1. The document discusses properties of coordination compounds including electronic spectra, hydration energies, lattice energies, and ionic radii as determined by crystal field theory.
2. Crystal field theory can be used to determine the splitting of d-orbital energies under the influence of ligands in an octahedral or tetrahedral field and explain observations from electronic spectra.
3. Hydration energies and lattice energies of metal ions can also be explained using crystal field theory calculations and comparisons to experimental data.
4. Crystal field theory further explains trends in ionic radii and the structures of spinel compounds containing mixed metal oxides.
2. 3 types of metal compounds
Electronic Spectra
Hydration energies
Lattice energies
Ionic radii
Spinels
3. Factors determining the type of bonds
(1) Oxidation number of the metal
ox.no. 0 indicates a
covalent M-L bond
(2) Coordination number
-> Coordination compounds show the
appropriate number of ligands, in contrast to
ionic compounds
Cu(SO4) as salt and as
coordination compound:
6. (1) Name of cation (potassium)
(2) Name of ligands in alphabetical order
with numbering “di” “tri” “tetra” “penta” “hexa”
Change anion ligand names to ending “-o”
(Cl- -> chloro, CN- -> cyano, C2O4
2- -> oxalato)
(3) Name the metal and add ox.number
Metal is in an anion -> add “ate”
(nickelate, cobaltate, cuprate, ferrate)
Metal is in cation or
neutral part -> metal
name
25. (1) Find the configuration (as tx
2gey
g / ext2
y), the number of
unpaired electrons and the LFSE (in terms of ∆o / ∆t and P)
(2) Which of the following complexes has higher LFSE:
(h) [MnF6]3- i) [NiBr4]2- j) [Fe(CN)6]4-
27. (1) Electronic Spectra (example)
Each peak in the spectrum (λ <-> ν ) corresponds to a change
in the electronic state (Shriver/Atkins p.576)
28. Find ∆o from Tanabe-Sugano diagrams
∆/B
E/B
V(H2O)6
2+ shows 2 peaks at
17200 and 25600 cm-1
(1) Get E-ratio: E2/E1 = 1.49
(2) Find by trial and error this ratio
in the diagram => ∆/B
we find here a value of 40/27 = 29
(3) From this we find B:
E2 = 25600 = 40 *B
E1 = 17200 = 27 *B => B = 640 cm-1
(4) When B is know, then
∆o = 29 * 640 cm-1 = 18600 cm-1
30. Energy ratios:
E2/E1 = 1.8
E3/E1 = 3.05
E3/E1 = 1.68
Now we move on the x-axis until
we find this ratio in the y values
again (approximately !)
=> ∆/B = 10
from that we get B:
E3/B = 29 (graph) => B = 896 cm-1
=> ∆o = 10 * 896 cm-1 = 8960 cm-1
(more exact: take the average for all 3 found B values)
36. M2+ ions in water are in a weak field => they are all high-spin
Then we get 2 minima in the radii at V2+ with 3 electrons in the t2g
levels and Ni2+ with 6 electrons in t2g
For low-spin there is only one minima at Fe2+ with 6 electrons in t2g
37. (5) Spinels (gemstones)
Spinels = crystals formed by mixed oxides,
originally used for MgAl2O4
In a closed packed solid, there are tetrahedral and
octahedral “holes” filled with a metal ion.
“ cubic closed packing”
(https://www.youtube.com/watch?v=U_n7DyCqv6U)
39. “normal”: A in Td , B in Oh holes
Fe3O4 is a spinel type : Fe2+ (Fe3+ )2 O4
“inverse”: A in Oh,
1/2B in Td and 1/2B in Oh
(neglectpairingenergyP)
http://www.everyscience.com/Chemistry/Inorganic/
Crystal_and_Ligand_Field_Theories/e.1016.php
“Magnetite”
40. If M3+ has a higher CFSE than M2+ in an octahedral
field, these ions will prefer octahedral holes and
forms a “normal” spinel.
Predict the structure for Mn3O4