Stability constants of complexes — chelate effect: Exercises

Everything here rests on three tools from the parent note, the topic note. Let me re-anchor them so no symbol is used before it is earned.

Here $R = 8.314 J K^{- 1} mol^{- 1}$ is the gas constant (the energy-per-degree "conversion rate" between temperature and free energy), $T$ is temperature in kelvin, and $2.303 = ln 10$ is just the factor that turns natural logs into base-10 logs (because $ln x = 2.303 lo g_{10} x$ ).

Level 1 — Recognition

Exercise 1.1

Which symbol refers to adding one ligand at a time, and which to forming the whole complex in one bookkeeping step: $K_{i}$ or $β_{n}$ ? State the definition of each in one line.

Recall Solution

$K_{i}$ is the stepwise constant: it describes the single equilibrium $M L_{i - 1} + L ⇌ M L_{i}$ , i.e. clipping on the $i$ -th ligand only.
$β_{n}$ is the overall (cumulative) constant: it describes $M + n L ⇌ M L_{n}$ all the way from bare metal to the finished complex. They are linked by T1: $β_{n} = \prod_{i} K_{i}$ . Recognition here is just matching the word to the symbol.

Exercise 1.2

For $M + 2 L ⇌ M L_{2}$ you are given $lo g K_{1} = 4.0$ and $lo g K_{2} = 3.0$ . Without a calculator, write $lo g β_{2}$ .

Recall Solution

By T1 the logs add (they add because the constants multiply): $lo g β_{2} = lo g K_{1} + lo g K_{2} = 4.0 + 3.0 = 7.0$

Level 2 — Application

Exercise 2.1

A metal binds four identical ligands with $lo g K_{1} = 8.0, lo g K_{2} = 6.5, lo g K_{3} = 5.0, lo g K_{4} = 3.5$ . Compute $lo g β_{4}$ and state $β_{4}$ as a power of ten.

Recall Solution

What: sum the stepwise logs (T1). $lo g β_{4} = 8.0 + 6.5 + 5.0 + 3.5 = 23.0 \Rightarrow β_{4} = 1 0^{23}$ Why it decreases each step: $K_{i}$ falls by 1.5 log units each time — the statistical + steric crowding described in the parent note. The trend is data, not error.

Exercise 2.2

At $T = 298 K$ , a complex has $lo g β = 18.0$ . Find $Δ G^{\circ}$ in kJ/mol.

Recall Solution

What tool and why: we want free energy from a tabulated $lo g_{10} β$ , so use T3 rearranged: $Δ G^{\circ} = - 2.303 R T lo g_{10} β$ . $Δ G^{\circ} = - 2.303 (8.314) (298) (18.0)$ $= - (2.303) (8.314) (298) (18.0) = - 1.027 \times 1 0^{5} J/mol \approx - 103 kJ/mol$ Negative and large — the complex is very stable, consistent with a large $β$ .

Exercise 2.3

Compare $[Ni(NH_{3})_{6}]^{2 +}$ ( $lo g β_{6} = 8.6$ ) with $[Ni(en)_{3}]^{2 +}$ ( $lo g β_{3} = 18.3$ ). How many times more stable (ratio of $β$ ) is the en complex?

Recall Solution

What: a ratio of constants means subtract the logs. $lo g \frac{β _{en}}{β _{NH_{3}}} = 18.3 - 8.6 = 9.7 \Rightarrow \frac{β _{en}}{β _{NH_{3}}} = 1 0^{9.7} \approx 5 \times 1 0^{9}$ The chelate wins by about five billion-fold — the bidentate en versus monodentate ammonia, same six N-donors.

Level 3 — Analysis

Exercise 3.1

Two reactions have equal $Δ H^{\circ}$ . Reaction A releases a net $0$ extra free particles; reaction B releases $+ 3$ extra free particles, giving $Δ S_{B}^{\circ} - Δ S_{A}^{\circ} \approx + 84 J K^{- 1} mol^{- 1}$ . At $298 K$ , by how many units does $lo g β$ change from A to B?

Recall Solution

What decides it: with equal $Δ H^{\circ}$ , the difference in $Δ G^{\circ}$ is purely the $- T Δ S^{\circ}$ term (T2). $Δ (Δ G^{\circ}) = - T Δ (Δ S^{\circ}) = - (298) (84) = - 2.50 \times 1 0^{4} J/mol$ Convert to log units with T3: $Δ (lo g β) = \frac{- Δ ( Δ G ^{\circ} )}{2.303 R T} = \frac{25000}{2.303 ( 8.314 ) ( 298 )} \approx 4.4$ A 4–5 unit jump in $lo g β$ from entropy alone — this is the chelate effect, seen numerically. See the bar chart below for how enthalpy and entropy split.

Figure — Stability constants of complexes — chelate effect

Exercise 3.2

Explain, using particle counting, why the reaction $[M(H_{2} O)_{6}] + EDTA^{4 -} ⇌ [M(EDTA)]^{2 -} + 6 H_{2} O$ has an even bigger $+ Δ S^{\circ}$ than the en case. Sketch the count.

Recall Solution

Count free particles on each side (the metal complex counts as one particle):

Left: $1$ (aqua complex) $+ 1$ (EDTA) $= 2$ .
Right: $1$ ([M(EDTA)]) $+ 6$ (waters) $= 7$ . So $2 \to 7$ : a gain of $+ 5$ free particles from one ligand. The en case was $4 \to 7$ (gain $+ 3$ ). More liberated waters per ligand consumed ⇒ larger $+ Δ S^{\circ}$ ⇒ larger $β$ . Hexadentate EDTA also closes five 5-membered chelate rings.

Level 4 — Synthesis

Exercise 4.1

A macrocyclic ligand and its open-chain analogue both wrap a metal with the same 4 N-donors. At $298 K$ the macrocycle's complex has $lo g β = 22.0$ ; the open-chain one has $lo g β = 18.5$ . (a) Find the extra stabilisation $Δ (Δ G^{\circ})$ in kJ/mol. (b) Name the effect and give the physical reason.

Recall Solution

(a) Difference of logs (T3), then convert: $Δ (lo g β) = 22.0 - 18.5 = 3.5$ $Δ (Δ G^{\circ}) = - 2.303 R T Δ (lo g β) = - 2.303 (8.314) (298) (3.5) \approx - 20 kJ/mol$ The macrocycle is stabilised by an extra $\sim 20$ kJ/mol. (b) This is the macrocyclic effect. The macrocycle is pre-organised — already shaped as a ring — so it pays no entropy/enthalpy penalty to fold up before binding, unlike the floppy open chain that must lose conformational freedom to close its rings.

Exercise 4.2

Using CFSE logic: for a $d^{4}$ ion (this chapter's tag), if an incoming stronger-field chelate pushes it from high-spin to low-spin, would you expect the corresponding stepwise $K_{i}$ to be anomalously high or low? Explain in words (no numbers needed).

Recall Solution

A geometry/spin switch to low-spin increases CFSE (extra ligand-field stabilisation), making that particular binding step release more energy — a more negative $Δ H^{\circ}$ for that step. So the affected $K_{i}$ is anomalously high, bucking the usual " $K_{i}$ decreases with $i$ " trend. This is exactly the "occasionally reversed by structural change" note (point 3) from the parent. So: anomalously high.

Level 5 — Mastery

Exercise 5.1

You must design a ligand to grip Ca²⁺ for a water-hardness titration. You may pick ring size (4-, 5-, or 6-membered chelate rings) and denticity (bidentate vs hexadentate). Justify each choice with the correct thermodynamic and geometric reasoning, and predict qualitatively how your best design's $lo g K$ compares to bidentate en.

Recall Solution

Ring size — pick 5-membered. Four-membered rings are strained (bad bite angle → poor $Δ H^{\circ}$ ); very large rings rarely close because the second donor has low probability of reaching the metal (low effective concentration → weak effect). 5-membered rings have unstrained ideal bite angles — maximal stability. Denticity — pick hexadentate (EDTA-like). One hexadentate molecule displaces six waters, giving the largest $+ Δ S^{\circ}$ (particle count $2 \to 7$ , a $+ 5$ gain from Exercise 3.2) and closing five 5-membered rings. Compared to bidentate en (which needs three molecules for six sites, a smaller $+ Δ S^{\circ}$ per ligand), the hexadentate design gives a much larger $lo g K$ . Indeed EDTA–Ca²⁺ has $lo g K \approx 10.7$ . Why this is the mastery answer: it correctly weighs both the geometric optimum (5-ring) and the entropy engine (maximise freed waters per ligand consumed), the two independent levers behind stability.

Exercise 5.2

At $T = 298 K$ , a Ca²⁺–EDTA complex has $lo g K = 10.7$ . Compute $Δ G^{\circ}$ . Then, if the reaction's $Δ H^{\circ}$ is only about $- 13 kJ/mol$ , estimate $Δ S^{\circ}$ and confirm it is positive (the entropy engine).

Recall Solution

Step 1 — $Δ G^{\circ}$ from $lo g K$ (T3): $Δ G^{\circ} = - 2.303 R T lo g K = - 2.303 (8.314) (298) (10.7) \approx - 61.1 kJ/mol$ Step 2 — solve $Δ S^{\circ}$ from T2 $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ : $Δ S^{\circ} = \frac{Δ H ^{\circ} - Δ G ^{\circ}}{T} = \frac{- 13000 - ( - 61100 )}{298} = \frac{48100}{298} \approx + 161 J K^{- 1} mol^{- 1}$ $Δ S^{\circ}$ is strongly positive — the six freed waters. Notice most of the $- 61$ kJ/mol driving force is entropic ( $- T Δ S^{\circ} \approx - 48$ kJ/mol) rather than enthalpic ( $- 13$ kJ/mol). This is the chelate/entropy engine, quantified.

Recall

Recall Quick self-test

β from three stepwise steps with log K = 8.0, 6.5, 5.0, 3.5? ::: log β₄ = 23.0, so β₄ = 10²³. ΔG° for log β = 18.0 at 298 K? ::: about −103 kJ/mol. How many times more stable is [Ni(en)₃]²⁺ than [Ni(NH₃)₆]²⁺? ::: 10^(18.3−8.6) = 10^9.7 ≈ 5×10⁹. ΔS° for a +3 particle chelate step (≈+84 J/K/mol) worth how many log β units at 298 K? ::: about +4.4. Best chelate ring size and why? ::: 5-membered — unstrained ideal bite angle.