Stability Constants and the Kinetics of Ligand Substitution · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Tight binding is not fast swapping

Science · Chemistry · ages 20-21
Name ______________________   Date ____________

  1. One bidentate ligand replaces two monodentate ligands. What can favor this exchange through entropy?

    • Stronger metal-ligand bonds lower the product enthalpy
    • Releasing two ligands while binding one increases free particles
    • Forming a chelate ring reduces the ligand's conformational freedom

  2. For successive formation steps with constants K1 and K2, what is beta2?

    • The product K1 × K2
    • The quotient K1/K2
    • The sum of K1 and K2

  3. At fixed complex concentration, doubling entering-ligand concentration doubles the substitution rate. What does this support?

    • Bond formation with Y dominates activation, favoring an associative pathway
    • Y affects binding equilibria but does not enter the kinetic rate expression
    • Y is kinetically involved in substitution

  4. Under the stated conditions, equilibrium favors a bound complex and its ligand-exchange half-time is 2 seconds. How is it classified?

    • Thermodynamically unfavored and kinetically labile
    • Thermodynamically favored and kinetically labile
    • Thermodynamically favored and kinetically inert

  5. At one temperature, K1 = 20 and K2 = 4. What is beta2?

    Answer: ______________

  6. A rate independent of entering-ligand concentration, by itself, proves a D mechanism with a separate dissociative intermediate.

    Circle one:   True   False

  7. If equilibrium disfavors a bound complex under stated conditions, its ligand substitution must be rapid.

    Circle one:   True   False

  8. A prior-association model gives kobs = kK[Y]/(1 + K[Y]). At K[Y] much greater than 1, kobs is nearly constant. What follows?

    • Prior association can produce saturation with kobs approaching k
    • Prior association must give a rate proportional to [Y] at high [Y]
    • Prior association makes the rate fall toward zero at high [Y]
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Tight binding is not fast swapping W1-mt_XiuL697fut-s1

  1. Releasing two ligands while binding one increases free particles · Consuming one free ligand while releasing two can increase the number of free particles. This is an entropic contribution, not a universal explanation.
  2. The product K1 × K2 · Multiplying the two stepwise expressions cancels the intermediate ML and gives the overall formation expression.
  3. Y is kinetically involved in substitution · The rate changes when [Y] changes, supporting kinetic involvement of Y. Examples include A substitution, preassociation and capture competing with return of a D intermediate. This observation does not establish activation mode or a unique mechanism.
  4. Thermodynamically favored and kinetically labile · Equilibrium favors the bound complex. Its 2-second exchange half-time is below the lesson's 1-minute lability threshold, so it is thermodynamically favored and kinetically labile.
  5. 80 · beta2 = K1K2 = 20 × 4 = 80.
  6. False · Concentration independence can arise from saturation in a prior-association model. It does not prove a D mechanism or a separate dissociative intermediate.
  7. False · Equilibrium preference does not specify the activation barriers or the substitution rate. An unfavored complex can persist through slow exchange.
  8. Prior association can produce saturation with kobs approaching k · At high K[Y], the denominator is approximately K[Y]. It cancels with the numerator factor, leaving kobs approximately k.
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