Ionic bonding — Born-Haber cycle, lattice energy (Kapustinskii equation): Question bank

True or false — justify

The lattice energy of formation is always negative.

True — forming a crystal from scattered gaseous ions releases energy (attraction wins), so

Δ H_{l a tt} < 0

; only the reverse (breaking the lattice) is positive.

A larger Madelung constant means a more stable lattice, all else equal.

True — a larger Madelung constant packs more net attractive geometry per ion, so the summed Coulomb energy is more negative and the crystal is bound more strongly.

Doubling both ionic charges roughly quadruples the lattice energy.

True in the leading term —

U \propto z_{+} z_{-}

, so

1 \times 1 \to 2 \times 2

multiplies the charge product by 4, though a smaller

r_{0}

for the higher-charged ions makes the real jump even larger.

If two salts have the same charges, the one with the larger ions has the larger lattice energy.

False — larger ions mean a larger separation

r_{0}

in the denominator, so

U

is smaller in magnitude (e.g. LiI weaker than LiF).

The Kapustinskii equation needs the crystal structure to work.

False — its whole point is to avoid the Madelung constant; it needs only charges, ion count

ν

, and ionic radii.

Electron affinity is always an exothermic step in a Born–Haber cycle.

False — the first Electron Affinity usually releases energy, but the second (e.g.

O^{-} \to O^{2 -}

) is endothermic because you push an electron onto an already-negative ion.

The Born exponent

n

correction makes the predicted lattice energy more negative.

False — the factor

(1 - \frac{1}{n})

is less than 1, so it reduces the magnitude, accounting for the short-range Pauli repulsion that opposes collapse.

Born–Haber and Born–Landé must give exactly the same lattice energy.

False — a real gap (Born–Haber more exothermic) reveals covalent character; agreement means the bonding is nearly purely ionic (Fajans' Rules).

Ionisation being endothermic means NaCl should not form.

False — ionisation costs energy, but the huge exothermic lattice energy pays it back and more, giving a negative overall Enthalpy of Formation.

MgO has a higher melting point than NaCl purely because Mg is a metal.

False — the cause is its far larger lattice energy (charge product 4 and small ions), which takes more thermal energy to break apart.

Spot the error

U = Δ H_{f} + Δ H_{a t o m} + I E_{1} + \frac{1}{2} D + E A

, so solving gives

U = Δ H_{f} + (sum)

The error is sign on rearranging — the cycle reads

Δ H_{f} = sum + U

, so

U = Δ H_{f} - Δ H_{a t o m} - I E_{1} - \frac{1}{2} D - E A

; every other term is subtracted.

"For CaF

_{2}

ν = 2

because there are two kinds of ion."

Wrong —

ν

counts ions per formula unit, not ion types; CaF

_{2}

has one Ca

^{2 +}

plus two F

^{-}

, so

ν = 3

"Net charge of NaCl is 0, so

z_{+} z_{-} = 0

in Kapustinskii."

Wrong —

z_{+}

and

z_{-}

are individual ion magnitudes (

1

and

1

), giving product

1

; you never use the neutral net charge.

^{-} \to O^{2 -}

releases energy since making an anion is favourable."

Wrong — this step is endothermic; a negative ion repels the incoming electron, so energy must be supplied.

"The half in

\frac{1}{2} D_{C l - C l}

means bond energy is halved because it's weak."

Wrong — the

\frac{1}{2}

comes from needing only one Cl atom, so you break half a mole of Cl

_{2}

bonds, not from bond weakness.

"Lattice energy is measured directly with a calorimeter."

Wrong — it cannot be measured directly; it is deduced by closing the Born–Haber loop via Hess's Law.

"Since

E A

of Cl is

- 349

, we write

U = Δ H_{f} - \dots - 349

Wrong — you subtract the signed value

- 349

, i.e.

- (- 349) = + 349

; failing to keep the sign flips the result by ~700 kJ/mol.

Why questions

Why is lattice energy so much larger than a single ion-pair Coulomb energy?

Because each ion is attracted to many neighbours in every direction across the infinite lattice, and summing these (Madelung sum) multiplies the pairwise value.

Why does Kapustinskii include the term

(1 - 34.5/ (r_{+} + r_{-}))

It mimics the Born short-range repulsion correction

(1 - \frac{1}{n})

, shrinking

U

slightly to account for electron-cloud repulsion at close range.

Why does the charge product matter more than ionic size in comparing MgO with NaCl?

Charge appears as a product (

z_{+} z_{-} = 4

1

), a fourfold lever, whereas radius changes

r_{0}

by only a modest fraction — so charge dominates the ranking.

Why can Born–Haber diagnose covalent character?

The ionic model assumes point charges; if experiment is more exothermic than the model predicts, extra bonding (electron sharing / polarisation) is present, exactly what Fajans' Rules describe.

Why is the Madelung constant not simply the number of nearest neighbours?

Because it sums all shells — nearest attract, next-nearest repel, and so on — as a converging alternating series, not just the first ring.

Why do small, highly-charged cations give exceptionally large lattice energies?

They increase

z_{+}

(numerator lever) and shrink

r_{0}

(denominator), so both factors in

U \propto z_{+} z_{-} / r_{0}

push the magnitude up together.

Why does a more negative lattice energy generally lower solubility in water?

A very stable crystal resists being pulled apart by hydration; if lattice energy outweighs hydration energy, dissolving is unfavourable (Lattice Energy → Solubility).

Edge cases

For a compound whose crystal structure is unknown, which lattice-energy route survives?

Kapustinskii — it needs no Madelung constant, only charges,

ν

, and radii, so it works where Born–Landé cannot.

In the oxide Born–Haber cycle, how many electron-affinity steps appear?

Two — EA₁ (

O \to O^{-}

, exothermic) and EA₂ (

O^{-} \to O^{2 -}

, endothermic); omitting the endothermic one is a classic error.

What happens to the

(1 - \frac{1}{n})

factor as the Born exponent

n

grows very large?

It tends to 1, meaning repulsion becomes negligibly soft and the lattice energy approaches the pure Coulomb (Madelung) value.

What if two salts have identical

r_{+} + r_{-}

but different

ν

and charges?

Kapustinskii separates them through

ν z_{+} z_{-}

in the numerator, so the one with more ions or higher charge wins the larger magnitude.

If experimental and calculated

U

agree almost perfectly, what does that tell you?

The bonding is essentially purely ionic — the point-charge model already captures it, leaving little room for covalent contribution.

Recall Rapid self-test

The three deadliest traps: (1) signs in the Born–Haber rearrangement, (2) second electron affinity being endothermic, (3) $ν$ vs net charge in Kapustinskii. Which trap flips your answer by ~700 kJ/mol if you miss it? ::: Dropping the sign on $E A$ (subtracting $- 349$ vs $+ 349$ ). Which trap makes you count CaF $_{2}$ as $ν = 2$ instead of $3$ ? ::: Confusing "types of ion" with "ions per formula unit".