Topic 17 · page 15 of 20

Stability and entropy

Replacing monodentate ligands with a bidentate or hexadentate ligand releases more particles into solution. That raises the entropy of the system, so the reaction is feasible and the new complex is more stable.

12 min read17.25

Read first: Bidentate and hexadentate ligands, Ligand exchange and colour

What "more stable" means 17.25

A complex ion is more stable than another if the ligand exchange that forms it is feasible and the equilibrium lies far to the right. This is a statement about the equilibrium, not about how fast it is reached.

In Topic 12 the total entropy change decides whether a reaction is feasible:

ΔStotal=ΔSsystem+ΔSsurroundings \Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surroundings}

where ΔSsurroundings=−ΔH/T\Delta S_\text{surroundings} = -\Delta H / T. A reaction is feasible when ΔStotal\Delta S_\text{total} is positive.

Counting particles 17.25

Consider replacing six ammonia ligands in [Ni(NH₃)₆]²⁺ with three bidentate 1,2-diaminoethane (en) ligands.

[Ni(NHX3)X6]X2++3 NHX2CHX2CHX2NHX2→[Ni(NHX2CHX2CHX2NHX2)X3]X2++6 NHX3 \begin{aligned} &\ce{[Ni(NH3)6]^2+ + 3NH2CH2CH2NH2} \\ &\ce{-> [Ni(NH2CH2CH2NH2)3]^2+ + 6NH3} \end{aligned}

Now replace six water ligands in [Cu(H₂O)₆]²⁺ with one hexadentate EDTA⁴⁻ ion.

[Cu(HX2O)X6]X2++EDTAX4−→[Cu(EDTA)]X2−+6 HX2O \ce{[Cu(H2O)6]^2+ + EDTA^4- -> [Cu(EDTA)]^2- + 6H2O}

In both reactions the number of particles in solution increases. More particles can share the energy in more ways, so ΔSsystem\Delta S_\text{system} is positive.

Exchange Particles on the left Particles on the right Change
6 NH₃ → 3 en 4 7 +3
6 H₂O → 1 EDTA⁴⁻ 2 7 +5
6 H₂O → 6 NH₃ (monodentate) 7 7 0

The third row shows the contrast. Replacing six monodentate ligands with six other monodentate ligands changes the number of particles by nothing, so there is no entropy gain to help.

Why the enthalpy change is small 17.25

Before and after, the metal ion is bonded to the same number of donor atoms of the same kind. In the nickel example six Ni–N dative bonds are broken and six Ni–N dative bonds are made. The bonds are very similar in strength, so ΔH\Delta H is close to zero.

Then ΔSsurroundings=−ΔH/T\Delta S_\text{surroundings} = -\Delta H / T is close to zero as well, and

ΔStotal≈ΔSsystem>0 \Delta S_\text{total} \approx \Delta S_\text{system} > 0

The ligand exchange is feasible and K is large: the complex with the polydentate ligand is more stable.

Two reactions drawn as particle counts. Top: one nickel ammine complex plus three en molecules, four particles, goes to one nickel en complex plus six ammonia molecules, seven particles, with delta S system positive. Bottom: one copper aqua complex plus one EDTA ion, two particles, goes to one copper EDTA complex plus six water molecules, seven particles, with delta S system positive.
Particles on each side for a bidentate and a hexadentate substitution. More particles on the right means a positive entropy change.

MethodExplaining why a polydentate ligand gives a more stable complex

  1. Write the ligand exchange equation.
  2. Count the particles on each side and say the number increases.
  3. State that ΔSsystem\Delta S_\text{system} is positive.
  4. Say that similar bonds are broken and made, so ΔH\Delta H is about zero.
  5. Conclude that ΔStotal\Delta S_\text{total} is positive, the exchange is feasible and the new complex is more stable.

Check yourself

  1. Explain, in terms of entropy, why [Cu(EDTA)]²⁻ is more stable than [Cu(H₂O)₆]²⁺.
  2. For the reaction [Co(H₂O)₆]²⁺ + EDTA⁴⁻ → [Co(EDTA)]²⁻ + 6H₂O, state the number of particles on each side and the sign of ΔSsystem.
  3. Why is ΔH for the reaction in question 2 close to zero?
  4. [Cu(NH₃)₄(H₂O)₂]²⁺ + 2en → [Cu(en)₂(H₂O)₂]²⁺ + 4NH₃. Count the particles and give the sign of ΔSsystem.
  5. A student says the EDTA complex is more stable because six bonds form. Explain why this is wrong.
  6. Would you expect much entropy gain when six water ligands are replaced by six ammonia ligands? Explain.

Answers cover these until you have tried the questions

  1. The reaction [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O has 2 particles on the left and 7 on the right. ΔSsystem is positive. Six dative bonds from water are broken and six dative bonds from EDTA⁴⁻ are made, so ΔH ≈ 0 and ΔStotal is positive. The reaction is feasible and K is large.
  2. Left: 1 + 1 = 2. Right: 1 + 6 = 7. The number of particles increases, so ΔSsystem is positive.
  3. Six dative bonds from water (O donors) are broken and six dative bonds from EDTA (N and O donors) are made. They are similar in strength, so the enthalpy change is about zero.
  4. Left: 1 + 2 = 3. Right: 1 + 4 = 5. The particles increase by two, so ΔSsystem is positive.
  5. Six bonds form, but six are also broken (to the water ligands), so the number of bonds does not change and ΔH is about zero. The reason for the stability is the increase in the number of particles, which makes ΔSsystem positive.
  6. No. There are 7 particles on each side (1 complex and 6 ligands), so ΔSsystem is about zero.