Stability of complexes

A coordination complex forming in solution is an equilibrium, not a one-way reaction:

Like any equilibrium, it has an equilibrium constant. Here it is called the stability constant or formation constant, written $K_f$ (sometimes $K$ or $\beta$):

where: $[ML_n]$ is the equilibrium concentration of the complex, $[M]$ is the free metal-ion concentration, $[L]$ is the free-ligand concentration, and $n$ is the number of ligands in the complex — the same $n$ that appears in the balanced equation, so the ligand concentration must be raised to that power. Because $K_f$ is just "products over reactants" for the forward (association) direction, its size tells you where the equilibrium sits. A large $K_f$ (complexes with $CN^-$ or chelating ligands are routinely $10^{10}$ to $10^{40}$) means the numerator dominates — almost all the metal ion is locked up as complex, and only a trace of free $M$ and $L$ survives. A small $K_f$ means the reverse: a comfortable equilibrium mixture of free ion and free ligand, with only some complex formed. The exact reverse reaction — a complex breaking back into its ions — is also an equilibrium, and its constant is simply the reciprocal:

$K_{inst}$ is called the instability constant (or dissociation constant). It carries exactly the same information as $K_f$, read the opposite way: a small $K_{inst}$, like a large $K_f$, signals a stable complex. Never let the two get mixed up — one grows with stability, the other shrinks. Three structural factors, all part of the syllabus's qualitative toolkit, let you predict which of two complexes has the larger $K_f$ without a table of numbers:

Charge density of the metal ion. A smaller ionic radius and/or higher oxidation state concentrates positive charge, pulling ligand lone pairs in more strongly through electrostatic attraction — so, other things equal, a smaller and more highly charged metal ion forms the more stable complex.

Field strength (Lewis basicity) of the ligand. Stronger-field, more strongly Lewis-basic ligands such as $CN^-$ bind the metal more tightly than weaker ones such as halide ions, so swapping in a stronger-field ligand at the same metal raises $K_f$.

The chelate effect. A polydentate ligand — one that attaches to the metal through two or more donor atoms at once, like ethylenediamine (en) or EDTA — forms a far more stable complex than the equivalent number of separate monodentate ligands providing the same donor atoms. This is the single most exam-tested idea in this topic, and it gets its own derivation-style reasoning below.

The stability constant of $[Cu(NH_3)_4]^{2+}$ is $K_f = 2.1 \times 10^{13}$. (a) Write its formation equilibrium and $K_f$ expression. (b) Comment on its stability. (c) Now compare it qualitatively with $[Cu(en)_2]^{2+}$, in which ethylenediamine (bidentate) replaces the four ammonia molecules with two chelating en ligands providing the same four nitrogen donors, and predict which has the larger $K_f$.

Step 1 — write the equilibrium and expression. The complex forms as $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$, so

Step 2 — interpret the magnitude. $K_f \sim 10^{13}$ is very large: the numerator (complex) vastly outweighs the denominator (free ion times free ligand raised to the 4th power) at equilibrium. Almost all copper(II) present is locked into the complex when excess ammonia is available; only a negligible free $Cu^{2+}$ concentration remains.

Step 3 — predict dilution behaviour. Diluting reduces all concentrations, and since the forward reaction converts five particles ($Cu^{2+}$ plus four $NH_3$) into one, Le Chatelier's principle says dilution nudges the equilibrium slightly toward more dissociation. Because $K_f$ is so large, though, the complex stays overwhelmingly intact even after substantial dilution.

Step 4 — compare with the chelate analogue. $[Cu(en)_2]^{2+}$ delivers the same four nitrogen donors to $Cu^{2+}$, but via two bidentate en molecules instead of four separate $NH_3$ molecules. By the chelate-effect reasoning above, dissociating $[Cu(en)_2]^{2+}$ requires both ends of an en ligand to detach essentially together and releases only two particles rather than four, both effects working against dissociation. $[Cu(en)_2]^{2+}$ therefore has a larger $K_f$ than $[Cu(NH_3)_4]^{2+}$ even though copper is bonded to the same number and type of donor atoms in each — the chelate effect alone accounts for the difference.

Competitive stretch

If asked to rank three complexes — $[Ni(NH_3)_6]^{2+}$, $[Ni(CN)_4]^{2-}$, $[Ni(en)_3]^{2+}$ — by expected stability, reason factor by factor: $CN^-$ beats $NH_3$ on ligand field strength/basicity alone (raising $[Ni(CN)_4]^{2-}$'s $K_f$ well above $[Ni(NH_3)_6]^{2+}$'s), while $[Ni(en)_3]^{2+}$ gains the chelate effect on top of nitrogen-donor bonding comparable to the ammine complex — in practice this typically makes $[Ni(en)_3]^{2+}$ the most stable of the three, since chelation is usually the dominant factor once several small chelate rings are involved.

• Confusing $K_f$ with $K_{inst}$ — remember $K_{inst} = 1/K_f$, so a large $K_f$ and a small $K_{inst}$ say the same thing (high stability); mixing them up flips every conclusion.
• Assuming more ligand molecules always means more stability — it is the number and delivery of donor atoms (chelation) that matters, not the raw count of separate ligand molecules, as the worked example shows.
• Treating a large $K_f$ as evidence of fast ligand exchange — $K_f$ is a thermodynamic (equilibrium-position) quantity, not a rate; a complex can be thermodynamically very stable yet still exchange ligands quickly, or vice versa (this is the separate idea of kinetic lability/inertness).
• Forgetting to raise $[L]$ to the power $n$ matching the stoichiometry when writing the $K_f$ expression, or writing the reverse (dissociation) expression when the question asks for the formation constant.

Ask the tutor: ask the AI tutor to give you two complexes with numeric $K_f$ values and walk you through ranking their stability using the charge-density, field-strength, and chelate-effect reasoning above.