Ionic bonding — Born-Haber cycle, lattice energy (Kapustinskii equation): Worked examples

Symbols & sign conventions (read this first)

Every example below uses these five arrows. Before you touch a number, know exactly what each one means and which sign it carries.

Definition The Born–Haber vocabulary

$Δ H_{f}$ — enthalpy of formation. Energy change when one mole of the compound forms from its elements in their standard states (e.g. $N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)$ ). For stable salts it is negative (exothermic). See Enthalpy of Formation.
$Δ H_{a t o m}$ — enthalpy of atomisation. Energy to turn one mole of an element into free gaseous atoms (e.g. $N a (s) \to N a (g)$ ). Always positive — you must pull atoms apart.
$I E$ — ionisation energy. Energy to remove one electron from a gaseous atom/ion ( $M (g) \to M^{+} (g) + e^{-}$ ). $I E_{1}$ is the first, $I E_{2}$ the second. Always positive (endothermic — electrons are held). See Ionisation Energy.
$E A$ — electron affinity. Energy change when a gaseous atom/ion gains one electron ( $X (g) + e^{-} \to X^{-} (g)$ ). $E A_{1}$ is the first, $E A_{2}$ the second. $E A_{1}$ for a neutral atom is usually negative (exothermic); a second $E A$ onto an already-negative ion is positive (endothermic). See Electron Affinity.
$D$ — bond dissociation energy. Energy to break one mole of a covalent bond into gaseous atoms ( $C l_{2} (g) \to 2 C l (g)$ ). Always positive. We often need only half a mole of $C l_{2}$ to make one $C l$ atom, hence the recurring $\frac{1}{2} D$ .
$U$ — lattice energy. The unknown we usually solve for; negative for lattice formation (parent-note convention).

Sign rule of thumb: endothermic (energy in, arrow up the diagram) $= +$ ; exothermic (energy out, arrow down) $= -$ .

Definition Where Kapustinskii's mystery numbers come from

$U \approx - \frac{1.202 \times 1 0 ^{5} ν z _{+} z _{-}}{r _{+} + r _{-}} (1 - \frac{34.5}{r _{+} + r _{-}}) kJ/mol$

$ν$ (Greek "nu") is the total number of ions in one formula unit — count every ion, not the number of element types. NaCl has one $N a^{+}$ and one $C l^{-}$ , so $ν = 2$ ; CaF $_{2}$ has one $C a^{2 +}$ and two $F^{-}$ , so $ν = 3$ . Writing $ν = 2$ for CaF₂ is a classic trap.
$z_{+}, z_{-}$ are the magnitudes of the individual ion charges (e.g. $C a^{2 +} \Rightarrow z_{+} = 2$ ), never the net formula charge (which would be $0$ ).
$r_{+}, r_{-}$ are the cation and anion radii, in picometres (pm).
$1.202 \times 1 0^{5}$ is not magic: it is all the physical constants of Born–Landé ( $N_{A}$ , $e^{2}$ , $4 π ε_{0}$ ) times the average Madelung-per-ion ratio, bundled into one number. Its units are chosen so that when $r$ is in pm and charges are pure numbers, $U$ comes out in kJ/mol. So $r_{+}, r_{-}$ must be in pm — feed it metres or Å and the answer is nonsense.
$34.5$ (pm) plays the role of the Born $(1 - 1/ n)$ repulsion correction: it is an empirical average distance that mimics how electron-cloud repulsion softens the attraction. It has units of pm so that $34.5/ (r_{+} + r_{-})$ is dimensionless.

The scenario matrix

Every problem in this topic is one of these cells. The label in the last column tells you which worked example nails it.

#	Case class	The twist it tests	Example
A	1:1 salt, both EA exothermic	baseline sign bookkeeping	Ex 1
B	Sign trap: EA₂ endothermic	$O^{2 -}$ needs two EAs, one positive	Ex 2
C	High charge product ( $z_{+} z_{-} = 4$ )	why MgO ≫ NaCl	Ex 3
D	Non-1:1 formula unit (CaF₂, $ν = 3$ )	counting ions, splitting stoichiometry	Ex 4
E	Kapustinskii vs Born–Landé	two routes, same salt, compare	Ex 5
F	Size limit (fix charge, vary $r$ )	$U \to 0$ as $r \to \infty$ ; LiF vs LiI	Ex 6
G	Covalent-character detective	experimental $U$ ≠ ionic prediction	Ex 7
H	Real-world word problem	melting point from $U$	Ex 8
I	Exam twist: solve for a missing step	unknown is EA, not $U$	Ex 9

Two "degenerate/limit" checks live inside the examples: the zero-charge case (a neutral atom pair gives $U = 0$ — Ex 6 discusses) and the infinite-separation limit (Ex 6 figure).

Ex 1 — Cell A: the baseline 1:1 salt (KCl)

Worked example Find

$U$ (KCl) from Born–Haber data Data (kJ/mol): $Δ H_{f} = - 437, Δ H_{a t o m} (K) = + 89, I E_{1} (K) = + 419, \frac{1}{2} D_{C l - C l} = + 122, E A_{1} (C l) = - 349.$

Forecast: guess the sign and rough size of $U$ before reading on. (Hint: it's a strong ionic solid — expect a big negative number, a bit weaker than NaCl's $- 788$ because $K^{+}$ is bigger than $N a^{+}$ .)

Step 1 — write the cycle equation. $Δ H_{f} = Δ H_{a t o m} + I E_{1} + \frac{1}{2} D + E A_{1} + U$ Why this step? This is Hess's Law: the total enthalpy going elements → compound is the same whether we go direct ( $Δ H_{f}$ ) or the long way through gaseous ions. Every step is one measurable arrow (all defined in the vocabulary box above).

Step 2 — rearrange for the unknown $U$ . $U = Δ H_{f} - Δ H_{a t o m} - I E_{1} - \frac{1}{2} D - E A_{1}$ Why this step? $U$ is the one thing we cannot measure directly, so we isolate it.

Step 3 — substitute with signs intact. $U = - 437 - 89 - 419 - 122 - (- 349)$ Why this step? Each number keeps its own sign: the " $- (- 349)$ " flips because subtracting an exothermic (already-negative) step.

Step 4 — add. $U = - 437 - 89 - 419 - 122 + 349 = - 718 kJ/mol$

Verify: Sign is negative ✔ (formation is exothermic). Magnitude $718 < 788$ (NaCl) ✔ — bigger cation $K^{+}$ means larger $r_{0}$ , so Coulomb's Law gives weaker attraction, exactly the forecast. Units: every term is kJ/mol, so $U$ is kJ/mol ✔.

Ex 2 — Cell B: the sign trap (MgO, two electron affinities)

Worked example Build

$U$ (MgO) with EA₂ endothermic Data (kJ/mol): $Δ H_{f} = - 602, Δ H_{a t o m} (M g) = + 148, I E_{1} (M g) = + 738, I E_{2} (M g) = + 1451, \frac{1}{2} D_{O = O} = + 249, E A_{1} (O) = - 141, E A_{2} (O) = + 798.$

Forecast: MgO has doubly charged ions — expect $U$ far more negative than NaCl, around $- 3800$ .

Step 1 — list every arrow to form $M g^{2 +}$ and $O^{2 -}$ . To make $M g^{2 +}$ you ionise twice: $I E_{1} + I E_{2}$ . To make $O^{2 -}$ you add two electrons: $E A_{1} + E A_{2}$ . Why this step? Missing the second ionisation or second EA is the classic error. Each charge unit is a separate physical step.

Step 2 — flag the sign of $E A_{2}$ . $E A_{1} (O) = - 141$ (exothermic: a neutral O happily grabs an electron). $E A_{2} (O) = + 798$ (endothermic!). Why this step? You are shoving a second electron onto an already-negative $O^{-}$ — like pushing a magnet's like-poles together, it costs energy. This positive term is the whole point of the cell.

Step 3 — assemble the cycle and solve for $U$ . $U = Δ H_{f} - Δ H_{a t o m} - I E_{1} - I E_{2} - \frac{1}{2} D - E A_{1} - E A_{2}$ $U = - 602 - 148 - 738 - 1451 - 249 - (- 141) - 798$ Why this step? This is the same Hess's Law loop as Ex 1, but with two extra arrows ( $I E_{2}$ and $E A_{2}$ ) because both ions are doubly charged. We isolate $U$ by moving every measurable step to the right-hand side; each term carries its own sign from Step 2.

Step 4 — add carefully. $U = - 602 - 148 - 738 - 1451 - 249 + 141 - 798 = - 3845 kJ/mol$

Verify: $- 3845$ is close to the accepted $\approx - 3800$ ✔ and $\sim 5 \times$ NaCl, matching the $z_{+} z_{-} = 4$ leverage ✔. If you had forgotten $E A_{2}$ 's $+ 798$ (i.e. dropped that $- 798$ from the sum), you'd get $- 3047$ — far too weak, a giveaway you left out the endothermic step.

Ex 3 — Cell C: charge product as the biggest lever (compare NaCl vs MgO)

Worked example Why is MgO's lattice energy ~5× NaCl's?

Use the Born–Landé ratio $U \propto \frac{z _{+} z _{-}}{r _{0}}$ (holding the other constants comparable).

Forecast: guess the ratio $U (MgO) / U (NaCl)$ before computing.

Step 1 — the charge factor. NaCl: $z_{+} z_{-} = 1 \times 1 = 1$ . MgO: $z_{+} z_{-} = 2 \times 2 = 4$ . Why this step? In Born–Landé the charge product multiplies the whole thing — quadrupling it is the dominant effect.

Step 2 — the size factor. $r_{0} (NaCl) = 102 + 181 = 283$ pm. $r_{0} (MgO) = 72 + 140 = 212$ pm. Why this step? $r_{0}$ sits in the denominator, so the smaller MgO separation adds to the charge boost.

Step 3 — combine into a ratio. $\frac{U ( MgO )}{U ( NaCl )} \approx \frac{z _{+} z _{-} ( MgO )}{z _{+} z _{-} ( NaCl )} \times \frac{r _{0} ( NaCl )}{r _{0} ( MgO )} = \frac{4}{1} \times \frac{283}{212} = 5.34$ Why this step? We ignore the small differences in Madelung constant and Born exponent, which nearly cancel, isolating the two dominant levers.

Verify: $- 788 \times 5.34 = - 4210$ ; the true MgO value ( $\approx - 3800$ ) is in this ballpark ✔. The charge product (factor 4) dominates; size adds another $\sim 33%$ ✔ — matching the parent note's "biggest lever" claim.

Ex 4 — Cell D: non-1:1 formula unit (CaF₂ by Kapustinskii)

Worked example Estimate

$U$ (CaF₂), $ν = 3$ Radii: $r (Ca^{2 +}) = 100$ pm, $r (F^{-}) = 133$ pm. Charges $z_{+} = 2, z_{-} = 1$ .

Forecast: two-plus cation → sizable $U$ ; but the anion charge is only 1, so expect between NaCl and MgO, roughly $- 2600$ .

Step 1 — count $ν$ . CaF₂ has one $C a^{2 +}$ and two $F^{-}$ → $ν = 3$ ions per formula unit. Why this step? As defined alongside the formula, $ν$ counts total ions, not atom types. Writing $ν = 2$ is the trap.

Step 2 — use ion charge magnitudes, not net. $z_{+} z_{-} = 2 \times 1 = 2$ . Not the net formula charge (which is 0). Why this step? The formula is built from pairwise Coulomb interactions between individual ions; net charge is meaningless here.

Step 3 — plug into Kapustinskii. $r_{+} + r_{-} = 100 + 133 = 233$ pm. $U \approx - \frac{1.202 \times 1 0 ^{5} ν z _{+} z _{-}}{r _{+} + r _{-}} (1 - \frac{34.5}{r _{+} + r _{-}})$ $U \approx - \frac{1.202 \times 1 0 ^{5} \times 3 \times 2}{233} (1 - \frac{34.5}{233})$ Why this step? We reach for Kapustinskii (not Born–Landé) precisely because we don't know CaF₂'s Madelung constant here. The formula asks only for what we do know — ion count $ν$ , charge magnitudes, and radii (in pm, as the constant demands) — and packs the crystal geometry into that $1.202 \times 1 0^{5}$ bundle.

Step 4 — evaluate. Front factor $= - \frac{721200}{233} = - 3095.3$ . Correction $= 1 - 0.1481 = 0.8519$ . $U \approx - 3095.3 \times 0.8519 = - 2637 kJ/mol$

Verify: Accepted $U$ (CaF₂) $\approx - 2630$ kJ/mol ✔ — within $0.3%$ . Sign negative ✔. Units kJ/mol ✔ (because $r$ was in pm). Forecast of " $\approx - 2600$ " spot on.

Ex 5 — Cell E: two routes, same salt (Kapustinskii vs Born–Haber for NaCl)

Worked example Do both methods agree for NaCl?

Kapustinskii inputs: $ν = 2, z_{+} = z_{-} = 1, r_{+} = 102, r_{-} = 181$ . Born–Haber gave $- 788$ (parent note).

Forecast: guess whether the two methods land within $\sim 10%$ .

Step 1 — Kapustinskii route. $r_{+} + r_{-} = 283$ . $U = - \frac{1.202 \times 1 0 ^{5} \times 2 \times 1}{283} (1 - \frac{34.5}{283}) = - 849.5 \times 0.8781 = - 746 kJ/mol$ Why this step? This is the purely ionic, structure-free estimate — no thermodynamic data needed.

Step 2 — compare to the Born–Haber (experimental) value $- 788$ . Difference $= ∣ - 788 - (- 746) ∣ = 42$ kJ/mol, i.e. $42/788 = 5.3%$ . Why this step? Born–Haber includes real bonding (any covalent character); Kapustinskii assumes pure ions. A small gap → NaCl is nearly perfectly ionic.

Verify: $5.3%$ agreement ✔ (matches parent note's "~5%"). Because Born–Haber is slightly more exothermic, there is a whisper of covalent character — but tiny, consistent with Fajans' Rules (Na⁺ has low polarising power).

Ex 6 — Cell F + limits: fix charge, vary size (LiF vs LiI, and $r \to \infty$ )

Worked example Same charges, different anion size

Kapustinskii for LiF and LiI. $r (Li^{+}) = 76$ . $r (F^{-}) = 133$ , $r (I^{-}) = 220$ . Both $ν = 2, z_{+} z_{-} = 1$ .

Forecast: which has the larger (more negative) $U$ ? Big anion I⁻ → larger $r_{0}$ → weaker. So LiF wins.

Step 1 — LiF. $r_{+} + r_{-} = 76 + 133 = 209$ . $U = - \frac{1.202 \times 1 0 ^{5} \times 2}{209} (1 - \frac{34.5}{209}) = - 1150.2 \times 0.8349 = - 960 kJ/mol$

Step 2 — LiI. $r_{+} + r_{-} = 76 + 220 = 296$ . $U = - \frac{1.202 \times 1 0 ^{5} \times 2}{296} (1 - \frac{34.5}{296}) = - 812.2 \times 0.8834 = - 718 kJ/mol$ Why these steps? Only $r_{0}$ changed; the drop from $- 960$ to $- 718$ isolates the size effect cleanly.

Step 3 — the limiting cases (degenerate inputs). The figure below plots the Kapustinskii $U$ (with charges fixed at $z_{+} z_{-} = 1$ ) against the ion-pair separation $r_{+} + r_{-}$ on the horizontal axis, with $U$ in kJ/mol on the vertical axis. Read it left-to-right:

The red curve starts steeply negative (small ions, strong attraction) and climbs toward zero as the ions get further apart.
The two black dots mark our two salts: LiF at $209$ pm sits lower (more negative) than LiI at $296$ pm — the geometric proof that the bigger anion weakens the bond.
Follow the curve to the far right: as $r_{+} + r_{-} \to \infty$ , the black arrow points to $U \to 0$ — infinitely separated ions feel no attraction. And if either charge were $0$ (a neutral atom), then $z_{+} z_{-} = 0$ and the whole formula gives $U = 0$ — no ionic bond at all, the zero-charge degenerate case.
Alt-text: graph, horizontal axis "ionic separation $r_{+} + r_{-}$ (pm)", vertical axis "lattice energy $U$ (kJ/mol)"; a red curve rising from strongly negative up toward the zero line; two black dots labelled LiF (209 pm, ≈−960) and LiI (296 pm, ≈−718); an arrow annotating "as separation grows, $U \to 0$ ".

Verify: LiF ( $- 960$ ) $>$ LiI ( $- 718$ ) in magnitude ✔ — the forecast. As the red curve shows, $U \to 0^{-}$ for large $r$ ✔, and the curve never crosses zero for finite $r$ (always attractive when charges non-zero) ✔.

Ex 7 — Cell G: covalent-character detective (AgCl)

Worked example Is AgCl "purely ionic"?

Born–Haber (experimental) $U (AgCl) = - 905$ kJ/mol. Kapustinskii (ionic model) gives: $ν = 2, z_{+} z_{-} = 1, r (Ag^{+}) = 115, r (Cl^{-}) = 181$ .

Forecast: Ag⁺ is a soft, polarising cation — expect the experimental value to be more exothermic than the ionic prediction.

Step 1 — ionic-model prediction. $r_{+} + r_{-} = 296$ . $U_{i o ni c} = - \frac{1.202 \times 1 0 ^{5} \times 2}{296} (1 - \frac{34.5}{296}) = - 812.2 \times 0.8834 = - 718 kJ/mol$ Why this step? This assumes pure point-charge ions — the pure-ionic benchmark.

Step 2 — compare to experiment. Experimental $- 905$ is more negative than predicted $- 718$ by $187$ kJ/mol. Why this step? Extra binding not explained by ionic Coulomb attraction must come from shared electrons — covalent character.

Verify: Gap $= 905 - 718 = 187$ kJ/mol, i.e. $\sim 26%$ extra ✔. By Fajans' Rules, small highly-charged or soft cations like Ag⁺ polarise the anion cloud, adding covalency — exactly what a $\sim 26%$ discrepancy signals. (Contrast NaCl's mere $5%$ in Ex 5.)

Ex 8 — Cell H: real-world word problem (which brick survives the furnace?)

Worked example Choosing a refractory lining

A furnace runs at $2000^{\circ}$ C. You must line it with either NaCl-type or MgO-type crystal. Melting points: NaCl $801^{\circ}$ C, MgO $2852^{\circ}$ C. Using lattice energies, justify the choice.

Forecast: melting = pulling the lattice apart, so higher $U$ magnitude → higher melting point → MgO.

Step 1 — link $U$ to melting. Melting overcomes the electrostatic grid holding ions in place; the energy needed scales with $∣ U ∣$ . Why this step? It turns an abstract enthalpy into a physical, testable property.

Step 2 — compare $∣ U ∣$ . $∣ U (MgO) ∣ \approx 3800$ vs $∣ U (NaCl) ∣ \approx 788$ — a factor $\sim 4.8$ . Why this step? From Ex 3, MgO's charge product (4) and small ions make it far harder to melt.

Step 3 — decide. $2000^{\circ}$ C exceeds NaCl's $801^{\circ}$ C (it would melt!) but is below MgO's $2852^{\circ}$ C. Choose MgO. Why this step? The engineering question is binary — "does it stay solid at operating temperature?" — so we test $2000$ against each melting point. A lining that melts fails catastrophically, so only the crystal whose melting point exceeds $2000^{\circ}$ C is safe, and the larger $∣ U ∣$ is exactly what buys MgO that margin.

Verify: MgO's melting point $2852 > 2000 > 801$ ✔ so only MgO stays solid ✔. The ratio $∣ U ∣_{MgO} /∣ U ∣_{NaCl} = 3800/788 = 4.82$ tracks the melting-point jump ✔.

Ex 9 — Cell I: exam twist — solve for the missing step (find EA of Br)

Worked example The unknown isn't

$U$ — it's the electron affinity For KBr the lattice energy is known from another method: $U = - 671$ kJ/mol. Also given (kJ/mol): $Δ H_{f} = - 392, Δ H_{a t o m} (K) = + 89, I E_{1} (K) = + 419, \frac{1}{2} D_{B r - B r} = + 97.$ Find $E A_{1} (Br)$ .

Forecast: Br readily gains an electron → expect a negative (exothermic) EA around $- 320$ .

Step 1 — write the full cycle. $Δ H_{f} = Δ H_{a t o m} + I E_{1} + \frac{1}{2} D + E A_{1} + U$ Why this step? Same Hess loop — but now $E A_{1}$ is the lone unknown.

Step 2 — isolate $E A_{1}$ . $E A_{1} = Δ H_{f} - Δ H_{a t o m} - I E_{1} - \frac{1}{2} D - U$ Why this step? Algebra: move everything else to the other side.

Step 3 — substitute. $E A_{1} = - 392 - 89 - 419 - 97 - (- 671) = - 392 - 89 - 419 - 97 + 671$ Why this step? Keep signs intact; the $- (- 671)$ becomes $+ 671$ because subtracting the negative (exothermic) lattice energy.

Step 4 — add. $E A_{1} = - 326 kJ/mol$

Verify: $- 326$ is close to the accepted $E A (Br) \approx - 325$ kJ/mol ✔, and negative as forecast ✔ (halogens release energy grabbing an electron). This shows the cycle is a solve-for-any-unknown machine — swap which arrow you don't know.

Recall checkpoints

Recall Why does subtracting an exothermic EA flip its sign in the algebra?

Because in $U = Δ H_{f} - \dots - E A$ , the $E A$ term is already negative, so $- (- 349) = + 349$ . ::: The double negative adds energy back when you isolate $U$ .

Recall In Kapustinskii, CaF₂ uses

$ν = ?$ and $z_{+} z_{-} = ?$ $ν = 3$ (three ions per formula unit), $z_{+} z_{-} = 2 \times 1 = 2$ (individual magnitudes, not net charge). ::: $ν = 3, z_{+} z_{-} = 2$

Recall What does a large gap between experimental

$U$ and Kapustinskii $U$ signal? Extra covalent character — the pure-ionic model under-binds, so shared-electron bonding makes up the difference (Fajans). ::: Covalent character.

See also: Ionic Radii, Ionisation Energy, Enthalpy of Formation, Lattice Energy → Solubility.