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