Inorganic Chemistry · Part 8 of 9

Reaction Mechanisms & Kinetics

Coordination Chemistry, Part 8 · 9 sections · about 26,616 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

A complex can be thermodynamically unstable and sit unchanged on the shelf for years, or be perfectly stable and exchange its ligands a million times a second. This part is about that second axis — rate rather than position — and it is where coordination chemistry becomes predictive. It covers why some centres are inert, how ligands are substituted in octahedral and square-planar complexes, the trans effect that lets a chemist choose which isomer to make, and the two mechanisms of electron transfer, including the theory that won Marcus a Nobel Prize and the experiment that won one for Taube. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 8

  • 1K.1 Stable is not the same word as inert Free below
  • 2K.2 Crystal field activation energy — why d³ and low-spin d6 are inert
  • 3K.3 D, A and I — naming the mechanism and proving which it is
  • 4K.4 Octahedral substitution — acid hydrolysis, base hydrolysis, anation and stereochemistry
  • 5K.5 Square-planar substitution — the two-term rate law
  • 6K.6 The trans effect — choosing which isomer you get
  • 7K.7 Outer-sphere transfer, the Franck–Condon restriction and Marcus theory
  • 8K.8 Inner-sphere transfer — Taube’s experiment
  • 9K.9 Photochemistry of coordination compounds — a short introduction

K.1 Stable is not the same word as inert

Free extract

Section 1 of Part 8, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

The single most important distinction in this entire part, made with two examples that cannot be argued with.

Two words are used loosely in conversation and precisely in exams, and the confusion between them costs more marks in this topic than any other single error.

Stable / unstable are thermodynamic words. They describe the position of equilibrium — how far downhill the products lie. They are quantified by ΔG°, by K, by the formation constants βn of Part 7.
Labile / inert are kinetic words. They describe the height of the barrier — how quickly the system gets to equilibrium. They are quantified by rate constants and by ΔG‡.

The two are set by different features of the same free-energy diagram: stability by the difference in height between the two ends, lability by the height of the hill in the middle. Nothing whatever connects them. You can have a deep valley behind a high wall, or a shallow one behind no wall at all.

Example 1 — stable but labile: [Ni(CN)₄]²⁻

Tetracyanidonickelate(II) is one of the most thermodynamically stable complexes in ordinary aqueous chemistry. The critically selected value for its overall (cumulative) formation constant is log β₄ = 30.2 at 25 °C and zero ionic strength. Read that symbol carefully: β₄ is the constant for the whole assembly Ni²⁺ + 4 CN⁻ ⇌ [Ni(CN)₄]²⁻, not the stepwise K₄ for adding the fourth ligand to [Ni(CN)₃]⁻. A β₄ of 10³⁰·² means that at equilibrium the concentration of free Ni²⁺ in a cyanide solution is vanishingly small. On any thermodynamic measure, the nickel is locked up.

Ni²⁺ + 4 CN⁻ ⇌ [Ni(CN)₄]²⁻   log β₄ = 30.2 (25 °C, I = 0)  —  the equilibrium lies almost entirely to the right
Note: A formation constant quoted without a temperature and an ionic strength is not really a constant, because both change it — often by a whole log unit or more. The 30.2 above is a thermodynamic constant, extrapolated to infinite dilution (I = 0). Individual experimental determinations of this particular system have been reported about an order of magnitude higher, near 10³¹, which is why a compilation value is worth more than any one paper. Whenever you quote a β, quote its conditions with it.

Now do the experiment that separates the two axes. Put ¹³C-labelled cyanide in the solution and watch, by NMR, how long a cyanide already bound to nickel stays bound before a labelled one takes its place.

It is fast — and this is the part worth getting right — it is fast in a way that has been measured, not merely observed to be over before anyone could look. The exchange obeys a clean second-order rate law, first order in the complex and first order in free cyanide, so the incoming ligand does the work: this is associative substitution at a square-planar centre, the subject of K.5. On the same measurement scale the platinum analogue [Pt(CN)₄]²⁻ has k₂ = 11 dm³ mol⁻¹ s⁻¹ at 298 K, and the exchange rate constants for Pt : Pd : Ni stand in the ratio 1 : 7 : 200 000 — which puts nickel at roughly 2 × 10⁶ dm³ mol⁻¹ s⁻¹.

[Ni(CN)₄]²⁻ + *CN⁻ → [Ni(CN)₃(*CN)]²⁻ + CN⁻   —  rate = k₂[complex][CN⁻],  k₂ ≈ 2 × 10⁶ dm³ mol⁻¹ s⁻¹ at 298 K

Put a number on what that means at the bench. In a solution 10⁻³ M in free cyanide the pseudo-first-order constant is k₂[CN⁻] ≈ 2 × 10³ s⁻¹, so the half-life of any particular Ni–CN bond is about 0.3 milliseconds. Pouring one solution into another and stirring takes of the order of a second. By the time the mixing is finished, every cyanide in the sample has been swapped some thousands of times over — which is why a hand-mixed radiolabel experiment can only ever report that the exchange was already complete. The reaction is far too fast for a stopwatch. It is not too fast for NMR, and the modern number is the one to quote.

The nickel system also shows you the intermediate that an associative path demands: [Ni(CN)₅]³⁻ is a real, characterised five-coordinate species rather than a hypothesis, and it is present in equilibrium in cyanide-rich solutions of the complex.

So this complex has essentially all of its nickel bound at any instant, and yet no individual cyanide stays bound for long. Stable and labile at the same time. There is no contradiction: the equilibrium constant says how much complex there is, the rate constant says how often a given ligand leaves. A crowded railway station can be permanently full and yet contain almost no-one who was there ten minutes ago.

Example 2 — unstable but inert: [Co(NH₃)₆]³⁺ in acid

Now the mirror image. Hexaamminecobalt(III) in 1 M acid is, thermodynamically, in serious trouble. Ammonia is a base; acid protonates it; the driving force for the reaction below is very large, with an equilibrium constant of the order of 10²⁵.

[Co(NH₃)₆]³⁺ + 6 H₃O⁺ → [Co(H₂O)₆]³⁺ + 6 NH₄⁺   K ≈ 10²⁵

A reaction with K = 10²⁵ is, for practical purposes, complete. And yet a solution of [Co(NH₃)₆]³⁺ in 1 M acid can be kept for days at room temperature without visible change. The barrier is high; the driving force is irrelevant to the rate. Unstable and inert at the same time.

The pair of examples is the standard exam illustration, and the standard exam trap is to be given one of them and asked whether the complex is “stable”. The correct answer always names the axis: thermodynamically unstable, kinetically inert.

Taube’s operational definition

“Fast” and “slow” are useless unless a line is drawn somewhere. Henry Taube drew it in a place that has nothing to do with theory and everything to do with what a chemist at a bench can see.

Taube’s criterion. A complex is labile if its substitution reactions are essentially complete within about one minute at 25 °C for solutions around 0.1 M. If they take appreciably longer, it is inert.

Three things about that definition deserve emphasis, because each is examinable.

  1. It is operational, not theoretical. The one-minute mark is not derived from anything; it is roughly the time in which a chemist mixing two solutions by hand can see whether something has happened. Everything faster looks instantaneous; everything slower can be followed.
  2. It fixes a temperature and a concentration. Without those the statement is meaningless, because rates depend on both. A complex that is inert at 25 °C may be labile at 80 °C.
  3. “Inert” is a relative term, not an absolute one. Roughly, it corresponds to a first-order rate constant below about 10⁻² s⁻¹, or a half-life above about a minute. Nothing is truly unreactive.
Note: Some authors, and some question papers, prefer “non-labile” to “inert”, on the reasonable ground that “inert” suggests chemical unreactivity of the noble-gas kind, which is not what is meant. Treat the two words as synonyms; the syllabus and most textbooks use inert.
Water exchange spans almost twenty orders of magnitude10-1010-810-610-410-21001021041061081010k(water exchange) / s⁻¹ at 298 K — logarithmic scaleIr³⁺Rh³⁺Cr³⁺Ru³⁺Ru²⁺Al³⁺V²⁺Fe³⁺Ni²⁺Ti³⁺Mg²⁺Co²⁺Fe²⁺Mn²⁺Cr²⁺Cu²⁺EXTREMELY INERTEXTREMELY LABILEd³ (Cr³⁺) and low-spin d⁶ (Rh³⁺, Ir³⁺, Ru²⁺) at the slow end; Jahn–Teller d⁹ (Cu²⁺) and high-spin d⁴ (Cr²⁺) at the fast end.
One ligand, one reaction, twenty powers of ten. Every point is the same elementary process — a water molecule in the first coordination sphere of an aqua ion trading places with one in the second. Iridium(III) does it about once every three hundred years; copper(II) does it several billion times a second. Nothing about the thermodynamic stability of these aqua ions varies over anything like that range, which is the whole point of K.1. Positions are plotted on a base-10 logarithmic scale from tabulated values; every individual value must be checked against Merbach’s compilations before publication.

The figure above is worth studying for a full minute. It plots one single reaction — a coordinated water swapping with a bulk water — for a set of ordinary metal aqua ions, on a logarithmic scale. The span from iridium(III) at the left to copper(II) at the right is close to twenty powers of ten. No other family of reactions in inorganic chemistry varies so widely while remaining, formally, the same reaction. K.2 is about why.

⚠ Common mistakes & exam traps

  • “Stable” does not mean “slow to react”. This is the single most common error in the whole of coordination kinetics. If a question gives you a large βn and asks about rate, the βn is a distractor. If it gives you a rate constant and asks about stability, the rate constant is a distractor.
  • Do not say “unstable” when you mean “labile”. In a written answer the marker is looking for the correct word, not the correct idea loosely expressed. Write thermodynamically unstable and kinetically inert in full when both axes are in play.
  • Inertness is not a property of the metal alone. It belongs to the complex — oxidation state, d-electron count, spin state and the ligands all matter. Cobalt(III) ammines are inert; cobalt(II) ammines are labile; the element is the same.
  • Taube’s one-minute line is a convention. A complex with a half-life of 90 s is not qualitatively different from one with a half-life of 30 s. Do not treat the boundary as physically meaningful.

[Fe(CN)₆]⁴⁻ is famously stable and is used in analysis; free cyanide is famously toxic. Explain, in kinetic and thermodynamic language, why potassium hexacyanidoferrate(II) can be handled as an ordinary laboratory salt. Medium

Thermodynamics. The formation constant of [Fe(CN)₆]⁴⁻ is very large, so at equilibrium in neutral water the free cyanide concentration is extremely small. That is a genuine part of the answer — the equilibrium simply does not release much CN⁻.
But that is not the whole answer, because equilibrium says nothing about how fast cyanide is released, and in an acidic environment such as the stomach the equilibrium is pushed the other way by protonation of CN⁻ to HCN.
Kinetics. Iron(II) with six strong-field cyanides is low-spin d6 — the archetypal inert configuration (K.2). Ligand release is slow, so even where the thermodynamics turn unfavourable the compound does not respond quickly.
The correct answer names both axes. Stability keeps the equilibrium concentration of free cyanide low; inertness keeps the approach to any new equilibrium slow. Neither alone would be sufficient, and this is exactly the pairing the examiner is testing.
Caveat worth stating. This is a chemical argument about lability, not a toxicological clearance. Prussian-blue and hexacyanidoferrate chemistry does release cyanide under photolysis and under strongly acidic conditions.
Advanced / reference layer

Where the two axes are formally connected — and where the connection is an illusion.

It is worth being precise about the sense in which stability and lability are “independent”, because a careful student will notice that they cannot be completely independent.

For any elementary step, the forward and reverse rate constants are tied to the equilibrium constant:

K = kf / kr

So the two rate constants and the equilibrium constant form a triangle: fix any two and the third follows. What is not fixed is the absolute magnitude of either rate constant. Multiplying both kf and kr by 10⁶ — which is what lowering the barrier does — leaves K untouched. In free-energy language: K is set by ΔG°, the difference between the two wells; the rates are set by ΔG‡, the height of the peak above the reactant well; and the peak can be raised or lowered without moving either well.

There is, however, a real and much-abused correlation lurking here. Within a closely related series — the same metal, the same mechanism, the same reaction type, only the leaving group varying — a linear free-energy relationship often holds:

log kaq = α log K + constant

For the acid hydrolysis of [Co(NH₃)₅X]²⁺ the slope α comes out close to 1.0, and that number is one of the sharpest pieces of mechanistic evidence in the subject (K.4). A slope of one means the transition state responds to a change of leaving group as strongly as the products do — that is, the Co–X bond is essentially completely broken at the transition state. This is a genuine kinetics–thermodynamics link, but note carefully what it is: it holds within a series, it is empirical, and it says something about the structure of one transition state. It does not license the general statement that stable complexes react slowly.

Why the confusion is so persistent

Two accidents of language keep the error alive. First, ordinary English uses “stable” to mean “does not change”, which is the kinetic sense. Second, in a great many familiar systems the two do coincide — strong bonds in simple molecules usually are both thermodynamically favourable and kinetically robust — so the habit is reinforced by experience before the exception is ever met. Coordination chemistry is precisely where the coincidence breaks down, because the barrier height is controlled by d-electron configuration (K.2) and the well depth by donor strength and chelation (Part 7), and those are two different things.

Med
[Cr(H₂O)₆]³⁺ and [Cr(H₂O)₆]²⁺ differ by one electron. Water exchange on the +3 ion has a rate constant of order 10⁻⁶ s⁻¹, on the +2 ion of order 10⁹ s⁻¹. Fifteen orders of magnitude for one electron. Is this a thermodynamic or a kinetic difference, and what is its origin?
Show solution
It is purely kinetic. Both ions are perfectly ordinary aqua ions with unremarkable formation constants; nothing about the position of any equilibrium changes by fifteen orders of magnitude. The difference is in the barrier. Cr³⁺ is d³ — a half-filled t2g set, the configuration with the largest crystal field stabilisation available in an octahedron for a high-spin ion (12 Dq, tied with d⁸), and one that loses a great deal of that stabilisation in any five- or seven-coordinate transition state (K.2). Cr²⁺ is high-spin d⁴ — strongly Jahn–Teller distorted, with two long axial Cr–O bonds that are already most of the way to being broken. The +2 ion starts halfway up the hill.
Easy
A student writes: ‘[Co(NH₃)₆]³⁺ does not decompose in acid, so the Co–N bonds must be very strong and the complex very stable.’ Criticise this sentence.
Show solution
Two errors, one of fact and one of logic. Of fact: the complex is thermodynamically unstable in acid — K for its conversion to the aqua ion and ammonium is of the order of 10²⁵, so the products are very much lower in free energy. Of logic: even if the bonds were strong, bond strength is a ground-state property and does not by itself determine a rate; what determines the rate is the energy of the transition state relative to the ground state. The correct sentence is: the complex is thermodynamically unstable in acid but kinetically inert, because low-spin d6 Co(III) has a large crystal field activation energy for any change of coordination number.
Easy
Give an example, other than the two in the text, of a species that is thermodynamically unstable with respect to a well-defined set of products and yet perfectly inert on a laboratory timescale — and say what the barrier is.
Show solution
Several will do. (i) [Co(CN)₆]³⁻ is unstable with respect to reaction with water but is low-spin d6 and inert. (ii) Almost any Cr(III) complex in a coordinating solvent — d³ centres are inert regardless of what the thermodynamics prefer. (iii) Outside coordination chemistry, diamond with respect to graphite, or any hydrocarbon with respect to CO₂ and H₂O in air. In every case the barrier is a high ΔG‡ for the only accessible pathway, and in the coordination examples that barrier is dominated by the crystal-field term of K.2. The general moral: a large negative ΔG° tells you a reaction can happen, never that it will.

On the stability/lability distinction, Taube’s criterion, and the two numbers quoted for [Ni(CN)₄]²⁻:

  • H. Taube, Chemical Reviews, 1952 — the review in which the correlation between electronic structure and substitution rate in octahedral complexes was set out. Volume, page and exact title to be confirmed.
  • The Nobel Prize in Chemistry 1983 was awarded to Henry Taube “for his work on the mechanisms of electron transfer reactions, especially in metal complexes”.
  • R. M. Smith and A. E. Martell, Critical Stability Constants, Vol. 4: Inorganic Complexes, Plenum Press, New York, 1976, continued as the NIST Critically Selected Stability Constants of Metal Complexes database (NIST Standard Reference Database 46) — the source of log β₄ = 30.2 for [Ni(CN)₄]²⁻ at 25 °C and zero ionic strength. The value is reproduced in US EPA SW-846 Method 9015, Revision 1 (July 2014), Table 3, Stability Constants of Metal Cyanide Complexes.
  • F. J. Monlien, L. Helm, A. Abou-Hamdan and A. E. Merbach, ‘Mechanistic diversity covering 15 orders of magnitude in rates: cyanide exchange on [M(CN)₄]²⁻ (M = Ni, Pd and Pt)’, Inorganic Chemistry, 2002, 41, 1717–1727 (DOI 10.1021/ic010917e) — the ¹³C NMR study that measures the cyanide-exchange rate law and rate constants used in K.1, and identifies the second-order, cyanide-dependent path.

Read the rest of Part 8

The remaining 8 sections of this part — K.2 Crystal field activation energy — why d³ and low-spin d6 are inert, K.3 D, A and I — naming the mechanism and proving which it is, K.4 Octahedral substitution — acid hydrolysis, base… — and all nine parts of Coordination Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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