Ligand substitution and lability

Rates of ligand exchange, half-lives, Taube's labile and inert classes and their link to $e_g^*$ electrons, dissociative and associative mechanisms, and the trans effect in the synthesis of cisplatin.

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1. What you will learn

By the end of this lesson you will be able to find the half-life of a ligand substitution, classify a complex as labile or inert, and use the trans effect to plan a synthesis.

2. What you already have

From general chemistry you know first-order kinetics: a rate constant $k$, and a half-life $0.693/k$ that does not depend on how much is present. From lesson 16 you know formation constants measure how far a complex forms; from lesson 26 that electrons in the antibonding $e_g^*$ orbitals weaken metal-ligand bonds. This lesson asks how fast ligands come and go, a separate question from how many stay.

3. Words for this lesson

4. How fast ligands come and go

Term	What it means
Ligand substitution	A reaction in which one ligand on a metal is replaced by another.
Water exchange	The replacement of a coordinated water by a water molecule from the solvent, the simplest substitution.
Labile	Exchanging ligands quickly; by Taube's definition, with a half-life under a minute at room temperature.
Inert	Exchanging ligands slowly; a half-life over a minute. A matter of rate, not of stability.
Dissociative mechanism, D	The leaving ligand departs first, making an intermediate of lower coordination number.
Associative mechanism, A	The entering ligand binds first, making an intermediate of higher coordination number.
Trans effect	The ability of a ligand to speed the substitution of the ligand across from it in a square-planar complex.

Put any metal ion in water and its coordinated water molecules are constantly being replaced by water from the solvent. The rate of this water exchange is measured as a first-order rate constant $k$, and its half-life is

The range is enormous. Copper(II) exchanges its water with $k$ about $4 \times 10^{9}$ s$^{-1}$, a half-life of a fraction of a nanosecond; nickel(II) with $k = 3.2 \times 10^{4}$ s$^{-1}$, about twenty microseconds; chromium(III) with $k = 2.4 \times 10^{-6}$ s$^{-1}$, a half-life of about three days; rhodium(III) with $k = 2.2 \times 10^{-9}$ s$^{-1}$, about ten years.

Henry Taube drew a practical line: a complex that exchanges its ligands with a half-life of less than a minute at room temperature is labile; one slower than that is inert. What decides which side a complex falls on is largely its electrons, as lesson 26 predicted. Complexes with *no electrons in the antibonding $e_g^$ set*, $d^3$ ions such as chromium(III) and low-spin $d^6$ ions such as cobalt(III), rhodium(III) and platinum(IV), are inert. Complexes with $e_g^$ electrons are labile, and the more weakly held their ligands, the faster: copper(II), with three $e_g^*$ electrons and a Jahn-Teller distortion that leaves two water molecules loosely bound, is among the fastest of all. Charge matters too: within similar ions, a higher charge holds the water more tightly.

Inert and labile describe rates. They are not the same as unstable and stable, which describe equilibrium. Hexaamminecobalt(III) is thermodynamically unstable in acid, where it would release its ammonia given the chance, but it is so inert that a solution of it survives for days. Tetracyanidonickelate(II) is very stable, with a huge formation constant, yet it exchanges cyanide within seconds.

Another way: picture

Think of the ligands as guests in chairs round a table. In a labile complex the guests keep getting up and others sitting down, so the seating changes every moment, even though the table is always full. In an inert complex the guests stay seated for hours. How long they stay says nothing about whether the table would rather have different guests; that is the equilibrium question, answered separately.

Another way: steps

Rate constant $k$ for the substitution, in s$^{-1}$.
Half-life $t_{1/2} = 0.693/k$.
Labile if $t_{1/2} < 60$ s; inert if longer.
Predict from electrons: empty $e_g^$ ($d^3$, low-spin $d^6$) inert; $e_g^$ electrons labile.
After $n$ half-lives, one part in $2^n$ remains.

5. How ligands take turns: the mechanisms

A substitution needs one ligand to leave and another to arrive, and the order matters. In a dissociative mechanism, D, the leaving ligand departs first, giving a short-lived intermediate with one fewer ligand, which the new ligand then fills. The rate depends on the complex but not on the incoming ligand: rate $= k[\mathrm{complex}]$. Octahedral complexes, already crowded with six ligands, usually react this way.

In an associative mechanism, A, the incoming ligand binds first, giving an intermediate with one more ligand, and then the leaving ligand departs. The rate depends on both: rate $= k[\mathrm{complex}][\mathrm{L}]$. Square-planar complexes, with open space above and below the plane, react this way, passing through a five-coordinate intermediate. Many real reactions lie between the two extremes, an interchange mechanism, I, in which the new ligand arrives as the old one leaves, and chemists label them $I_d$ or $I_a$ by which bond matters more.

Telling the mechanisms apart takes experiments. The rate law is the first clue: if doubling the concentration of the entering ligand doubles the rate, that ligand takes part in the slow step, pointing to association. A second clue comes from squeezing the solution. Running the reaction under high pressure speeds up an associative reaction, whose crowded intermediate takes up less room, and slows a dissociative one, whose intermediate has lost a ligand and takes up more. Measurements of this kind, made at pressures of a few thousand atmospheres, showed that water exchange shifts from associative for the early first-row ions such as vanadium(II) to dissociative for the late ones such as nickel(II).

6. The trans effect and the synthesis of cisplatin

In square-planar complexes of platinum(II), some ligands make the ligand directly across from them, trans, leave much faster. This trans effect follows a sequence that runs roughly $\mathrm{CN^-}$, $\mathrm{CO}$ $>$ $\mathrm{PR_3}$ $>$ $\mathrm{I^-}$ $>$ $\mathrm{Br^-}$ $>$ $\mathrm{Cl^-}$ $>$ $\mathrm{NH_3}$ $>$ $\mathrm{H_2O}$: ligands that are strong $\pi$ acceptors or strong $\sigma$ donors labilize the position across from them.

Chemists use it to make a particular isomer on purpose. To make cisplatin, $\mathrm{cis\text{-}[PtCl_2(NH_3)_2]}$, start from $\mathrm{[PtCl_4]^{2-}}$ and add ammonia. The first ammonia replaces any chloride. Now there are two kinds of chloride left: the one trans to ammonia and the two trans to other chlorides. Chloride has the larger trans effect, so the second ammonia replaces a chloride trans to another chloride, and the two ammonias end up cis. Starting instead from $\mathrm{[Pt(NH_3)_4]^{2+}}$ and adding chloride gives the trans isomer by the same logic. The order of addition decides the product.

7. Why the electrons control the rate

Ligand field theory explains Taube's pattern. For a dissociative reaction, a bond must break, and the energy needed depends on how strong the metal-ligand bonds are and how much ligand field stabilization is lost on the way to the five-coordinate intermediate. A $d^3$ or low-spin $d^6$ ion has all its d electrons in the nonbonding $t_{2g}$ set, its bonds are at full strength, and it loses a large amount of stabilization when a ligand leaves; the barrier is high and the reaction slow.

An ion with $e_g^*$ electrons starts with weakened bonds and loses little or no stabilization, so its barrier is low. High-spin $d^5$ manganese(II), with no ligand field stabilization to lose at all, exchanges water about as fast as a simple ion of its size and charge. Chromium(III) and cobalt(III), by contrast, are inert enough that Werner could separate and study their isomers in the 1900s, which is why so much of early coordination chemistry was built on them.

8. Checking a kinetics answer

Three checks catch most slips. First, the half-life formula: $0.693$ divided by $k$, not multiplied, and not $k$ divided by $0.693$. A larger rate constant must give a shorter half-life. Second, the units: $k$ in s$^{-1}$ gives a half-life in seconds, and converting to hours means dividing by $3{,}600$, not $60$.

Third, the classification must match the electrons: an ion with an empty $e_g^$ set that comes out labile, or a high-spin ion with $e_g^$ electrons that comes out inert, deserves a second look at the arithmetic. And remember the powers of ten: a rate constant of $10^{-6}$ s$^{-1}$ means days, one of $10^{6}$ s$^{-1}$ means a fraction of a microsecond. After $n$ half-lives, one part in $2^n$ of the starting complex remains: ten half-lives leave about one part in a thousand.

Finally, keep kinetics and thermodynamics apart when checking. A question about how fast a ligand leaves is answered with a rate constant or a half-life; a question about how much complex is present at equilibrium is answered with a formation constant. Using one to answer the other is the mistake the misconception below describes, and it is easy to make when both are called stability in everyday speech.

9. In the world: how cisplatin works

Cisplatin, approved in the United States in 1978, remains one of the most widely used cancer drugs, especially against testicular, ovarian and bladder cancers. It works because of substitution kinetics. In the bloodstream the chloride concentration is high, about $0.1$ M, which holds back the loss of chloride. Inside a cell the chloride concentration is much lower, and one chloride is replaced by water with a half-life of about two hours.

The aquated complex is more reactive, and it binds to nitrogen atoms of guanine bases in DNA, linking neighboring bases and kinking the double helix, which the cell cannot repair. Its partner drug carboplatin replaces the two chlorides with a chelating dicarboxylate ring that hydrolyzes far more slowly, over about a day, which lowers the dose reaching the kidneys at once and is a large part of why it has milder side effects. The rate of one substitution step shapes a whole treatment.

10. In the world: why chromium(III) is used and chromium(VI) is feared

Chromium(III) is inert, and that is why it is useful and comparatively safe. In leather tanning, chromium(III) binds to the collagen of animal hides and stays bound for the life of the leather, because its ligands exchange over days, not microseconds. The same inertness keeps it from moving easily into cells.

Chromium(VI), as chromate or dichromate, is a different species altogether: an oxyanion that passes into cells through channels meant for sulfate and phosphate, then is reduced inside to reactive intermediates that damage DNA. It is a recognized carcinogen, and U.S. agencies regulate it strictly in drinking water and in workplaces such as plating shops. Treating chromium(VI) waste usually means reducing it to chromium(III), turning a mobile, toxic species into an inert one that can be precipitated and removed.

11. Inert does not mean stable

Because inert complexes last, it is tempting to call them stable. The two words answer different questions. Stability is thermodynamic: how far the equilibrium lies toward the complex, measured by a formation constant. Inertness is kinetic: how fast the complex reacts, measured by a rate constant. Hexaamminecobalt(III) is thermodynamically unstable in acid but survives for days because it is inert; tetracyanidonickelate(II) is highly stable but labile.

A second error is to think a labile complex falls apart. A labile complex keeps its full set of ligands at every moment; it simply swaps them rapidly for identical ones. Lability describes the traffic, not the occupancy.

12. Chromium(III) in water

13. Nickel(II) in water

14. Making cisplatin, not transplatin

15. Your turn: iron(III) in water exchanges with $k = 1.6 \times 10^{2}$ s$^{-1}$. What is its half-life, and is it labile?

16. Guided practice

17. Guided practice

Complete the worked solution: decide whether $\mathrm{[Ni(H_2O)_6]^{2+}}$, whose water exchange has a rate constant of $3.2 \times 10^{4}$ s$^{-1}$, is labile or inert.

18. Guided practice

$\mathrm{[Cr(H_2O)_6]^{3+}}$ exchanges its water ligands with a rate constant of $2.4 \times 10^{-6}$ s$^{-1}$. By Taube's definition, is it labile or inert?

19. Practice

$\mathrm{[Cr(H_2O)_6]^{3+}}$ ($d^{3}$) exchanges water with $k = 2.4 \times 10^{-6}$ s$^{-1}$ and $\mathrm{[Ni(H_2O)_6]^{2+}}$ ($d^{8}$) with $k = 3.2 \times 10^{4}$ s$^{-1}$; both are high spin except low-spin $d^6$. For each, in that order, fill in its $e_g^*$ electrons, its half-life in seconds, and $1$ if it is labile or $0$ if inert.

20. Practice

$\mathrm{[Rh(H_2O)_6]^{3+}}$ exchanges its water ligands with a first-order rate constant of $2.2 \times 10^{-9}$ s$^{-1}$. What is the half-life of the exchange, in seconds?

21. Practice

The water ligands of $\mathrm{[Ni(H_2O)_6]^{2+}}$ exchange with a first-order rate constant of $3.2 \times 10^{4}$ s$^{-1}$, a half-life of $0.0000217$ s. How many seconds pass before only one part in $2^{2}$ of the original water ligands is left unexchanged?

22. Somewhere new

A pharmacologist compares how quickly three metal drugs are activated in the body, each by a first-order reaction. cisplatin loses a chloride ligand to water with $k = 0.0000963$ s$^{-1}$; carboplatin loses its dicarboxylate ring to water with $k = 0.00000770$ s$^{-1}$; a ruthenium(III) anticancer candidate loses a chloride ligand to water with $k = 0.0000385$ s$^{-1}$. For each, in that order, fill in the half-life in seconds and in hours.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

$\mathrm{[V(H_2O)_6]^{2+}}$ ($d^{3}$) exchanges water with $k = 8.7 \times 10^{1}$ s$^{-1}$ and $\mathrm{[Ni(H_2O)_6]^{2+}}$ ($d^{8}$) with $k = 3.2 \times 10^{4}$ s$^{-1}$; both are high spin except low-spin $d^6$. For each, in that order, fill in its $e_g^*$ electrons, its half-life in seconds, and $1$ if it is labile or $0$ if inert.

25. What you can do now

You can reason about substitution rates. Explain why chromium(III) keeps its water ligands for days while nickel(II) swaps them in microseconds, and why inert is not the same as stable.

Working for the steps left to you

15. Your turn: iron(III) in water exchanges with $k = 1.6 \times 10^{2}$ s$^{-1}$. What is its half-life, and is it labile?, step 3

	$2.2 \times 10^{-9}$ s⁻¹	$1.8 \times 10^{-2}$ s⁻¹	$3.2 \times 10^{6}$ s⁻¹	$4.4 \times 10^{9}$ s⁻¹
$\mathrm{[Rh(H_2O)_6]^{3+}}$
$\mathrm{[Ru(H_2O)_6]^{2+}}$
$\mathrm{[Co(H_2O)_6]^{2+}}$
$\mathrm{[Cu(H_2O)_6]^{2+}}$

	eg* electrons	half-life (s)	labile (1) or inert (0)
the first ion
the second ion

	half-life (s)	half-life (h)
the first drug
the second drug
the third drug

	eg* electrons	half-life (s)	labile (1) or inert (0)
the first ion
the second ion