Lecture 10

Question 1

Give the first and second stepwise constant and overall second constant for equation;
M + L = ML

Answer 1

First stepwise formation constant;
K1 = [ML]/[M][L]

Second stepwise formation constant;
ML + L = ML2
K2= [ML2]/[ML][L]

Second overall formation constant;
M + 2L= ML2
B2 = [ML2]/[M][L]^2

Question 2

How do you get the overall formation constant?

Answer 2

B can be obtained by multiplying the stepwise formation constants (Kn)
K1 x K2 = B2
units= mol-2 dm6
or the sum of stepwise constant log values
logBn = logK1 + logK2 + logK3…
no units here

Question 3

Describe the trend in successive formation constants

Answer 3

As stepwise constant/ number of ligands increases, Kn value decreases
It becomes less favourable to displace one ligand for another

Question 4

Explain the trend in successive formation constants

Answer 4

decreases due to;
- statistics (the fewer ligands there are
to replace, the less likely the reaction)
- steric hindrance ( (increases with the
number of ligands - if they are larger than
original ligand)
- electrostatics (important if the ligands are
charged)

Question 5

What does it mean if the general trend in decreasing formation constants isnt observed?

Answer 5

indication of structural change, e.g. change in
coordination number or geometry

Question 6

How do we know if a complex is stable?

Answer 6

large overall formation constant

Question 7

How do we know if a complex has thermodynamic stability?

Answer 7

Kf > 1, ∆G < 0

Question 8

Give the factors affecting stabilities of complexes

Answer 8

Enthalpic effects;
- bond strength
- effects of the crystal field (CFSE)
- steric repulsion between ligands
- electrostatic repulsion between ligands

Entropic effects;
- changes in solvation
- number and size of chelate rings

Question 9

What is the chelate effect?

Answer 9

Complexes with chelating multidentate ligands are more stable than equivalent complexes with monodentate ligands. i.e. they have a bigger overall formation constant, beta

Question 10

What are reasons for the increase in stability?

Answer 10

Effective concentration
1. Chelate ligand- the 2nd N-donor is held close to the metal, allowing the second bond to form easily
2. Monodentate case Ni-NH3 + NH3;
the 2nd N-donor must keep movng until it comes close enough to the metal to bond, i.e. the second bond is less likely to form

Entropy
e.g. no change in number of molecules ∆S=0, 2 molecules produce 3 molecules ∆S=+ve
when ∆S is positive, ∆G becomes more negative. Hence, the complex with the didentate en ligand is more stable

Question 11

Place the chelate rings in order of stability

Answer 11

4-membered ring- strained &laquo_space;5-membered- most stable > 6-membered- increase in bond strain

when rings become larger than 6 the enhancement of the local concentration is lost

Question 12

How does entropy influence formation constants?

Answer 12

• the larger the increase in the number of molecule, the higher the entropy of the reaction
• more negative ∆G
• higher formation constant

Question 13

How can you determine a formation constant?

Answer 13

1. potentiometrically
   if complexation involves pH change e.g. ligand has to deprotonate to bind
2. spectroscopically
   UV/vis absorption, proton NMR etc.
   (have to be able to quantify signal, e.g.
   through Beer-Lambert law or integration of
   NMR resonances)

Question 14

Why are hexadentate ligands useful?

Answer 14

form very stable 1:1 complexes- h=have very high formation constants- stable
increased holding power at low iron concentrations