Chemistry · Lesson

Ligand substitution, electron transfer and metal-complex reactivity

Distinguish ligand replacement, electron transfer and equilibrium preference. Test simple rate laws and conserved redox ledgers without treating one observed dependence as a complete mechanism.

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By the end, you can

Compare associative, dissociative and interchange substitution descriptions.
Calculate a stipulated first-order conversion and half-time.
Separate formal electron bookkeeping from electron-transfer pathway evidence.
Explain how forward and reverse rates can change while an equilibrium ratio stays fixed.

Ligand replacement has several possible pathways

A substitution changes which ligand is bound to a metal. In an associative limiting description, entry of the new ligand precedes departure of the old one through a higher-coordinate intermediate. In a dissociative limiting description, departure precedes entry through a lower-coordinate intermediate. Interchange describes coupled entry and departure without a separately resolved intermediate; associative or dissociative character can differ within that class.

The starting coordination number, electron structure, solvent, ligand sterics and other interactions influence the possibilities. A schematic sequence is a model to test. Changes in rate with entering ligand, activation parameters and evidence for intermediates can help discriminate pathways, but one trend may admit several explanations. Solvent participation and pre-equilibria can modify a simple interpretation.

A first-order toy makes the timescale explicit

In an invented irreversible conversion with constant observed first-order coefficient k_obs=0.0200 s⁻¹, the fraction of starting entity remaining is exp(−k_obs t). At 60.0 s it is exp(−1.20)≈ 0.301194; converted fraction is 0.698806. The model half-time ln 2/k_obs≈ 34.6574 s.

This equation assumes a single monitored conversion, constant coefficient and no appreciable reverse reaction or other products in the stated pool. An observed first-order coefficient can arise under maintained excess conditions or from a more involved mechanism; the exponential shape alone does not identify the ligand’s entry/departure sequence.

Fit the dependence, then limit the mechanistic claim

A separate invented data pair at maintained free entering-ligand concentrations 0.0100 and 0.0500 M has observed coefficients 0.00300 and 0.0110 s⁻¹. Fit k_obs=k₀+k₂[L]. The slope is (0.0110−0.00300)/(0.0500−0.0100)=0.200 M⁻¹ s⁻¹, and intercept 0.00100 s⁻¹. At 0.100 M the selected linear model predicts 0.0210 s⁻¹.

These two inputs exactly determine a line but do not establish its validity outside the stated range, uncertainty or unique mechanism. A ligand-dependent channel plus a ligand-independent channel is one interpretation; pre-equilibria, solvent routes and interchange possibilities require their own evidence. A rate law constrains hypotheses rather than naming an intermediate by itself.

Redox bookkeeping and transfer pathway are separate

Consider the symbolic net [Fe(CN)₆]⁴⁻+[Co(NH₃)₆]³⁺→[Fe(CN)₆]³⁻+[Co(NH₃)₆]²⁺. Using the selected ligand assignments, Fe changes +2→+3 and Co +3→+2, exchanging one formal electron. Each side has total charge −1, and the ligand inventories are unchanged in this net. No practical redox conditions or actual potential are supplied.

Outer-sphere descriptions retain the relevant coordination shells during the transfer step; inner-sphere descriptions involve a bridging interaction connecting centres in the pathway. A balanced equation with unchanged net ligand inventory cannot by itself distinguish them. Molecular electronic coupling and reorganization require evidence beyond an oxidation-state ledger.

For a separate stipulated thermodynamic calculation at 298.15 K, assign this one-electron net a standard cell potential +0.200 V with consistent activity standard states. F=N_Ae≈ 96485.3321 C mol⁻¹ gives Δ_rG°=−FE°≈−19297.1 J mol⁻¹. Using R=N_Ak_B with exact k_B=1.380649× 10⁻²³ J K⁻¹ gives lnK=FE°/(RT)≈ 7.78435 and K≈ 2402.70. This invented potential is not a measured potential for the named complexes or a prediction of their actual rates.

Speed can change while a reversible ratio stays fixed

In another invented closed two-state pool, A⇌B has k_f=0.00300 s⁻¹ and k_r=0.00100 s⁻¹. Its stationary ratio B/A=k_f/k_r=3 gives B fraction 0.750. Starting entirely in A, f_B(t)=0.750[1−exp(−0.00400t)], with relaxation time 1/(k_f+k_r)=250 s. At 60.0 s, f_B≈ 0.160029.

Multiplying both coefficients by 100 preserves ratio 3 and equilibrium fraction 0.750, but changes the relaxation time to 2.50 s. At 60.0 s the faster model is effectively at its stationary fraction. This explicitly preserves the same forward/reverse ratio. It illustrates the difference between rate scale and equilibrium preference; it is not evidence that every real intervention scales both rates by the same amount.

Thermodynamic stability and substitution inertness answer different questions. Inert and labile refer to kinetic behavior on a specified timescale and conditions, rather than unconditional labels for every compound of one metal. Connect a persistence claim to the observed process and environment.

Two rate scales with the same stationary composition

Fraction of the stated pool in B

Model time / s

Slower reversible model
Both coefficients multiplied by 100
— Slower two-state approach

Underlying data

Model time / s	Fraction of the stated pool in B
0	0 · Slower reversible model
60	0.16002910420008498 · Slower reversible model
100	0.2472599654732705 · Slower reversible model
0	0 · Both coefficients multiplied by 100
60	0.7499999999716864 · Both coefficients multiplied by 100
100	0.75 · Both coefficients multiplied by 100

Original closed A⇌B model starting entirely in A. Slow coefficients k_f=0.00300 s⁻¹ and k_r=0.00100 s⁻¹ are both multiplied by 100 for the faster case, preserving B/A=3 and stationary B fraction 0.750. Markers select three times for each case; the dashed curve follows the slower equation. These are invented rates, not a measured metal-complex reaction or proof of a catalyst mechanism.

Worked examples

Separate remaining and converted fractions

Use k_obs=0.0200 s⁻¹ and t=60.0 s in the irreversible toy.

The dimensionless exponent is −0.0200× 60.0=−1.20.
Remaining fraction≈ 0.301194.
Converted fraction=1−0.301194≈ 0.698806.
Half-time=ln 2/0.0200≈ 34.6574 s.

About 69.9% converted within the selected single-conversion model, without determining its molecular pathway.

Retain slope units in the two-point rate fit

Use the stated concentration/coefficient pairs 0.0100 M/0.00300 s⁻¹ and 0.0500 M/0.0110 s⁻¹.

Slope 0.00800/0.0400=0.200 M⁻¹ s⁻¹.
Intercept 0.00300−0.200× 0.0100=0.00100 s⁻¹.
At 0.100 M, k_obs=0.00100+0.0200=0.0210 s⁻¹.

The fitted linear dependence, with a unique associative, dissociative or interchange mechanism unproved.

Check the one-electron and whole-net constraints

Use the stated iron/cobalt net, then the separate invented E°+0.200 V.

Fe loses one formal electron; Co gains one.
Net charge −4+3=−1 and −3+2=−1.
At 298.15 K, Δ_rG°≈−19.2971 kJ mol⁻¹ and lnK≈ 7.78435.

One conserved redox ledger and an invented standard-state K≈ 2402.70; neither identifies an actual transfer pathway.

Compare a ratio with a relaxation time

Use k_f=0.00300 and k_r=0.00100 s⁻¹, then scale both by 100.

Ratio remains 3 and stationary B fraction remains 3/4.
Initial relaxation time is 1/0.00400=250 s.
Scaled time is 1/0.400=2.50 s.

A hundredfold faster approach with the same stationary fraction in the explicitly consistent two-state model.

Try it yourself

In the same irreversible k_obs=0.0200 s⁻¹ model, what fraction remains after one model half-time?
Show the answer and reasoning
One half.
The time ln 2/k_obs gives exp(−ln 2)=1/2; this does not identify a molecular mechanism.
Use the fitted k_obs=0.00100+0.200[L], with [L] in M. Predict k_obs at 0.0200 M within the selected linear model.
Show the answer and reasoning
0.00500 s⁻¹.
0.00100+0.200× 0.0200=0.00500; the fit alone does not prove a unique intermediate.
If both reversible coefficients are multiplied by 10, how do the stationary B fraction and relaxation time change?
Show the answer and reasoning
B fraction remains 0.750; relaxation time becomes 25.0 s.
The ratio stays 3, while the coefficient sum grows tenfold.
Why does an unchanged net ligand inventory not prove an outer-sphere transfer pathway?
Show the answer and reasoning
A net equation does not reveal every intermediate interaction.
A bridge can participate in a pathway without being a different final ligand inventory; structural and kinetic evidence is needed.

What to carry forward

Separate a substitution sketch from mechanism evidence.
State the assumptions behind an observed rate coefficient.
A fitted dependence constrains but does not uniquely identify a pathway.
Conserve charge and formal electron transfer in redox nets.
Keep equilibrium preference distinct from the approach timescale.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.