Factors that determine the stability of tranistion metal complexes

1. Properties of coordination
compounds
Part 3 (3/3)
Stability of coordination
compounds

2. Review questions
Give the Oxidation number, d-orbital occupation, co-ordination
number and expected magnetic moment of the central metal ion
in the following complexes.
Draw the expected structure.
(i) K3[Co(C2O4)3]
(ii) (NH4)2[CoF4]
(iii) diamagnetic [NiCl2{P(C6H5)3}2]
(iv) cis-[CrCl2(bipy)2]Cl
(v) [Mn(H2O)6]SO4
where C2O4
2- is the oxalate ion and bipy is 2,2'-bipyridine.
http://wwwchem.uwimona.edu.jm/tutorials/tic10k1.html

3. Answers
1). (a) Give the Oxidation Number, d-orbital occupation, co-ordination number
and expected magnetic moment of the central metal ion in the following
complexes.
Draw the expected structure.
(i) K3[Co(C2O4)3]
This octahedral Co(III) complex can display optical isomerism.
The CN = 6, the OS=3+, the d-orbital occupation is that of a LOW spin Co(III)
complex ie t2g
6 eg
0. All Co(III) are treated as LOW spin for CHEM1902 (C10K).
The magnetic moment is therefore 0 B.M.
(ii) (NH4)2[CoF4]
This Co(II) complex is tetrahedral (for CHEM1902 (C10K) we have said that
square planar complexes will only be seen for d8 configurations and Co(II) is d7.
The d-orbital configuration is e4 t2
3 and the magnetic moment is 3.87 B.M {
sqrt(15) }.

4. (iii) diamagnetic [NiCl2{P(C6H5)3}2]
The indication that the compound is diagmagnetic means it must be square
planar since a Ni(II) d8 configuration in a tetrahedral shape would be
paramagnetic.
Examples with this formulation in BOTH tetrahedral and square planar shapes
have been found, generally the tetrahedral are green/blue while the square
planar are red.
The triphenyl phosphine ligand is a neutral monodentate ligand.
(iv) cis-[CrCl2(bipy)2]Cl
Cr(III) has a d3 configuration (t2g
3).
In this octahedral complex we do not need to worry about high/low spin
since we always fill from the lower level and there are 3 t2g orbitals and 3
electrons and no electrons left to occupy the eg level.
(v) [Mn(H2O)6]SO4
The aqua group gives rise to HIGH spin complexes so this octahedral Mn(II)
d5 complex is paramagnetic with 5 unpaired electrons (t2g
3 eg
2).

5. http://wwwchem.uwimona.edu.jm/courses/IC10Kstability.html

6. Thermodynamic stability
(equilibrium constant)
http://chimge.unil.ch/En/complexes/ressources/cpxenc.pdf

7. Equilibrium constant K and β
http://www.youtube.com/watch?v=LydEaN8-WJ8

8. Structural effect

9. Factors affecting stability of complexes
https://www.youtube.com/watch?v=LydEaN8-WJ8&t=1s

10. Example
Assume that in the reaction of Cu2+ with ammonia,
the only complex ion to form is the tetra-ammine
species, [Cu(NH3)4]2+.
Given a solution where the initial [Cu2+] is 0.10 M,
and the initial [NH3] is 1.0 M and that β4 = 2.1 x 1013,
calculate the equilibrium concentration of the
Cu2+ ion.
http://wwwchem.uwimona.edu.jm/courses/IC10K1.html

13. Examples
Calculate the equilibrium concentration of the Fe3+ ion in a
solution that is initially 0.10 M Fe3+ and 1.0 M SCN-, given that
β2 for Fe(SCN)2
+ = 2.3 x 103
http://aris.gusc.lv/ChemFiles/ComplexCompounds/WestIndiMona/tutorials/tic10k1.html

14. (1) Irving-Williams series
M-L bonds become more
covalent

16. Hydration energy
gets higher if LFSE is
more negative !
(2) Ligand Field Stabilization Energy

17. (3) Jahn-Teller Effect
Estimate which d-orbital occupation(s) can cause a JT
effect ….. (distinguish between high- and low-spin)
https://www.youtube.com/watch?v=Omy3NTu7pf8

18. Example Cr(II)(H2O)6
Why is a
distortion to D4h
preferred over
regular Oh ?

19. Solution:
estimate the LFSE for both symmetries
For both symmetries, the 3 electrons in the lower
levels have LFSE = 3 (-3/5 ∆o)
But the one electron in the upper level is lower in
energy for D4h
therefore the total LFSE is more negative (more stable)

20. Cr(2+) Cu(2+)

23. (4) HSAB principle

24. Definitions
Pearson's Hard Lewis Acids
Pearson's features: Small, positively charged species
which are difficult to polarise and do not contain
electron pairs in their valence shells.
Klopman's features: High energy LUMO. Form
charge-controlled complexes with hard Lewis bases.
Soft Lewis Acids:
Has low lying LUMO

25. http://www.meta-synthesis.com/webbook/12_lab/lab.html

26. https://en.wikipedia.org/wiki/HSAB_theory

27. https://en.wikipedia.org/wiki/HSAB_theory

28. “Hard acids”
Small metal ions with
high charge
(Fe 3+, Co 3+, Ni 3+)
“Soft acids”
 Bigger metal ions with
low charge
(Cu 2+, Ag +, Au +, Pt 2+)
“Hard bases”
ligands with low polarization
(F -, OH -, Cl -, NH3)
“Soft bases”
 ligands with high electron density
(I -, SCN -, CN -, CO, PR3)

31. “basicity”  high electron density
(5) Basicity of Ligands

33. Estimate basicity of molecules
Order these ligands from lowest to highest basicity
NH3 PH3 P(CH3)3 SR2
F- Br- OH- CN-
Main effect is the electron density !
Determined by:
electronegativity of atom with lone pair
and/or electron-pull or donate effect of neighbor atoms

34. (7) Chelate effect
https://www.youtube.com/watch?v=iiR75lB9PKo
Example:

36. Example:
the driving force is the increase in entropy

38. Conclusion
Complexes formed by multidentate ligands are much
more stable than those formed by “normal” ligands !

39. (8) Bulk and Size of Ligands

40. (9) Macrocyclic effect

41. Review Questions

44. Review Questions
(1) Explain why:
(2) Explain why:

45. (3)
(4)  see explanation below !

46. Orbit Angular Momentum
Contribution to Magnetism
In order for an electron to have orbital angular momentum, it
must be possible to transform the orbital it occupies into an
entirely equivalent and degenerate orbital by rotation. The
electron is then effectively rotating about the axis used for the
rotation of the orbital.
In an octahedral complex, for example, the three t2g orbitals can
be interconverted by rotations through 90 deg; thus, an electron
in a t2g orbital has orbital angular momentum. The eg orbitals,
having different shapes, cannot be interconverted and so
electrons in eg orbitals never have angular momentum.
(The equivalent rotated orbital must NOT have already a single
electron, so in t2g (3) there is NO orbital contribution to m !)

47. (5)
(6)

48. (2)
(1)
(3)
(4)
(6)
(5*)
* 1 / 3 / 4 / 5 / 0 / 3 / 2
unpaired el.

50. (7)
(8)