JEE Mains Chemistry · Chemical Bonding and Molecular Structure

Ionic Bonding, Lattice Enthalpy and Fajans' Rules

An ionic solid is held by its lattice enthalpy, found from a Born-Haber cycle, and Fajans' rules say how far a small or highly charged cation distorts the anion and gives the bond covalent character.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Nine PYQs, seven of them multiple choice, and one from 2026. Four use the Born-Haber cycle or lattice enthalpy: for a lattice enthalpy, a bond enthalpy, a melting-point order or an order of ionic character. Five rank compounds by covalent character with Fajans' rules. Two ideas cover the page.

Concept 1 of 2: The Born-Haber cycle and lattice enthalpy

You cannot measure a lattice enthalpy directly, but Hess's law lets you walk round it. Take the metal and non-metal from their standard states to gaseous ions step by step, then let the ions fall together into the solid. The total must equal the enthalpy of formation.

Definition

Steps for $\mathrm{M(s) + \tfrac{1}{2}X_2(g) \rightarrow MX(s)}$ : sublimation of M, ionisation of M, half the X–X bond enthalpy, electron gain by X, then the ions forming the lattice.
$\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H + \Delta_{lattice}H$ , with the lattice term for ions coming TOGETHER (negative).
If the data give the lattice enthalpy as the solid breaking into ions (positive), subtract it instead.
Lattice enthalpy grows with the ion charges and falls as the ions get larger: $|\Delta_{lattice}H| \propto \dfrac{z^+ z^-}{r^+ + r^-}$ . Melting points follow it.
For one cation with several partners, the partner with the more negative electron gain enthalpy forms the more ionic compound.

Born-Haber cycle

\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H + \Delta_{lattice}H

Worked example

For NaCl: sublimation of Na

= 108

, ionisation enthalpy of Na

= 496

, bond enthalpy of

\mathrm{Cl_2}

= 242

, electron gain enthalpy of Cl

= -349

and

\Delta_f H(\mathrm{NaCl}) = -411

, all in kJ mol

^{-1}

. Find the lattice enthalpy of NaCl.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q47Moderate

Example 1 · Chemical Bonding and Molecular Structure · Ionic Bonding, Lattice Enthalpy and Fajans' Rules

If the enthalpy of sublimation of Li is

155\text{ }kJ{\text{ }mol}^{- 1}

, enthalpy of dissociation of

F_{2}

150\text{ }kJ{\text{ }mol}^{- 1}

, ionization enthalpy of Li is

520\text{ }kJ{\text{ }mol}^{- 1}

, electron gain enthalpy of

F

- 313\text{ }kJ{\text{ }mol}^{- 1}

, standard enthalpy of formation of LiF is

- 594\text{ }kJ{\text{ }mol}^{- 1}

. The magnitude of lattice enthalpy of LiF is

\_\_\_\_

kJ{mol}^{- 1}

(Nearest integer).

Half the bond enthalpy, not all of it

One formula unit of MX needs one X atom, which is half an

\mathrm{X_2}

molecule. Adding the whole bond enthalpy shifts the answer by half of it.

Check which way the lattice step runs

Lattice enthalpy is quoted both for the solid breaking into ions (positive) and for ions forming the solid (negative). Write the step in the direction of the cycle and give it the matching sign, then answer with the magnitude if that is what is asked.

Drill 3 more on the born-haber cycle and lattice enthalpy

Concept 2 of 2: Fajans' rules and covalent character

A cation pulls on the electron cloud of the anion next to it. A small, highly charged cation pulls hard, and a large anion's cloud is easy to pull. The more the cloud is drawn in between the two ions, the more the bond is shared, so it becomes more covalent.

Definition

Covalent character rises as the cation gets smaller: $\mathrm{Li^+ > Na^+ > K^+ > Cs^+}$ .
It rises with the cation's charge: $\mathrm{Sn^{4+}}$ compounds are more covalent than $\mathrm{Sn^{2+}}$ ones.
It rises as the anion gets larger: $\mathrm{I^- > Br^- > Cl^- > F^-}$ .
A cation with an 18-electron outer shell ( $\mathrm{Cu^+}$ , $\mathrm{Ag^+}$ , $\mathrm{Zn^{2+}}$ ) polarises more than a noble-gas cation of the same size and charge, because its d electrons shield the nucleus poorly.
Between two atoms, ionic character grows with their electronegativity difference.

Rule	Order of covalent character	Why
Smaller cation	$\mathrm{LiCl > NaCl > KCl > CsCl}$	$\mathrm{Li^+}$ is the smallest and most polarising
Higher cation charge	$\mathrm{AlCl_3 > MgCl_2 > NaCl}$ ; $\mathrm{SnCl_4 > SnCl_2}$	More charge on a smaller ion
Larger anion	$\mathrm{CaI_2 > CaBr_2 > CaCl_2 > CaF_2}$ ; $\mathrm{KI > KF}$	$\mathrm{I^-}$ has the largest, softest cloud
18-electron cation	$\mathrm{CuCl > NaCl}$ ; $\mathrm{AgCl > KCl}$	d electrons shield the nuclear charge poorly
Electronegativity difference	Ionic character: $\mathrm{N_2 < ClF_3 < SO_2 < K_2O < LiF}$	$\Delta\chi$ is 0 for $\mathrm{N_2}$ , about 0.8 for Cl–F, 0.9 for S–O

The same polarisation that adds covalent character lowers the melting point and the solubility in water.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q45Moderate

Example 2 · Chemical Bonding and Molecular Structure · Ionic Bonding, Lattice Enthalpy and Fajans' Rules

Order of covalent character: A.

KF > KI;LiF > KF

KF < KI;LiF > KF

SnCl_{4}>SnCl_{2};CuCl > NaCl

LiF > KF;CuCl < NaCl

KF < KI;CuCl > NaCl

Choose the correct answer from the options given below:

A bigger cation means LESS covalent

Size works in opposite directions for the two ions. A large anion raises covalent character; a large cation lowers it. So KF is less covalent than LiF, and KI is more covalent than KF.

Rank electron gain by magnitude

For one metal bonded to several non-metals, the most ionic product comes from the partner that releases the most energy on gaining an electron, the most negative value. Ranking the values as signed numbers puts the order backwards.

Drill 4 more on fajans' rules and covalent character

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

The Born-Haber cycle and lattice enthalpy
Born-Haber cycle
$\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H + \Delta_{lattice}H$

Reference tables (1)

Fajans' rules and covalent character5 rows

Rule	Order of covalent character	Why
Smaller cation	$\mathrm{LiCl > NaCl > KCl > CsCl}$	$\mathrm{Li^+}$ is the smallest and most polarising
Higher cation charge	$\mathrm{AlCl_3 > MgCl_2 > NaCl}$ ; $\mathrm{SnCl_4 > SnCl_2}$	More charge on a smaller ion
Larger anion	$\mathrm{CaI_2 > CaBr_2 > CaCl_2 > CaF_2}$ ; $\mathrm{KI > KF}$	$\mathrm{I^-}$ has the largest, softest cloud
18-electron cation	$\mathrm{CuCl > NaCl}$ ; $\mathrm{AgCl > KCl}$	d electrons shield the nuclear charge poorly
Electronegativity difference	Ionic character: $\mathrm{N_2 < ClF_3 < SO_2 < K_2O < LiF}$	$\Delta\chi$ is 0 for $\mathrm{N_2}$ , about 0.8 for Cl–F, 0.9 for S–O

The same polarisation that adds covalent character lowers the melting point and the solubility in water.

Watch out for (4)

Half the bond enthalpy, not all of it
→ The Born-Haber cycle and lattice enthalpy
Check which way the lattice step runs
→ The Born-Haber cycle and lattice enthalpy
A bigger cation means LESS covalent
→ Fajans' rules and covalent character
Rank electron gain by magnitude
→ Fajans' rules and covalent character

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