Lecture 5

Question 1

Describe the crystal- field theory

Answer 1

• Makes the simplifying assumption that the bonding between the metal and the ligands is purely ionic:
• The metal is represented as a point positive charge (the central ion)
• surrounded by a set of point negative charges (the ligands L).
• Bonding energy is produced through ionic
  electrostatic forces.

Question 2

What other effects are there as part of the crystal-field theory

Answer 2

• Electrons in different d-orbitals interact with
  the ligands to different extents
• Result: splitting of the d-orbitals into
  groups of different energies

Question 3

What are the interactions between the d-orbitals and ligands in an octahedral complex?

Answer 3

• electron clouds point straight at ligands so large repulsion between d-electrons and ligands- dz2 and dx2-y2
• electron clouds between ligands so repulsions are smaller- dyz, dzx and dxy

Question 4

Whats the effect of the ligands on the d-orbital energies?

Answer 4

• For a free transition metal ion (spherical
  environment), all d-orbitals are at equal energy
• For an octahedral complex, the dxy, dxz, dyz
  orbitals have a lower energy than the dz2, dx2-y2 orbitals
• If the d-orbitals are not equally occupied, most electrons are in the lower energy d-orbitals.
• This extra stabilisation of complexes is called
  the ‘crystal field stabilisation energy’ CFSE.
  NB: The CFSE is generally only a few
  tenths of % of the total energy of complex
  formation

Question 5

Calculating CFSE- d0-d3 configurations

Answer 5

For a metal ion with n electrons in dxy, dxz and dyz;
CFSE= n(-0.4∆o)

Question 6

Calculating CFSE- d4-d7 configurations

Answer 6

high spin; CFSE= n(-0.4∆o) + m(0.6∆o)
low spin; CFSE= n(-0.4∆o) + P

m= electrons in dz2 and dx2-y2

Question 7

What is a high/ low spin configuration?

Answer 7

For d4-d7;
Low spin= The 4th electron enters one of the
lower orbitals. It pairs up with the electron
already there and experiences an electrostatic
repulsion, the pairing energy P.
High spin= The 4th electron occupies one
of the orbitals of higher energy.

Question 8

How do we known whether low or high spin is adopted?

Answer 8

which of -1.6 ∆o+ P and -0.6 ∆o is
energetically more favourable
weak field; high spin
strong field; low spin

Question 9

What is the special rule for low/high spin complex for 3d6 complexes?

Answer 9

(see page 32)
Low spin CFSE= -2.4∆o +2P
All 3 electrons are paired here but only contains 2P
High spin CFSE= -0.4∆o
High spin has to have 1 electron paired as well so cancels 1P from low spin

Question 10

Calculating CFSE for d8-d10 configurations

Answer 10

only 1 possible configuration- no high/low spin
CFSE= n(-0.4∆o) + m(0.6∆o)

Question 11

Give the overall equation for calculating CFSE

Answer 11

n(-0.4∆o) + m(0.6∆o) + P

Question 12

Describe the trend in hydration enthalpies of 3d-metal M2+ ions

Answer 12

M2+ + 6H2O = [M(H2O)6]2+ (high spin)
- The increase of the predicted values reflects the increasing effective nuclear charge
- The wave like deviation of the experimental
values reflects the variation in CFSE:
increase from d1 to d3, decrease to d5 and rise to d8
- d4 and d9 special cases