Ionic bonding — Born-Haber cycle, lattice energy (Kapustinskii equation): Visual walkthrough

Everything here builds on Coulomb's Law; we will meet it in Step 2.

Step 1 — Two ions, one ruler: what does "charge" even mean here?

WHAT. We start with the smallest possible ionic world: one positive ion and one negative ion floating in empty space, a distance apart.

WHY. A whole crystal is millions of ions, but every one of those interactions is just this two-ball picture repeated. If we cannot handle two, we have no hope with millions. So we earn the pieces here first.

PICTURE. Look at the figure. The red ball carries a charge we write as $z_{+} e$ ; the blue ball carries $z_{-} e$ .

Let us decode those symbols slowly:

So $z_{+} e$ literally reads " $z_{+}$ bricks of charge, each of size $e$ ". Nothing mysterious — it is counting.

Step 2 — Coulomb's Law: the energy stored in that gap

WHAT. We attach an energy number to the pair using Coulomb's Law.

WHY THIS TOOL. We could ask many questions about two charges — the force between them, the field, the energy. We want energy because lattice energy is an energy (kJ per mole), and because energies simply add up when we later combine many pairs. Forces are vectors and point in directions — a nightmare to sum. Energies are single numbers. That is why we pick potential energy, not force.

PICTURE. The figure shows the pair with a spring-like "energy in the gap". The formula sits right on the arrow between them:

$E_{p ai r} = nature’s constant \frac{1}{4 π ε _{0}} \cdot charges \div distance \frac{( z _{+} e ) ( z _{-} e )}{r}$

Term by term, right where each lives:

$\frac{1}{4 π ε _{0}}$ — a fixed constant of the universe. $ε_{0}$ (the permittivity of free space) measures "how much empty space resists an electric field". You never change it; it just converts our charges-and-distance into joules.
$(z_{+} e) (z_{-} e)$ — the product of the two charges. One is $+$ , one is $-$ , so this product is negative. A negative energy means "released / stable / bound" — the ions want to be together.
$\div r$ — divide by distance. Push the ions apart (large $r$ ) and the energy shrinks toward zero. Pull them close (small $r$ ) and the attraction energy grows enormous.

Step 3 — Why one pair is not enough: the crowd around each ion

WHAT. In a real crystal each ion is surrounded by many neighbours, not one partner.

WHY. If NaCl were just isolated $Na^{+} Cl^{-}$ pairs, lattice energy would be small. The reason it is huge is that every ion is hugged by six opposite ions, then repelled by twelve same-charge ions further out, then attracted by eight even further... on and on through the crystal.

PICTURE. The figure shows a central red $Na^{+}$ and rings of neighbours. The nearest ring (blue, $-$ ) pulls it in — energy negative. The next ring (red, $+$ ) pushes — energy positive. Each ring is at a different distance $r$ , so each contributes a different-sized term.

First, name the yardstick we will measure every ring against:

Now we can add up every ring:

$E_{t o t a l} = \frac{e ^{2} z _{+} z _{-}}{4 π ε _{0}} [- \frac{6}{r _{0}} + \frac{12}{r _{0} 2} - \frac{8}{r _{0} 3} + \frac{6}{2 r _{0}} - \dots]$

Reading the bracket:

$- 6/ r_{0}$ : the 6 nearest opposite ions, at distance $r_{0}$ — attraction (minus).
$+ 12/ (r_{0} 2)$ : the 12 next same-charge ions, further out — repulsion (plus).
$- 8/ (r_{0} 3)$ : the 8 next-next opposite ions — attraction again.
...and so on forever, signs flipping, terms shrinking.

Notice the leading sign is negative (attraction wins, as it must for a stable crystal). Keep an eye on that minus — it is the whole reason $A$ comes out positive in Step 4.

This alternating sum is the whole difficulty. Step 4 tames it.

Step 4 — The Madelung constant: packing the whole crowd into one number

WHAT. Factor $- 1/ r_{0}$ out of that infinite bracket. What remains is a pure positive number that depends only on the crystal's shape, not its size.

WHY. The bracket has no $r$ in it once we pull $1/ r_{0}$ out — it is pure geometry. Every NaCl-shaped crystal, big or small, has the same bracket value. So we compute it once, name it, and never sum again.

Watch the sign carefully. The Step 3 bracket (with $1/ r_{0}$ already pulled out) is $[- 6 + \frac{12}{2} - \frac{8}{3} + \dots] = - (6 - \frac{12}{2} + \frac{8}{3} - \dots) = - A .$ That is: the Step 3 bracket equals $- A$ . We define $A$ as the version with a leading $+$ so that $A$ is a tidy positive number, and the crystal's overall minus sign lives out front where it clearly means "attraction / released energy". So the bracket is negative, but $A$ itself is positive — the minus was simply moved outside.

PICTURE. The figure shows the infinite sum for $A$ converging (bouncing up and down but settling) onto a single dashed line — the ==Madelung constant $A$ ==.

Recall How fast does this sum actually converge?

Slowly — and delicately. Taken shell by shell as written, the series is only conditionally convergent (it settles to different values if you rearrange the terms!), because far-away shells contain more and more ions that nearly cancel. To compute $A$ honestly from scratch you use convergence-accelerating tricks — most famously Ewald summation, which splits the sum into a fast-settling real-space part and a fast-settling reciprocal-space part. We do not do that here; we just quote the settled value $1.748$ . Just know that " $A = 1.748$ " hides a genuinely subtle numerical calculation.

Now our lattice-wide attraction energy per mole is clean:

$E_{a tt r} = - \frac{N _{A} A z _{+} z _{-} e ^{2}}{4 π ε _{0} r _{0}}$

$N_{A}$ = Avogadro's number ( $6.022 \times 1 0^{23}$ ), because we want energy per mole of formula units, not per single ion pair.
$A$ = the (positive) geometry we just packaged.
The minus sign out front = net attraction = released energy — this is the minus we relocated from the bracket.

Step 5 — Ions are not points: the short-range push-back

WHAT. Add the missing physics: when two ions get too close, their electron clouds refuse to overlap and shove back hard.

WHY. Look at $E_{a tt r}$ : as $r_{0} \to 0$ it goes to $- \infty$ . That predicts the crystal collapses to a point and releases infinite energy — obviously false. Something stops the squeeze. That something is Pauli repulsion: electron clouds cannot sit on top of each other. Max Born modelled it as a steeply rising energy.

PICTURE. The figure plots two curves against distance $r$ : the attractive $- 1/ r$ (dipping down) and a very steep repulsive $+ B / r^{n}$ (shooting up at small $r$ ). Their sum is a valley. The bottom of the valley is the real, stable spacing $r_{0}$ .

Where the $(1 - 1/ n)$ actually comes from — the calculus, one line at a time. Write the total energy of one mole as attraction plus wall, as a function of the variable spacing $r$ : $V (r) = - \frac{M}{r} + \frac{B}{r ^{n}}, M \equiv \frac{N _{A} A z _{+} z _{-} e ^{2}}{4 π ε _{0}} (> 0) .$ The crystal settles where $V (r)$ is lowest — the valley floor. "Lowest" means the slope is zero there, so differentiate and set to zero at $r = r_{0}$ : $\frac{d V}{d r}_{r_{0}} = \frac{M}{r _{0}^{2}} - \frac{n B}{r _{0}^{n + 1}} = 0 ⟹ B = \frac{M r _{0}^{n - 1}}{n} .$ That is the key: solving the "zero slope" condition gives $B$ with an $n$ in the denominator. Now put that $B$ back into $V (r_{0})$ : $U = V (r_{0}) = - \frac{M}{r _{0}} + \frac{B}{r _{0}^{n}} = - \frac{M}{r _{0}} + \frac{M}{n r _{0}} = - \frac{M}{r _{0}} (1 - \frac{1}{n}) .$ And there it is: the wall did not vanish, it left behind exactly a $\frac{1}{n}$ slice — because "set the derivative to zero" forced the repulsion at the minimum to be exactly $1/ n$ of the attraction.

$U = E_{a tt r} \times (1 - \frac{1}{n})$

$(1 - \frac{1}{n})$ is always slightly less than 1 (since $n$ is a positive number bigger than 1). It says: "the true bound energy is a little smaller than pure attraction, because some was spent fighting the electron-cloud wall."
If $n = 9$ , the factor is $1 - \frac{1}{9} \approx 0.889$ — we keep about $89%$ of the raw attraction.

Step 6 — Assemble the Born–Landé equation

WHAT. Multiply Step 4's attraction by Step 5's correction. That is the whole result.

WHY. Every piece is now earned: the pair energy (Coulomb), the crowd (Madelung), per mole ( $N_{A}$ ), and the electron-cloud wall ( $1 - 1/ n$ ). Nothing left to invent.

PICTURE. The figure lays out the final formula with every symbol colour-tagged to the step where we built it.

Reading the levers, term by term:

$z_{+} z_{-}$ (the charge product) — the biggest lever. Double both charges and $U$ roughly quadruples. This is why $Mg^{2 +} O^{2 -}$ ( $z_{+} z_{-} = 4$ ) crushes $Na^{+} Cl^{-}$ ( $z_{+} z_{-} = 1$ ). (See Ionic Radii and Fajans' Rules for how charge and size interplay.)
$1/ r_{0}$ — smaller ions sit closer, larger $U$ . Uses Ionic Radii.
$A$ — crystal geometry.
$(1 - 1/ n)$ — the electron-cloud tax.

Step 7 — When we don't know the crystal shape: Kapustinskii

WHAT. Sometimes we do not know the structure, so $A$ is unknown. Kapustinskii's trick removes the need for it.

WHY. Kapustinskii noticed that $A$ divided by $ν$ (the number of ions in one formula unit) is nearly the same for every structure. So he folded $A$ , $N_{A}$ , $e^{2}$ , and $4 π ε_{0}$ all into a single pre-baked number, and swapped the true spacing $r_{0}$ for the sum of tabulated radii $r_{+} + r_{-}$ .

PICTURE. The figure shows the transformation: Born–Landé's structure-dependent parts collapsing into one constant, leaving only charges, ion count, and radii as inputs.

Intuition Where does

$1.202 \times 1 0^{5}$ come from? It is all the fixed physics bundled together. Start from Born–Landé's prefactor $\frac{N _{A} A e ^{2}}{4 π ε _{0}}$ and make three moves:

Average out the geometry. Replace $A$ by $ν \times (average A / ν)$ ; Kapustinskii used $A / ν \approx 0.874$ , roughly constant across structures. This turns the structure-specific $A$ into a plain factor of $ν$ .
Fix the units. Radii are tabulated in picometres ( $1 pm = 1 0^{- 12} m$ ), not metres, so a factor of $1 0^{12}$ is absorbed to let you plug pm straight in.
Convert to kJ per mole. Multiply by $N_{A}$ (per mole) and divide by $1000$ (J → kJ).

Grind $\frac{N _{A} e ^{2}}{4 π ε _{0}} \times 0.874 \times 1 0^{12} /1000$ through and you land on $\approx 1.202 \times 1 0^{5} kJ\cdotpm/mol$ . That is why the boxed constant carries hidden units of "kJ·pm per mole" — it has already eaten $N_{A}$ , $e^{2}$ , $ε_{0}$ , the average geometry, the pm-to-m conversion, and the kJ conversion.

Step 8 — Edge and degenerate cases (never get surprised)

WHAT. Check what the formula does at its extremes, so no scenario ambushes you.

WHY. A formula you trust must behave sanely in every limit. Here are the corners.

PICTURE. The figure sketches $U$ against $r_{0}$ and against charge, marking each special case.

$r_{0} \to \infty$ (ions dragged far apart): $1/ r_{0} \to 0$ , so $U \to 0$ . Correct — infinitely separated ions have no lattice, no released energy.
$r_{0} \to 0$ (ions on top of each other): here we must be careful about which expression we mean. The full energy function $V (r) = - M / r + B / r^{n}$ does not run to $- \infty$ : because $n > 1$ , the repulsive $+ B / r^{n}$ term rises faster than the attraction falls, so $V (r) \to + \infty$ as $r \to 0$ . That climbing wall is exactly what carves out the finite valley floor at $r_{0}$ . The boxed $U$ is $V$ evaluated at that floor, so it is a plain finite number — you never actually "plug in $r_{0} = 0$ ". The Born wall guarantees the minimum sits at a real, non-zero spacing.
$z_{+} z_{-} = 0$ (a neutral atom, no charge): $U = 0$ . No ions, no ionic lattice — the formula correctly says "nothing here". This is the boundary where ionic bonding hands over to covalent (see Fajans' Rules).
$n \to \infty$ (a perfectly hard wall): $(1 - 1/ n) \to 1$ . The correction disappears and $U$ equals pure Madelung attraction — the idealised, no-give crystal.
Kapustinskii with tiny $(r_{+} + r_{-})$ : the term $(1 - 34.5/ (r_{+} + r_{-}))$ could go negative if $r_{+} + r_{-} < 34.5$ pm — but no real ion pair is that small, so we never hit it. The correction stays between $0$ and $1$ for all real ions.

Every corner behaves. The formula is safe to use.

The one-picture summary

One figure, the whole journey: one pair (Coulomb $1/ r$ ) → a crowd of rings (alternating sum) → package the geometry ( $A$ , and relocate its minus sign out front) → stop the collapse (Born wall, $1 - 1/ n$ from zeroing the slope) → scale to a mole ( $N_{A}$ ) = the Born–Landé equation, with Kapustinskii as its structure-free cousin.

Recall Feynman: the walkthrough in plain words

Picture two magnetic marbles, one red-plus, one blue-minus. Bring them together and energy comes out — that's Coulomb's law, and because energies just add, we use energy not force. But one marble isn't alone: in a crystal it's hugged by six opposite marbles, shoved by twelve same ones further out, hugged by eight beyond that — forever. Add all those pulls and pushes and, amazingly, the endless (and delicately converging) sum settles on one positive number that depends only on the shape of the packing — the Madelung constant, with the overall attraction-minus pulled out to the front. Multiply by Avogadro's number to get a mole's worth. But marbles aren't points — squeeze them and their electron fuzz shoves back, so the crystal settles at a comfy spacing where pull and push balance; demanding that comfy point (zero slope) trims the energy by exactly $(1 - 1/ n)$ . Put it together and you have Born–Landé — read as the depth of the valley, one number, not a curve. And when you don't know the crystal's shape, Kapustinskii quietly averages the geometry away into one baked-in constant, so you need nothing but the charges, the ion count, and the sizes off a table.

Recall

Why do we use potential energy (not force) to build lattice energy? ::: Energies are single numbers that simply add over all pairs; forces are vectors with directions and are far harder to sum. What does the Madelung constant capture, and why is it positive? ::: The infinite alternating attraction/repulsion sum with $1/ r_{0}$ factored out; the Step 3 bracket equals $- A$ , so pulling the attraction-minus out front leaves $A$ as a positive geometric number. Why is the $(1 - 1/ n)$ factor always less than 1, and where does it come from? ::: Setting $d V / d r = 0$ at the valley floor gives $B = M r_{0}^{n - 1} / n$ ; substituting back leaves the repulsion as exactly $1/ n$ of the attraction, so $U = - \frac{M}{r _{0}} (1 - 1/ n)$ with $n > 1$ . What stops $V (r) \to - \infty$ as $r \to 0$ ? ::: The Born wall $+ B / r^{n}$ with $n > 1$ rises faster than the attraction, so $V \to + \infty$ ; the crystal settles at the finite valley floor $r_{0}$ , and the boxed $U$ is that floor's depth, not a function of $r$ . When does $U \to 0$ , and why is that physically right? ::: When $r_{0} \to \infty$ (ions infinitely apart) or $z_{+} z_{-} = 0$ (no charge) — no lattice, no released energy. What hides inside the Kapustinskii constant $1.202 \times 1 0^{5}$ ? ::: $N_{A}$ , $e^{2} / (4 π ε_{0})$ , the average $A / ν \approx 0.874$ , the pm→m factor, and the J→kJ conversion, all baked into one number with units kJ·pm/mol.