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

Level 1 — Recognition

Goal: read the definitions and signs correctly, no arithmetic.

Which of these processes has enthalpy equal to the lattice energy of KBr (formation convention)? (a) $KBr (s) \to K^{+} (g) + Br^{-} (g)$ (b) $K^{+} (g) + Br^{-} (g) \to KBr (s)$ (c) $K (s) + \frac{1}{2} Br_{2} (l) \to KBr (s)$

Recall Solution 1.1

The parent defines lattice energy (formation convention) as gaseous ions collapsing into the solid. That is exactly option (b), and it is exothermic (negative).

(a) is the reverse — breaking the lattice — so it is positive, the "lattice dissociation" convention.
(c) is the enthalpy of formation $Δ H_{f}$ (from elements, not ions) — a different quantity entirely. Answer: (b).

Q1.2

Classify each step as endothermic (+) or exothermic (–): Ionisation of Na · First electron affinity of Cl · Atomisation of Na(s) · Lattice formation of NaCl.

Recall Solution 1.2

Ionisation $Na (g) \to Na^{+} (g) + e^{-}$ : you rip off an electron ⇒ endothermic (+).
First EA of Cl, $Cl (g) + e^{-} \to Cl^{-} (g)$ : a neutral atom wants the electron ⇒ exothermic (–).
Atomisation $Na (s) \to Na (g)$ : breaking metallic bonds to make gas ⇒ endothermic (+).
Lattice formation: scattered ions snap together ⇒ exothermic (–).

Level 2 — Application

Goal: plug numbers into one formula.

Q2.1 — Kapustinskii for KCl

Given $r_{+} (K^{+}) = 138$ pm, $r_{-} (Cl^{-}) = 181$ pm, $ν = 2$ , $z_{+} = z_{-} = 1$ . Estimate $U$ .

Recall Solution 2.1

WHAT: use the Kapustinskii equation because we only have radii and charges — no crystal structure. Sum of radii: $r_{+} + r_{-} = 138 + 181 = 319$ pm. First factor: $- \frac{1.202 \times 1 0 ^{5} ( 2 ) ( 1 ) ( 1 )}{319} = - 753.6 kJ/mol$ Repulsion correction: $(1 - \frac{34.5}{319}) = 1 - 0.1082 = 0.8918$ Multiply: $U = - 753.6 \times 0.8918 = - 672 kJ/mol$ Answer: $\approx - 672$ kJ/mol (experiment $\approx - 701$ ; within ~4%). Uses Ionic Radii.

Q2.2 — Solve Born–Haber for LiF

Data (kJ/mol): $Δ H_{f} = - 616, Δ H_{a t o m} (Li) = + 161, I E_{1} (Li) = + 520, \frac{1}{2} D_{F - F} = + 79, E A (F) = - 328$ . Find $U$ .

Recall Solution 2.2

WHY: rearrange Hess's Law around the cycle (parent §4) to isolate the one unknown $U$ : $U = Δ H_{f} - Δ H_{a t o m} - I E_{1} - \frac{1}{2} D - E A$ Substitute, tracking every sign: $U = - 616 - 161 - 520 - 79 - (- 328)$ $U = - 616 - 161 - 520 - 79 + 328 = - 1048 kJ/mol$ Answer: $U (LiF) = - 1048$ kJ/mol. Uses Hess's Law, Ionisation Energy, Electron Affinity.

Level 3 — Analysis

Goal: compare two systems and explain the direction of the trend.

Q3.1 — LiF vs LiI, quantitatively

Using Kapustinskii ( $ν = 2$ , $z_{+} z_{-} = 1$ ) with $r (Li^{+}) = 76$ , $r (F^{-}) = 133$ , $r (I^{-}) = 220$ pm, compute both and explain the trend.

Recall Solution 3.1

LiF: $r_{+} + r_{-} = 76 + 133 = 209$ pm. $- \frac{1.202 \times 1 0 ^{5} ( 2 )}{209} (1 - \frac{34.5}{209}) = - 1150.2 \times 0.8349 = - 960 kJ/mol$ LiI: $r_{+} + r_{-} = 76 + 220 = 296$ pm. $- \frac{1.202 \times 1 0 ^{5} ( 2 )}{296} (1 - \frac{34.5}{296}) = - 812.2 \times 0.8834 = - 717 kJ/mol$ Trend: same charge product, but LiI has the larger $r_{+} + r_{-}$ because I $^{-}$ is much bigger. Since $U \propto 1/ (r_{+} + r_{-})$ , LiI has the weaker (less negative) lattice energy. Answer: $U (LiF) \approx - 960$ , $U (LiI) \approx - 717$ kJ/mol. This is the parent's §5 mistake made rigorous.

Q3.2 — MgO vs NaCl charge lever

Both have similar ionic sizes. Using only the proportionality $U \propto z_{+} z_{-} / (r_{+} + r_{-})$ , estimate roughly how many times larger $U (MgO)$ should be than $U (NaCl)$ if sizes were identical, then say why the real ratio is even bigger.

Recall Solution 3.2

Charge lever alone: NaCl has $z_{+} z_{-} = 1$ ; MgO has $z_{+} z_{-} = (2) (2) = 4$ . So purely from charge, $U (MgO)$ is $\approx 4 \times$ that of NaCl. WHY real ratio is bigger: Mg $^{2 +}$ (86 pm) and O $^{2 -}$ (140 pm) are smaller than Na $^{+}$ (102) + Cl $^{-}$ (181). A smaller $r_{+} + r_{-}$ in the denominator further raises $U$ . So the observed ratio $\approx 3800/788 \approx 4.8$ , above the pure charge factor of 4. Answer: factor ≈ 4 from charge, ≈ 4.8 observed — the extra comes from smaller ions.

Level 4 — Synthesis

Goal: build a whole cycle, including a compound with two electron affinities or a divalent metal.

Q4.1 — Born–Haber for CaO (double ionisation, double EA)

Find $U$ from: $Δ H_{f} (CaO) = - 635$ ; $Δ H_{a t o m} (Ca) = + 178$ ; $I E_{1} (Ca) = + 590$ ; $I E_{2} (Ca) = + 1145$ ; $\frac{1}{2} D_{O = O} = + 249$ ; $E A_{1} (O) = - 141$ ; $E A_{2} (O) = + 798$ (all kJ/mol).

Recall Solution 4.1

WHAT changes vs NaCl: Ca must lose two electrons ( $I E_{1} + I E_{2}$ ) to reach Ca $^{2 +}$ , and O must gain two ( $E A_{1}$ negative, $E A_{2}$ positive/endothermic — parent §5 warns of this). Generalised Hess's Law: $U = Δ H_{f} - Δ H_{a t o m} - (I E_{1} + I E_{2}) - \frac{1}{2} D - (E A_{1} + E A_{2})$ Sum the ionisations: $590 + 1145 = 1735$ . Sum the EAs: $- 141 + 798 = + 657$ . $U = - 635 - 178 - 1735 - 249 - 657$ $U = - 3454 kJ/mol$ Answer: $U (CaO) \approx - 3454$ kJ/mol — enormous, consistent with a $z_{+} z_{-} = 4$ oxide. Uses Ionisation Energy, Electron Affinity.

Q4.2 — CaF₂: the $ν = 3$ subtlety

Estimate $U (CaF_{2})$ by Kapustinskii. $r (Ca^{2 +}) = 100$ , $r (F^{-}) = 133$ pm. Here $z_{+} = 2$ , $z_{-} = 1$ , $ν = 3$ .

Recall Solution 4.2

WHY $ν = 3$ : one Ca $^{2 +}$ + two F $^{-}$ = three ions per formula unit (parent §5 mistake). $r_{+} + r_{-} = 100 + 133 = 233$ pm. Use $z_{+} z_{-} = (2) (1) = 2$ . $- \frac{1.202 \times 1 0 ^{5} ( 3 ) ( 2 )}{233} = - 3094.4 kJ/mol$ Correction: $(1 - \frac{34.5}{233}) = 1 - 0.1481 = 0.8519$ . $U = - 3094.4 \times 0.8519 = - 2636 kJ/mol$ Answer: $U (CaF_{2}) \approx - 2636$ kJ/mol (experiment $\approx - 2630$ — excellent).

Figure — Ionic bonding — Born-Haber cycle, lattice energy (Kapustinskii equation)

Level 5 — Mastery

Goal: use the cycle to diagnose covalent character and reason backwards.

Q5.1 — Covalent character of AgCl

Kapustinskii predicts $U_{ionic} (AgCl) \approx - 734$ kJ/mol. A Born–Haber cycle using experimental thermochemistry gives $U_{exp} = - 905$ kJ/mol. Interpret the gap.

Recall Solution 5.1

WHAT: the experimental lattice energy is more negative (stronger) than the pure-ionic model predicts, by $∣ - 905 ∣ - ∣ - 734 ∣ = 171 kJ/mol extra binding .$ WHY it matters: the ionic model treats ions as hard charged spheres. Extra binding beyond that means bonding electrons are partly shared — i.e. covalent character. Ag $^{+}$ is polarising and Cl $^{-}$ is polarisable, so the electron cloud distorts (Fajans' rules). Answer: AgCl shows significant covalent character (~171 kJ/mol of extra binding). This is the diagnostic in parent §6. See Fajans' Rules.

Q5.2 — Reason backwards to an unknown $Δ H_{f}$

For MgCl₂ you know: $Δ H_{a t o m} (Mg) = + 148$ ; $I E_{1} = + 738$ ; $I E_{2} = + 1451$ ; $D_{C l - C l} = + 244$ (full bond, since two Cl atoms are needed); $E A (Cl) = - 349$ (per Cl); $U = - 2526$ kJ/mol. Predict $Δ H_{f} (MgCl_{2})$ .

Recall Solution 5.2

WHAT is different: MgCl₂ needs two chlorines. So we take the full $Cl_{2}$ bond dissociation ( $D = 244$ , i.e. $2 \times \frac{1}{2} D$ ) and two electron affinities ( $2 \times E A$ ). Rearranged Hess's Law (now solving for $Δ H_{f}$ ): $Δ H_{f} = Δ H_{a t o m} + (I E_{1} + I E_{2}) + D + 2 E A + U$ Sum ionisations: $738 + 1451 = 2189$ . Two EAs: $2 \times (- 349) = - 698$ . $Δ H_{f} = 148 + 2189 + 244 + (- 698) + (- 2526)$ $Δ H_{f} = - 643 kJ/mol$ Answer: $Δ H_{f} (MgCl_{2}) \approx - 643$ kJ/mol (literature $\approx - 641$ — spot on). Uses Hess's Law, links to Lattice Energy → Solubility.

Recall Self-check checklist

Signs tracked in brackets before simplifying? ::: Yes — prevents the double-negative slip. Used ion charge magnitudes, not net charge, in Kapustinskii? ::: Yes. Counted all ions for $ν$ and all atoms for $D$ / $E A$ ? ::: Yes — stoichiometry drives the counts. Compared experimental vs ionic-model $U$ to judge covalency? ::: Yes — bigger gap ⇒ more covalent.

Exercises — Ionic bonding — Born-Haber cycle, lattice energy (Kapustinskii equation)