Ionic Bonding

Master ionic bond formation, lattice energy calculations, Born-Haber cycle, and Fajan's rules for JEE Main & Advanced.

12 min read

Prerequisites

Before diving into ionic bonding, make sure you understand:

The Hook: Salt to Circuit Boards

Connect: From Kitchen to Technology

Ever wondered why salt dissolves in water but oil doesn’t? Why do surgeons use saline solution (salt water) for medical procedures? Why can salt water conduct electricity but pure water can’t?

The answer lies in ionic bonding — one of nature’s most powerful forces! From the salt in your food to the lithium-ion battery in your smartphone, ionic compounds are everywhere. In Oppenheimer (2023), the atomic bomb’s trigger mechanism used ionic compounds that conduct electricity. Understanding ionic bonds isn’t just chemistry—it’s understanding the technology that powers modern life!

JEE Perspective: 3-4 questions appear every year on lattice energy, Born-Haber cycle, and ionic vs covalent character. Master this, and you’ve secured 12-16 marks!

The Core Concept

What is an Ionic Bond?

An ionic bond is the electrostatic force of attraction between oppositely charged ions formed by complete transfer of electrons from a metal to a non-metal.

graph LR
    A[Metal Atom
Na] -->|loses 1e⁻| B[Cation
Na⁺]
    C[Non-metal Atom
Cl] -->|gains 1e⁻| D[Anion
Cl⁻]
    B -.->|electrostatic
attraction| D
    B --> E[Ionic Compound
NaCl]
    D --> E

    style B fill:#ff6b6b
    style D fill:#4ecdc4
    style E fill:#95e1d3

In simple terms: Think of it like a magnetic attraction. Metals lose electrons to become positive (like losing negative charge makes you positive), non-metals gain electrons to become negative. Opposites attract — that’s your ionic bond!

Formation of Ionic Bonds

Step-by-Step Process

For NaCl formation:

Step 1: Ionization of Metal

$$\text{Na}(g) \rightarrow \text{Na}^+(g) + e^- \quad \Delta H = +496 \text{ kJ/mol}$$

Step 2: Electron Affinity of Non-metal

$$\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g) \quad \Delta H = -349 \text{ kJ/mol}$$

Step 3: Lattice Formation

$$\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s) \quad \Delta H = -788 \text{ kJ/mol}$$
Energy Insight

Notice that steps 1 and 2 require/release energy but don’t make the process favorable. The huge energy release in step 3 (lattice formation) is what makes ionic bond formation energetically favorable!

Net Energy: +496 - 349 - 788 = -641 kJ/mol (exothermic, hence favorable!)

Interactive Demo: Visualize Ionic Bond Formation

Watch electron transfer from sodium to chlorine and ionic lattice formation.

Lattice Energy: The Powerhouse

Definition

Lattice Energy (U) is the energy released when one mole of an ionic compound is formed from gaseous ions under standard conditions.

$$M^+(g) + X^-(g) \rightarrow MX(s) + \text{Lattice Energy}$$

Alternative Definition: Energy required to completely separate one mole of ionic solid into gaseous ions.

$$MX(s) + \text{Energy} \rightarrow M^+(g) + X^-(g)$$
Sign Convention Alert!

Formation definition: Lattice energy is negative (exothermic) Dissociation definition: Lattice energy is positive (endothermic)

JEE problems can use either! Read carefully. NCERT uses formation (negative), but many books use dissociation (positive).

Born-Landé Equation

The theoretical calculation of lattice energy:

$$\boxed{U = -\frac{N_A M z^+ z^- e^2}{4\pi\epsilon_0 r_0}\left(1 - \frac{1}{n}\right)}$$

Where:

  • $N_A$ = Avogadro’s number
  • $M$ = Madelung constant (geometry dependent)
  • $z^+, z^-$ = charges on cation and anion
  • $r_0$ = interionic distance
  • $n$ = Born exponent (8-12, usually 9)

Simplified for JEE:

$$\boxed{U \propto \frac{z^+ \times z^-}{r_0}}$$

In simple terms: Lattice energy depends on charge (more charge = more energy) and size (smaller ions = more energy).

Factors Affecting Lattice Energy

1. Charge on Ions

Higher charge → Higher lattice energy

CompoundChargesLattice Energy (kJ/mol)
NaCl+1, -1788
MgO+2, -23850
Al₂O₃+3, -2~15,000

Memory Trick: “Double charge, quadruple the attraction!” (because charge product increases)

2. Size of Ions

Smaller ions → Higher lattice energy (closer approach = stronger force)

CompoundCation Radius (pm)Lattice Energy (kJ/mol)
LiFLi⁺: 761037
NaFNa⁺: 102923
KFK⁺: 138821
RbFRb⁺: 152785
JEE Shortcut

Compare lattice energies instantly:

Same anion, different cations: Smaller cation wins Same cation, different anions: Smaller anion wins Different charges: Higher charge wins (dominates over size!)

Example: MgO > NaCl (charge dominates) > KCl (size effect)

Born-Haber Cycle: The Masterpiece

What is it?

A thermochemical cycle that applies Hess’s Law to calculate lattice energy indirectly using other measurable quantities.

The Cycle for NaCl

graph TD
    A[Na(s) + ½Cl₂(g)] -->|ΔHf = -411| B[NaCl(s)]
    A -->|ΔHsub = +108| C[Na(g) + ½Cl₂(g)]
    C -->|IE = +496| D[Na⁺(g) + ½Cl₂(g) + e⁻]
    D -->|½D = +121| E[Na⁺(g) + Cl(g) + e⁻]
    E -->|EA = -349| F[Na⁺(g) + Cl⁻(g)]
    F -->|U = ?| B

    style A fill:#e8f5e9
    style B fill:#c8e6c9
    style F fill:#ffccbc

The Equation

$$\boxed{\Delta H_f = \Delta H_{sub} + \frac{1}{2}D + IE + EA + U}$$

Where:

  • $\Delta H_f$ = Enthalpy of formation (negative for stable compounds)
  • $\Delta H_{sub}$ = Sublimation energy of metal
  • $D$ = Bond dissociation energy of non-metal (½D for Cl₂, O₂)
  • $IE$ = Ionization energy of metal
  • $EA$ = Electron affinity of non-metal (negative)
  • $U$ = Lattice energy (to be calculated)

Rearranging for Lattice Energy

$$\boxed{U = \Delta H_f - \Delta H_{sub} - \frac{1}{2}D - IE - EA}$$

Memory Tricks & Patterns

Born-Haber Mnemonic

“Super Ionized Elephants Eat Lettuce”

  • Sublimation: Take metal from solid to gas
  • Ionization: Remove electron from metal
  • Electron gain (EA): Non-metal gains electron
  • Electrostatic attraction: Formation (lattice energy)
  • Leads to formation enthalpy

Pattern Recognition

Lattice Energy Trends

Down a group: LiF > NaF > KF > RbF > CsF (decreasing U) Reason: Increasing cation size

Across a period: MgO > MgS > MgSe > MgTe (decreasing U) Reason: Increasing anion size

Charge effect dominates: Al₂O₃ > MgO > NaCl > LiF Reason: Charge product (3×2 > 2×2 > 1×1)

Ionic vs Covalent Character

When Does Ionic Character Dominate?

An ionic compound forms when:

  1. Large difference in electronegativity (ΔEN > 1.7)
  2. Low ionization energy of metal
  3. High electron affinity of non-metal
  4. Large lattice energy (makes process favorable)

Fajan’s Rules: When Ionic Becomes Covalent

Even in “ionic” compounds, there’s some covalent character due to polarization:

Fajan's Rules for Covalent Character

Covalent character INCREASES when:

  1. Cation is small (high charge density)

    • Li⁺ > Na⁺ > K⁺ (more covalent)

  2. Cation has high charge

    • Fe³⁺ > Fe²⁺ (more covalent)

  3. Anion is large (easily polarizable)

    • I⁻ > Br⁻ > Cl⁻ > F⁻ (more covalent)

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

    • Ag⁺ (4d¹⁰) more polarizing than Na⁺ (noble gas)
    • Cu⁺, Hg²⁺ are highly polarizing

Memory: “Small Cation, Large Anion → Covalent fashion!”

Examples

CompoundCharacterReason
NaFHighly ionicLarge size difference, F⁻ small
LiICovalent characterLi⁺ small, I⁻ large, polarization
AgClCovalent characterAg⁺ has 18e⁻ config, polarizing
AlCl₃CovalentAl³⁺ small with high charge

When to Use Ionic Bonding Concepts

Decision Tree

Use lattice energy when:

  • Comparing stability of ionic compounds
  • Calculating enthalpy changes using Born-Haber
  • Predicting solubility (high U → less soluble in water)
  • Explaining melting points (high U → high MP)

Use Fajan’s rules when:

  • Predicting bond character (ionic vs covalent)
  • Explaining solubility in non-polar solvents (AlCl₃, BeCl₂)
  • Understanding color of compounds (covalent character → colored)
  • Predicting hydration vs hydrolysis

JEE Tip: If question asks “which is more ionic?” → think size and charge. If it asks “which has covalent character?” → apply Fajan’s rules!

Common Mistakes to Avoid

Trap #1: Sign of Lattice Energy

Mistake: Confusing formation vs dissociation definitions

Correct Approach:

  • Formation: $M^+(g) + X^-(g) \rightarrow MX(s)$ → Negative (energy released)
  • Dissociation: $MX(s) \rightarrow M^+(g) + X^-(g)$ → Positive (energy required)

In Born-Haber cycle, if calculating U from formation data, expect negative value!

Trap #2: Bond Dissociation Energy

Mistake: Forgetting to divide by 2 for diatomic molecules

Correct: For Cl₂: Use ½D in Born-Haber (only ½ mole Cl₂ needed) For O₂: Use ½D for oxide formation

Don’t use full D value — you’ll get the wrong answer!

Trap #3: Comparing Lattice Energies

Mistake: Ignoring charge when comparing sizes

Example: Which has higher lattice energy: MgO or LiF?

Wrong thinking: “Li⁺ smaller than Mg²⁺, so LiF wins” Correct thinking: MgO has 2×2 charge product, LiF has 1×1. Charge dominates!

Answer: MgO » LiF (3850 vs 1037 kJ/mol)

Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1

Question: Arrange the following in order of increasing lattice energy: NaCl, MgO, KCl, CaO

Solution: Strategy: Compare charges first, then sizes

  • MgO and CaO: 2×2 charge → higher U than 1×1 compounds
  • Between MgO and CaO: Mg²⁺ smaller than Ca²⁺ → MgO > CaO
  • Between NaCl and KCl: Na⁺ smaller than K⁺ → NaCl > KCl

Answer: KCl < NaCl < CaO < MgO

Actual values: KCl (717) < NaCl (788) < CaO (3520) < MgO (3850) kJ/mol

Problem 2

Question: Why is MgCl₂ more ionic than AlCl₃?

Solution: Apply Fajan’s rules!

Al³⁺ vs Mg²⁺:

  • Al³⁺ has higher charge (+3 vs +2)
  • Al³⁺ is smaller (67 pm vs 86 pm)
  • Both have same anion Cl⁻

According to Fajan’s rules: Smaller cation with higher charge → more polarizing → more covalent character

Therefore: AlCl₃ has more covalent character, MgCl₂ is more ionic.

Proof: AlCl₃ is covalent (sublimes at 180°C), MgCl₂ is ionic (melts at 714°C)

Level 2: JEE Main Type

Problem 3

Question: Calculate the lattice energy of NaCl using the following data:

  • ΔH_f(NaCl) = -411 kJ/mol
  • ΔH_sub(Na) = +108 kJ/mol
  • IE(Na) = +496 kJ/mol
  • ½D(Cl₂) = +121 kJ/mol
  • EA(Cl) = -349 kJ/mol

Solution: Using Born-Haber cycle:

$$\Delta H_f = \Delta H_{sub} + IE + \frac{1}{2}D + EA + U$$ $$-411 = 108 + 496 + 121 + (-349) + U$$ $$-411 = 376 + U$$ $$U = -411 - 376 = -787 \text{ kJ/mol}$$

Answer: Lattice energy = -787 kJ/mol (or +787 kJ/mol for dissociation)

JEE Tip: Always check if your answer is reasonable. U for NaCl should be around -788 kJ/mol.

Problem 4

Question: Among the following, which has the maximum covalent character? (A) NaF (B) NaCl (C) NaBr (D) NaI

Solution: Applying Fajan’s rules — cation same (Na⁺), compare anions:

  • F⁻ smallest → least polarizable → least covalent
  • I⁻ largest → most polarizable → most covalent

Order of covalent character: NaI > NaBr > NaCl > NaF

Answer: (D) NaI

Practical proof: NaF highly ionic (MP 993°C), NaI has more covalent character (MP 661°C)

Level 3: JEE Advanced Type

Problem 5

Question: The lattice energy of an ionic compound MX (where M²⁺ and X²⁻) is four times that of NaCl. If the interionic distance in MX is the same as NaCl, identify the possible compound.

Solution: From Born-Landé equation:

$$U \propto \frac{z^+ \times z^-}{r_0}$$

For NaCl: $z^+ \times z^- = 1 \times 1 = 1$, $U_{NaCl} = U_0$

For MX: $z^+ \times z^- = 2 \times 2 = 4$, $U_{MX} = 4U_0$ (given same r₀)

This matches! Charge product is 4 times.

Possible compounds with M²⁺ and X²⁻:

  • MgO (Mg²⁺, O²⁻) ✓
  • CaO (Ca²⁺, O²⁻) ✓
  • SrO (Sr²⁺, O²⁻) ✓

But we need same interionic distance as NaCl (r₀ = 282 pm):

  • MgO: r₀ = 212 pm ✗
  • CaO: r₀ = 240 pm ✗

Trick: This is a theoretical problem. In reality, no M²⁺X²⁻ compound has exactly same r₀ as NaCl. The concept being tested is the charge effect on lattice energy.

Answer: The compound would be any M²⁺X²⁻ oxide like MgO (conceptually), but strictly speaking, the interionic distance condition cannot be met perfectly.

Problem 6: Multi-concept

Question: The electron affinity of chlorine is 3.7 eV. Using the following data, calculate the lattice energy of NaCl:

  • ΔH_f(NaCl) = -4.24 eV/molecule
  • ΔH_sub(Na) = 1.12 eV/atom
  • IE(Na) = 5.14 eV/atom
  • D(Cl₂) = 2.52 eV/molecule

Solution: Born-Haber cycle (per molecule):

$$\Delta H_f = \Delta H_{sub} + IE + \frac{1}{2}D + EA + U$$ $$-4.24 = 1.12 + 5.14 + \frac{2.52}{2} + (-3.7) + U$$ $$-4.24 = 1.12 + 5.14 + 1.26 - 3.7 + U$$ $$-4.24 = 3.82 + U$$ $$U = -8.06 \text{ eV/molecule}$$

Converting to kJ/mol:

$$U = -8.06 \times 96.5 = -778 \text{ kJ/mol}$$

Answer: Lattice energy = -778 kJ/mol (close to actual value of -788)

JEE Advanced Twist: This tests unit conversion (eV → kJ/mol) and careful arithmetic!

Quick Revision Box

SituationFormula/Approach
Compare lattice energiesHigher charge > Smaller size
Born-Haber calculation$U = \Delta H_f - \Delta H_{sub} - IE - \frac{1}{2}D - EA$
Ionic vs covalentΔEN > 1.7 → ionic, ΔEN < 1.7 → covalent
Covalent characterSmall cation + large anion + high charge
Melting point trendHigher U → Higher MP
Solubility in waterLower U → More soluble (easier to overcome)

Important Values to Remember

CompoundLattice Energy (kJ/mol)
LiF1037
NaCl788
MgO3850
CaO3520

Real-World Applications

Where Ionic Bonding Matters

1. Batteries: Lithium-ion batteries use Li⁺ ions moving between electrodes. The small size and high charge density of Li⁺ makes it perfect for energy storage!

2. Medicine: Saline solution (NaCl in water) conducts electricity and matches body fluid composition — essential for surgeries and IVs.

3. Water Softeners: Replace Ca²⁺ and Mg²⁺ (hard water ions) with Na⁺ using ion exchange resins.

4. Cement: CaO (quicklime) is crucial in cement. Its high lattice energy means it releases huge energy when hydrated — that’s what makes concrete harden!

5. Fertilizers: K⁺, NO₃⁻, PO₄³⁻ ions from ionic salts dissolve in soil water and are absorbed by plants.

Teacher’s Summary

Key Takeaways

1. Ionic bonding = complete electron transfer + electrostatic attraction

  • Metals lose, non-metals gain electrons
  • Large ΔEN (>1.7) favors ionic bonding

2. Lattice energy is the KEY to understanding ionic compounds

  • Formula: $U \propto \frac{z^+ \times z^-}{r_0}$
  • Higher charge and smaller size → Higher U → Higher MP, lower solubility

3. Born-Haber cycle connects thermodynamics to ionic bonding

  • Apply Hess’s Law to calculate lattice energy
  • Remember: $U = \Delta H_f - \Delta H_{sub} - IE - \frac{1}{2}D - EA$

4. Fajan’s rules explain covalent character in ionic compounds

  • Small cation + large anion + high charge → polarization → covalent character
  • AgCl, AlCl₃, BeCl₂ are “ionic” but behave covalently

5. For JEE:

  • Master Born-Haber calculations (100% expected!)
  • Compare lattice energies quickly (charge first, then size)
  • Apply Fajan’s rules for bond character questions

“In ionic bonding, charge rules the game — but size plays when charge is the same!”

Within Chemical Bonding

Foundation Topics

Cross-Subject: Physics