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Properties of Coordination Compounds
Dr.Christoph UP 2014
Coordination Compounds
Example
A) [Cr(NH3)3Cl3] B) [Cr(NH3)6]Cl3 C) Na3[Cr(CN)6] D) Na3[CrCl6]
Which of these compounds will form a precipitation with AgNO3 ?
Review: CFT
Applications of Crystal Field Theory
Crystal Field Theory Review
http://wwwchem.uwimona.edu.jm/courses/CFT.html
http://www.youtube.com/watch?v=LydEaN8-WJ8
Spectrochemical series
Conclusions from CFT
Extension: LFT
pi-donor ligands => low splitting
pi-acceptor ligands => high splitting
Symmetry Symbols
http://www.webqc.org/symmetrypointgroup-td.html
Which symmetry do the 5 d-orbitals in a tetrahedron have ?
Conclusions from LFT
(1) Find the configuration (as tx2ge
yg / 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-
Applications of CFT
(1) Electronic spectra (2) Hydration energies (3) Lattice energies (4) Ionic radii (5) Spinel types
(1) Electronic Spectra (example)
Each peak in the spectrum (λ <-> ν ) corresponds to a change in the electronic state (Shriver/Atkins p.576)
Find ∆o from Tanabe-Sugano diagrams
∆/B
E/B
V(H2O)62+ 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
http://www.chem.uky.edu/courses/che610/JPS_Spring_2007/PS2_2007key.pdf
Self-study:
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)
(2) Hydration energies
http://www.youtube.com/watch?v=awD1qa7TF4A
(3) Lattice energies
Experimental data vs. calculated for MCl2 high-spin compounds according to the Born-Haber Cycle
(4) Ionic Radii M(2+) --- O
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
(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)
Example
What is the oxidation number of Fe in Fe3O4 ?
“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
(neg
lect
pa
irin
g e
ner
gy
P)
http
://ww
w.everyscien
ce.com
/Ch
emistry/In
organ
ic/C
rystal_and
_Ligand
_Field_Th
eories/e.1
01
6.p
hp
“Magnetite”
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
Applications for Spinels
Spinel Ferrites M2+ Fe3+2 O4
Fe-O framework
Spinel Aluminates M2+ Al3+2 O4
Al-O framework
https://www.youtube.com/watch?v=U_n7DyCqv6U
Visualization of molecules and orbitals www.chemtube3d.com
3D structures of crystals
Fe(2+) Fe(3+)
Magnetism
(extensive explanations: https://www.youtube.com/watch?v=U_n7DyCqv6U)
https://www.youtube.com/watch?v=U_n7DyCqv6U
Ferro-, Ferri-, Antiferro-, Para-Magnetism
http://en.wikipedia.org/wiki/Curie_temperature
Para-magnetism Atoms that have unpaired electrons (also metals) and therefore a total electron-spin S. These individual magnetic moments can be oriented by an outer magnetic field to line up. Paramagnetic substances are not magnetic by themselves but can become magnetic when an outer field is applied. Every ferromagnetic material has a Curie-Temperature Tc where it loses its permanent magnets and becomes para-magnetic.
Curies Law Describes the “magnetic susceptibility” of a material dependent on the outer field B and Temperature T
=> High magnetization M:
(a) High external field B (b) Low temperature
Or: ‘chi” = magn.
susceptibility
Curies Temperature Tc
Heating up a permanent magnet brings the spins to become randomly oriented (at temperature Tc) -> the material loses its magnetism but can still become magnetic again in an external field
https://www.youtube.com/watch?v=1W7dou4kAU8
http://www.youtube.com/watch?v=u36QpPvEh2c
Dia- and Para-Magnetism
Magnetic Properties
Paramagnetism arises from unpaired electrons. Each electron has a magnetic moment with one component associated with the spin angular momentum of the electron and (except when the quantum number l ¼0) a second component associated with the orbital angular momentum.
(p.579)
Where does magnetism come from ?
Effect of unpaired electrons
http://www.youtube.com/watch?v=qfooM_Gl69k
Gouy Balance
http://wwwchem.uwimona.edu.jm/utils/gouy.html
Spin-only formula
Examples
Conclusions from magn. susceptibility
Spin Cross Over SCO
Some complexes can change from low-spin to high-spin at higher temperatures: => low magnetic moment -> high m.m. => M-L bonds short -> longer (why ?) This can happen quickly in a small T-range
http://www.youtube.com/watch?v=e9SMMA9Xe9c
SCO applications
Spin and orbital contributions to the magnetic moment
If S-L coupling is weak
Deviations from spin-only formula
Example: [Fe(CN)6]3- has μ = 2.3μB which is between low- and high-spin calculated (check this out)
Russell-Saunders Coupling (L-S Coupling) Important esp. for second and third row metal compounds There we cannot use the simple spin-only formula anymore
https://www.youtube.com/watch?v=1W7dou4kAU8
Spin
-orb
it c
ou
plin
g
Is higher than spin-only – typical for d-complexes with more than half-filled d-shell
Info to solve problems:
Spin-Orbit coupling
Russell-Saunders Coupling
Electronic states review
Alternative to state the electron configuration of an ion as 4s2 3d6, we can express this configuration as “microstates”:
Ti(3+): 4s0 3d1
Microstates: S = ½ L = 2 (“d”) => J = 5/2, 3/2 ( L+S),(L+S-1),…(L-S)
Ground State with lowest J: 2D3/2
Examples
Attkins p.505
Stability of coordination compounds
http://wwwchem.uwimona.edu.jm/courses/IC10Kstability.html
Thermodynamic stability (equilibrium constant)
http://chimge.unil.ch/En/complexes/ressources/cpxenc.pdf
Equilibrium constant K and β
http://www.youtube.com/watch?v=LydEaN8-WJ8
Entropy effect
Example 1
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 0.10M, and the initial [NH3] is 1.0M and that β4 = 2.1 x 1013, calculate the equilibrium concentration of the Cu2+ ion.
http://wwwchem.uwimona.edu.jm/courses/IC10K1.html
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
(1) Irving-Williams series
M-L bonds become more covalent
Hydration energy gets higher if LFSE is more negative !
(2) Ligand Field Stabilization Energy
(3) Jahn-Teller Effect
Estimate which d-orbital occupation(s) can cause a JT effect ….. (distinguish between high- and low-spin)
Example Cr(II)(H2O)6
Why is a distortion to D4h preferred over regular Oh ?
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)
(4) HSAB principle
“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)
“basicity” high electron density
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
(7) Chelate effect
Example the driving force is the increase in entropy
Conclusion
Complexes formed by multidentate ligands are much more stable than those formed by “normal” ligands !
(8) Bulk and Size of Ligands
(9) Macrocyclic effect
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