on bond strength!

4. Medicine: Drug molecules bind to receptors via weak covalent-like interactions (hydrogen bonds, van der Waals). Understanding polarity helps design better drugs!

5. Graphene & Carbon Nanotubes: Strongest materials known! All due to sp² covalent bonding in carbon sheets.

6. Fuel Combustion: Breaking C-H and C-C bonds releases energy. Bond energy calculations predict fuel efficiency!

Teacher’s Summary

Key Takeaways

1. Covalent bonding = electron sharing for mutual stability

  • Octet rule guides formation (8 electrons in valence shell)
  • Exceptions: H (2), Be/B (<8), Period 3+ (>8 with d-orbitals)

2. Lewis structures + formal charge = stability predictor

  • Draw all possible structures, calculate FC
  • Lowest FC, negative on EN atom = most stable
  • Resonance: multiple structures → hybrid reality

3. Bond parameters are interconnected

  • Higher bond order → Shorter length, Higher energy, More stable
  • Triple > Double > Single (in strength and shortness)

4. Electronegativity determines polarity

  • ΔEN: 0 (nonpolar) → 0.4-1.7 (polar covalent) → >1.7 (ionic)
  • Polar bonds ≠ polar molecule (check symmetry!)
  • Dipole moment = vector sum of bond dipoles

5. Fajan’s rules explain covalent character in “ionic” bonds

  • Small cation + large anion + high charge → polarization → covalent character
  • Explains why AgCl, BeCl₂, AlCl₃ behave covalently

6. For JEE:

  • Master Lewis structures (appears in 50% of bonding questions!)
  • Remember electronegativity scale (F > O > N ≈ Cl)
  • Know bond parameter correlations (saves calculation time)
  • Apply Fajan’s rules for tricky character prediction

“In covalent bonds, sharing is caring — but electronegativity determines who gets the bigger share!”

Within Chemical Bonding

Atomic Structure Foundation

Organic Chemistry Applications

Physical Chemistry Connections

Cross-Subject: Physics

Cross-Subject: Math