1. Dear Students of Inorganic Chemistry 2,
Please note that due to the holy month of Ramadan, on Sunday, 21 June 2015 classes
begin at 11:20 and ends at 12:40
(meeting room 33-29)
Have a Happy and peaceful Ramadan

2. Chapter 3.

3. Bonding in coordination compounds
Nobel prize 1913
Alfred Werner
VBT
Crystal Field Theory (CFT)
Modified CFT, known as Ligand Field Theory
MOT

4. How & Why?

5. 3.1. Valance Bond Theory Linus Carl Pauling (1901-1994) Nobel prizes:
Basic Principle
A covalent bond forms when the orbtials of two atoms overlap and are occupied by a pair of electrons that have the highest probability of being located between the nuclei.
Linus Carl Pauling
( )
Nobel prizes:
1954, 1962

7. Valance Bond Model Octahedral Complex e.g. [Cr(NH3)6]3+
Ligand = Lewis base
Metal = Lewis acid
s, p and d orbitals give hybrid orbitals with specific geometries
Number and type of M-L hybrid orbitals determines geometry of the complex
Octahedral Complex
e.g. [Cr(NH3)6]3+

8. Limitations of VB theory Cannot account for colour of complexes
Tetrahedral e.g. [Zn(OH)4]2-
Square Planar e.g. [Ni(CN)4]2-
Limitations of VB theory
Cannot account for colour of complexes
May predict magnetism wrongly
Cannot account for spectrochemical series

9. 3.2.Crystal Field Theory
400
500
600
800
The relationship between colors and complex metal ions

11. Crystal Field Model
A purely ionic model for transition metal complexes.
Ligands are considered as point charge.
Predicts the pattern of splitting of d-orbitals.
Used to rationalize spectroscopic and magnetic properties.

12. d-orbitals: look attentively along the axis
Linear combination of
dz2-dx2 and dz2-dy2
d2z2-x2-y2

13. We assume an octahedral structure of negative charges placed around the metal ion (which is positive).
The ligand and orbitals on the same axes as negative charges.

15. In Octahedral Field
dx2-y2
dz2
dxy
dxz
dyz eg
t2g

17. How to calculate crystal field stabilisation energy?
I have [Co(F)6]3− now see cobalt has 3d7, 4s2 system in ground state but in excited state it loses three electrons in the formation of ions and two elctrons from 4s and one from 3d orbital
so thus Cobalt gets 3d6 configuration.
Now it is of high spin so 4 electrons go to t2g orbital and 2 electrons go to eg orbital. By applying formula,
4(−0.4)+2(0.6)=− =−0.4

19. In Tetrahedral Field

22. consider the compound [Ni(CO)4] and how to calculate CFSE for it?

23. Step 2: Find the appropriate crystal field splitting diagram
Step 1: Look up Nickel Carbonyl and find out what geometry it has.
We need the geometry to know how the d orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.
                                         
Step 2: Find the appropriate crystal field splitting diagram
for this geometry.
                                                                     
Step 3: Figure out how many d electrons there are.
Nickel  is in Group 10, with a configuration of [Ar]4s23d8 or [Ar]4s13d9. In coordination compounds we will consider this configuration to become [Ar]4s23d8⟹[Ar]3d10. The selectrons are considered to move to the d subshell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations.

24. eg (dx2−y2, dz2) E=−3/5Δtet (stabilized)
Step 4: Fill in the d electrons.
You can do this part yourself. How do you put 10 d electrons in the orbital diagram?
Step 5: Determine ΔE.
Δ is the energy different between the eg and t2g sets of orbitals. In tetrahedral complexes, the energy of the energy of the egorbitals is lower and the energy of the t2g orbitals is higher. The energies are:
eg (dx2−y2, dz2) E=−3/5Δtet (stabilized)
t2g (dxy, dxz, dyz) E=2/5Δtet (destabilized)
Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.
ΔE=4×(−3/5Δtet)+6×(2/5Δtet)=?

26. 3.3. Consequence of CFT

27. When the 4th electron is assigned it will either go into the higher energy eg orbital at an energy cost of Do or be paired at an energy cost of P, the pairing energy. d4
Strong field =
Low spin
(2 unpaired)
Weak field =
High spin
(4 unpaired)
P < Do
P > Do
Coulombic repulsion energy and exchange energy

28. [Mn(H2O)6]3+
Weak Field Complex
the total spin is 4  ½ = 2 High Spin Complex
[Mn(CN)6]3-
Strong field Complex
total spin is 2  ½ = 1
Low Spin Complex

29. Spectrochemical Series: An order of ligand field strength based on experiment:
Weak Field
I-  Br- S2- SCN- Cl- NO3- F-  C2O42- H2O NCS- CH3CN NH3 en  bipy phen NO2- PPh3 CN- CO
Strong Field

30. Placing electrons in d orbitals
1 u.e.
5 u.e. d5
0 u.e.
4 u.e. d6
1 u.e.
3 u.e. d7
2 u.e. d8
1 u.e. d9
0 u.e.
d10

31. What is the CFSE of [Fe(CN)6]3-?
CN- = s.f.l.
C.N. = 6  Oh
Fe(III)  d5
l.s.
h.s. 3- eg
t2g
+ 0.6 Doct
- 0.4 Doct
CFSE = 5 x Doct + 2P =
- 2.0 Doct + 2P
If the CFSE of [Co(H2O)6]2+ is -0.8 Doct, what spin state is it in?
C.N. = 6  Oh
Co(II)  d7
l.s.
h.s. eg
t2g 2+
+ 0.6 Doct
- 0.4 Doct
CFSE = (5 x Doct)
+ (2 x 0.6 Doct) +2P = Doct+2P
CFSE = (6 x Doct)
+ (0.6 Doct) + 3P= Doct + P

32. The origin of the color of the transition metal compounds
Ligands influence O, therefore the colour

33. The colour can change depending on a number of factors e.g.
1. Metal charge
2. Ligand strength

34. Assigned transition: eg t2g This corresponds to the energy gap
The optical absorption spectrum of [Ti(H2O)6]3+
Assigned transition: eg t2g
This corresponds to the energy gap
O = 243 kJ mol-1

35. absorbed color
observed color

36. [CrF6]3-
[Cr(H2O)6]3+
[Cr(NH3)6]3+
[Cr(CN)6]3-
As Cr3+ goes from being attached to a weak field ligand to a strong field ligand,  increases and the color of the complex changes from green to yellow.

37. 3.4. Regular and distorted geometries on the Scope
(Janh – Teller effect)

38. Lecture 30 Electronic Spectra of Coordination Compounds 1) Jahn-Teller effect
Octahedral complexes can be a subject to tetragonal or trigonal distortions leading to less symmetrical but more stable structures.
According to the Jahn-Teller theorem, a non-linear molecule with a not-completely filled degenerate electronic levels can undergo a (vibrational) distortion lowering its symmetry and energy.
In the case of octahedral complexes the tetragonal distortion reduces Oh symmetry of a complex to D4h producing either elongated or compressed tetragonal bipyramid and reduces degeneracy of eg and t2g orbitals.
The most pronounced stabilization due to the tetragonal distortion is expected for the following configurations: d1 (compression), d4, d7, d9 (various types), d2 (elongation).
No stabilization and thus no distortion is expected for d3, d5 (high spin), d6 (low spin) and d8 configurations.

39. 3.5. σ and π bonding ineraction

41. Limitations of CFT
Considers Ligand as Point charge/dipole only
Does not take into account of the overlap of ligand and metal orbitals
Consequence
e.g. Fails to explain why CO is stronger ligand than CN- in complexes having metal in low oxidation state

42. 3.6. LCAO – MO diagrams for some selected
complexes
Linear Combination of Atomic Orbitals
A linear combination of atomic orbitals, or LCAO, is a quantum superposition of atomic orbitals and a technique for calculating molecular orbitals in quantum chemistry. In quantum mechanics, electron configurations of atoms are described as wave functions.

43. Metals in Low Oxidation States
In low oxidation states, the electron density on the metal ion is very high.
To stabilize low oxidation states, we require ligands, which can simultaneously bind the metal center and also withdraw electron density from it.

44. Stabilizing Low Oxidation State: CO Can Do the Job

46. Stabilizing Low Oxidation State: CO Can Do the Job
Ni(CO)4], [Fe(CO)5], [Cr(CO)6], [Mn2(CO)10], [Co2(CO)8], Na2[Fe(CO)4], Na[Mn(CO)5]

47. O C M
 orbital serves as a very weak donor to a metal atom