Ionic (Electrovalent) Bond & Lattice Enthalpy

What Makes a Bond Ionic?

We met the ionic bond briefly in Section 1 with $NaCl$. Now let's understand when and why atoms choose electron transfer over sharing.

Definition: An ionic (electrovalent) bond is the electrostatic force of attraction that holds together oppositely charged ions formed by the complete transfer of one or more electrons from a metal atom to a non-metal atom.

For this transfer to be favourable, three things must line up:

1. Low ionisation enthalpy of the metal — it should give up its electron(s) easily. That's why Group 1 and 2 metals (Na, K, Ca, Mg) form ionic bonds readily.
2. High (negative) electron gain enthalpy of the non-metal — it should release a lot of energy when it accepts the electron. Halogens and oxygen are champions here.
3. High lattice enthalpy of the product — the ions formed must pack into a crystal that releases a large amount of energy. This is often the deciding factor.

[JEE Tip] Forming an isolated ion pair $Na^+Cl^-$ in the gas phase is actually endothermic on its own. It is the huge lattice enthalpy released when billions of ions assemble into a 3-D crystal that makes the overall process exothermic. Never forget the role of the lattice.

Factors Favouring the Formation of Ionic Compounds

Putting the three conditions together, ionic character is favoured by:

1. Large difference in electronegativity

The bigger the EN gap between the two atoms, the more complete the electron transfer. A difference greater than about 1.7 on the Pauling scale usually signals a predominantly ionic bond.

2. Low ionisation enthalpy of the cation-forming element

Down a group, IE decreases, so $Cs$ forms ionic bonds more readily than $Li$. Across a period IE rises, so metals on the left are the best cation formers.

3. High negative electron gain enthalpy of the anion-forming element

Halogens (especially $Cl$) and oxygen accept electrons with a large energy release.

4. Small ionic sizes and high charges → high lattice enthalpy

Smaller, more highly charged ions sit closer together and attract more strongly.

Key Point: The single most important quantity deciding whether an ionic solid forms — and how stable it is — is the lattice enthalpy. Everything else feeds into it.

Lattice Enthalpy & the Born-Haber Cycle

Definition: The lattice enthalpy of an ionic solid is the energy required to completely separate one mole of the solid into its gaseous ions. $$NaCl(s) \rightarrow Na^+(g) + Cl^-(g) \qquad \Delta_{lattice}H = +788\ \text{kJ mol}^{-1}$$

A large positive lattice enthalpy means a very stable crystal. Lattice enthalpy increases with higher ionic charge and smaller ionic radius (force $\propto q_1 q_2 / r^2$). This is why $MgO$ ($+2,-2$, small ions) has a far larger lattice enthalpy than $NaCl$ ($+1,-1$).

The Born-Haber Cycle

We cannot measure lattice enthalpy directly, so we calculate it using Hess's law in a thermochemical cycle. For $NaCl$, the formation enthalpy is broken into measurable steps:

$$\Delta_f H = \Delta_{sub}H(Na) + \tfrac{1}{2}\Delta_{diss}H(Cl_2) + \Delta_i H(Na) + \Delta_{eg}H(Cl) + (-\Delta_{lattice}H)$$

where $\Delta_{sub}H$ = sublimation, $\Delta_{diss}H$ = bond dissociation of $Cl_2$, $\Delta_i H$ = ionisation enthalpy of $Na$, and $\Delta_{eg}H$ = electron gain enthalpy of $Cl$.

[JEE Tip] The Born-Haber cycle is essentially Hess's law applied to ionic-solid formation. In a numerical, write every step with the correct sign, then solve for the unknown — usually the lattice enthalpy.

Properties of Ionic Compounds

The strong, non-directional electrostatic forces in an ionic lattice explain all their characteristic properties:

• High melting and boiling points — large lattice enthalpy means a lot of energy is needed to pull the ions apart ($NaCl$ melts at $801\,^{\circ}C$).
• Hard but brittle — the lattice is rigid, but when a layer is pushed so like charges align, repulsion shatters the crystal.
• Conduct electricity in molten state or aqueous solution, not when solid — ions are locked in place in the solid; in melt or solution they become mobile charge carriers.
• Soluble in polar solvents (water), insoluble in non-polar solvents — polar water molecules surround and stabilise the ions (hydration).
• Non-directional bond — the electrostatic field of an ion acts equally in all directions, so ionic compounds do not have discrete 'molecules'; the formula $NaCl$ is just the simplest whole-number ratio.

Key Point: An ionic 'molecule' does not really exist in the solid — what exists is a giant 3-D lattice. The formula unit only gives the ratio of ions.

[NEET Important] Solid $NaCl$ does NOT conduct electricity; molten or aqueous $NaCl$ does. This is a frequently tested fact.

Solved Examples

Example 1: Reading a lattice enthalpy value

The lattice enthalpy of $KCl$ is $+715$ kJ mol$^{-1}$. What does the positive sign tell you?

Solution:

1. Definition: Lattice enthalpy is the energy needed to separate the solid into gaseous ions.
2. Positive sign: energy must be supplied — the process is endothermic, confirming the ions are strongly bound.
3. Interpretation: The larger this value, the more stable the crystal.

Takeaway: A bigger (more positive) lattice enthalpy = stronger ionic bonding = higher melting point.

Example 2: Comparing lattice enthalpies by charge

Arrange $NaF$, $MgO$ and $NaCl$ in increasing order of lattice enthalpy.

Solution:

1. Charges: $NaF$ and $NaCl$ have $+1,-1$; $MgO$ has $+2,-2$.
2. Force $\propto q_1 q_2 / r^2$: doubling both charges makes $MgO$'s attraction far stronger → highest lattice enthalpy.
3. Between $NaF$ and $NaCl$: $F^-$ is smaller than $Cl^-$, so $r$ is smaller in $NaF$ → larger lattice enthalpy.
4. Order (increasing): $NaCl < NaF < MgO$.

Takeaway: Charge dominates over size; for the same charge, smaller ion → larger lattice enthalpy.

Example 3: A Born-Haber calculation

For $NaCl$: $\Delta_{sub}H(Na)=108$, $\Delta_i H(Na)=496$, $\tfrac{1}{2}\Delta_{diss}H(Cl_2)=121$, $\Delta_{eg}H(Cl)=-349$, $\Delta_f H(NaCl)=-411$ (all kJ mol$^{-1}$). Find the lattice enthalpy.

Solution:

1. Born-Haber: $\Delta_f H = \Delta_{sub}H + \Delta_i H + \tfrac{1}{2}\Delta_{diss}H + \Delta_{eg}H - \Delta_{lattice}H$.
2. Substitute: $-411 = 108 + 496 + 121 - 349 - \Delta_{lattice}H$.
3. Simplify: $-411 = 376 - \Delta_{lattice}H$.
4. Solve: $\Delta_{lattice}H = 376 + 411 = 787$ kJ mol$^{-1}$.

Takeaway: Keep track of signs; the lattice term is subtracted because lattice formation releases energy.

Example 4: Why solid NaCl doesn't conduct but molten NaCl does

Explain the difference in electrical conductivity.

Solution:

1. In the solid: $Na^+$ and $Cl^-$ are locked in fixed lattice positions — no mobile charges, so no conduction.
2. On melting: the lattice collapses; ions become free to move.
3. Result: molten (or aqueous) $NaCl$ conducts electricity because the now-mobile ions carry charge.

Takeaway: Conductivity needs mobile charge carriers — present in melt/solution, absent in the solid.

Example 5: Predicting ionic vs covalent from electronegativity

Given EN values $Na = 0.9$, $Cl = 3.0$, $H = 2.1$, decide which of $NaCl$ and $HCl$ is more ionic.

Solution:

1. EN difference $NaCl$: $3.0 - 0.9 = 2.1$.
2. EN difference $HCl$: $3.0 - 2.1 = 0.9$.
3. Rule: larger $\Delta EN$ → more ionic character. $2.1 > 0.9$.
4. Conclusion: $NaCl$ is far more ionic; $HCl$ is a polar covalent molecule.

Takeaway: $\Delta EN > 1.7$ usually means predominantly ionic; smaller values mean polar covalent.

Example 6: Brittleness of ionic crystals

Why do ionic crystals shatter when struck, despite being hard?

Solution:

1. Hardness: strong electrostatic forces hold ions firmly → hard.
2. On striking: one layer of ions slides over another.
3. Like charges align: cations come opposite cations, anions opposite anions → strong repulsion.
4. Result: the layers fly apart along that plane — the crystal cleaves/shatters.

Takeaway: Rigidity + charge-alignment repulsion = hard but brittle.

Example 7: Effect of ion size on melting point

Why does $LiF$ have a higher melting point than $LiI$?

Solution:

1. Same charges ($+1,-1$) for both.
2. Anion size: $F^-$ is much smaller than $I^-$, so inter-ionic distance $r$ is smaller in $LiF$.
3. Lattice enthalpy $\propto 1/r$: smaller $r$ → larger lattice enthalpy in $LiF$.
4. Conclusion: $LiF$ needs more energy to melt → higher melting point.

Takeaway: For equal charges, the compound with smaller ions melts higher.

Example 8: Identifying the most ionic compound

Among $AlCl_3$, $MgCl_2$ and $NaCl$, which is the most ionic?

Solution:

1. Fajans' preview: higher cation charge and smaller size → more covalent character (more polarising).
2. Cation charges: $Al^{3+} > Mg^{2+} > Na^{+}$ in charge; $Al^{3+}$ is also smallest → most polarising → most covalent.
3. Conclusion: $NaCl$ (largest, lowest-charge cation) is the most ionic; $AlCl_3$ is the least.

Takeaway: A small, highly charged cation polarises the anion and adds covalent character (Fajans' rules).

Example 9: Why $MgO$ is used as a refractory

$MgO$ has a melting point of about $2850\,^{\circ}C$. Relate this to its bonding.

Solution:

1. Ion charges: $Mg^{2+}$ and $O^{2-}$ — both doubly charged.
2. Lattice enthalpy: force $\propto (2)(2)/r^2 = 4/r^2$ — about four times that of a $+1,-1$ pair at similar distance.
3. Consequence: enormous lattice enthalpy → extremely high melting point → ideal as a heat-resistant (refractory) lining.

Takeaway: Doubly charged small ions give exceptionally strong, high-melting lattices.

Example 10: Solubility reasoning

Why is $NaCl$ soluble in water but not in petrol (a non-polar solvent)?

Solution:

1. Water is polar: its $\delta^-$ oxygen surrounds $Na^+$ and its $\delta^+$ hydrogens surround $Cl^-$ (hydration), releasing hydration energy that overcomes the lattice enthalpy.
2. Petrol is non-polar: it cannot stabilise ions, so it cannot supply enough energy to break the lattice.
3. Conclusion: dissolves in polar water, insoluble in non-polar petrol.

Takeaway: 'Like dissolves like' — polar/ionic solutes dissolve in polar solvents.

Example 11: Sign convention in electron gain enthalpy

In a Born-Haber cycle, why is $\Delta_{eg}H(Cl) = -349$ kJ mol$^{-1}$ written with a negative sign?

Solution:

1. Electron gain by Cl: $Cl(g) + e^- \rightarrow Cl^-(g)$ releases energy.
2. Sign convention: energy released → exothermic → negative $\Delta H$.
3. Use in cycle: it is added as $-349$, lowering the overall enthalpy of formation.

Takeaway: Energy released = negative sign; energy absorbed (sublimation, ionisation) = positive sign.

Example 12: Charge balance in an ionic formula

Write the formula of calcium phosphate from $Ca^{2+}$ and $PO_4^{3-}$.

Solution:

1. Charges: $Ca^{2+}$ and $PO_4^{3-}$.
2. LCM of 2 and 3 = 6: need 3 $Ca^{2+}$ ($+6$) and 2 $PO_4^{3-}$ ($-6$).
3. Formula: $Ca_3(PO_4)_2$ (neutral overall).

Takeaway: Cross-over and reduce the charges to get the simplest neutral ratio.