Exam 2 Fall 2012

1.In a few brief sentences, explain concisely the connection between the following pairs of terms:

a) CFT and symmetry

The symmetry of the arrangement of ligands about a metal center determines the degeneracy of the d-orbitals and who much the 5-fold degenerate set spllits into 2 or more groups of d-orbitals. Symmetry doesn't however say anything about the energies of these groups of orbitals. In general the higher the symmetry the more degeneracy in the groups.
b) Spectroscopy and symmetry

The specroscopic parameters of a complex are determined in part by its symmetry. What we covered was the absorptivity or the extinction coefficient and how its magnitude was determined by a selection rule, the LaPorte selection rule, that specifically deals with centrosymmetric complexes. Those with inversion symmetry will have forbideen, or small extinction coefficients for electronic transition between like symmetry orbitals, such as g--> g d-orbitals.
c) Spin degeneracy and symmetry

Spin degeneracy is forbidden by quantum mechanics, meaning it is a high energy, unstable situation. This is 'relaxed' by lowering the symmetry, distortion, so that the degeneracy that causes the spin degeneracy is removed.
d) Spin state and crystal field

The spin state of the electrons in s complex can vary in terms of the degree of paired electrons or unpaired electrons, 'low' vs 'high' spin states. The crystal field strength mainly determines this, since a stronger crystal field makes paired electrons configurations relatively more stable.
e) Spectrochemical series and pi-bonding

The sequence of ligands in the spectrochemical series can be explained by incorpating pi-bonding. Ligands that are pi-donating, pi-bases, cause a destabilization of previously non-bonding d-orbitals thereby decreasing the delta E. Ligands that are pi-accepting, pi-acids, cause a stabilization of those d-orbitals, increasing the delta E.

f) Hardness and covalency

Hardness = non-polarizable and non-polarizable ions will have mainly electrostatic, or ionic bonding interactions.

Softness is the coverse and is equivalent to covalency in a MX type material.

2.a) octahedral: Oh (of course!)

trigonal prismatic: D3h

b) The d-orbtials in Oh transform as: Eg [d(z2), d(x2-y2)] and T2g [d(xy), d(xz), d(yz)] .

The d-orbtials in D3h transform as: A1' [d(z2)] , E' [d(xy), d(x2-y2)] and E" [d(xz), d(yz)] .

c) d(xz) and d(yz) orbitals will point most closely to the ligands, while d(xy) and d(x2-y2) will have the least interaction. d(z2) is in the middle. Note although the point group is D3h, the CF diagram is not the same as D3h trigonal bipyramid which has a different geometry. This is a goodillustration of 1a) that the symmetry does not determine the energies.

d) The doubly degenerate set Eg [d(z2), d(x2-y2)] will participate in sigma bonding while the triply degerate T2g set [d(xy), d(xz), d(yz)] will be non-bonding.

e) No, because there are no triply degenerate symmetry types (irreducible representations) in D3h (as determined by surveying the character table). While it is true that there are no triply degenrate d-orbitals, that is an insufficient answer: there could be other atomic orbitals in triply degenerate groups. However not in D3h for reason above.

3. The simplest anwer is that Pd has a much stronger crystal field die to its 2nd row position, so that even if it has weak field ligands like Cl- it still prefers the low spin-favored d8 configuration in the square planar geometry. Ni having a weaker field and with Cl- ligands woudl still be high spin, so that the Td configuration is lower in energy than putting two unpaired electrons in the highest two d-orbitals of a D4h sq.pl.

4. The lambda max = 450 nm for  [Cr(NH3)6]3+. 
The lambda max = 420 nm for  [Co(NH3)6]3+. 
Assume P= 18,000 cm-1 for both complexes.

(a)  Draw CFT diagrams for each complex.

CFT
(b) Calculate the delta(o) and CFSE in kJ/mol for each complex.
(Recall the relationship between wavelength, nm, and wavenumber, cm-1  and that
 1.1962 x 10-2  x E (cm-1)= E (kJ/mol).

Cr: 450 nm corresponds to 22,222 cm-1; using the conversion relationship, delta = 266 kJ/mol.

Cr(3+) is d3, therefore CFSE = 1.2 delta or CFSE = 319 kJ/mol

Co: 420 nm corresponds to 23,800 cm-1; using the conversion, delta = 285 kJ/mol.

Co(3+) is d6 and low spin, therefore CFSE = 2.4 delta - 2P or CFSE = 684 kJ/mol - 2P

(c) If the cobalt complex is low spin, estimate what the total pairing energy must be.

Since delta = 285 kJ/mol for the Co(3+) complex, 2P must be less than 285 kJ/mol (or P< 143 kJ/mol) for the low spin state to be energetically favorable.

5. Consider the reactions below to answer questions (a) and (b).

(i) 3 phenanthroline   +  [Ru(H2O)6]2+ ---> 6 H2O + [Ru(phen)3] 2+

(ii) 3 oxalate    +  [Ru(H2O)6]2+ ---> 6 H2O + [Ru(oxalate)3]4-

(a) Which reaction, (i) or (ii), is predicted to have larger value for its Kf and why?

Reaction (i) will have a larger Kf. While both reactions use bidentate ligands and so benefit from the chelate effect, the first reaction uses ligands with N-atom donors. Since Ru(2+) is a soft ion it will prefer teh N atoms of phen ligands to the harder O atoms of oxalate.

(b) What is the point symmetry of the Ru product?

D3

6.Using CFT applied to coordination complexes, explain why the two very strong field ligands CO and CN- on nickel make Ni(CO)4 and [Ni(CN)4] 2-, 4-coordinate complexes that have two different geometries.

The short answer is: the difference in oxidation states.

Ni(CO)4 is Ni(O) and has a d10 configuration. With a full d-shell, regardless of geometry, there will be no CFSE (LFSE). The complex therefore adopts the geometry that places the carbonyl ligands as far away from each other as possible.

[Ni(CN)4]2- is Ni(II) and is a d8 ion. For Ni, especially with strong ligand fields, a square planar geometry is favored. In this D4h symmetry, all 8 e- are in the lowest energy d orbitals and the one very high energy x2-y2 orbital is empty

7.Chromium is a metal of many colors! [chroma, Greek for “color”] One green pigment used in paints as well as in eyeshadows is chrome green, Cr2O3.  (Note that chromium in chrome green adopts an octahedral site surrounded by oxide ions in the solid state).   Chromium takes on a completely different hue in the yellow pigment chrome yellow, or lead chromate, whose empirical formula is [Pb+2][CrO4-2]. Is the green color of chrome green produced by the same type of electronic transition that causes the yellow color of chrome yellow

No, the colors do not originate from the same type of electronic transision. The green color of chrome green is due to a d-d transition in Cr(3+) in an octahedral field whereas the yellow of chrome yellow must be due to a LMCT since that pigment has Cr as a d0 C4(6+) ion.

Predict the relative saturation (i.e., absorptivity or extinction coefficient) of chrome green versus chrome yellow. Defend your answers with detailed explanations.

Since chrome green is due to a d-d transition in an octahedral field, it is LaPorte forbidden and is expected to be weak (an extinction coefficient less than 100). The LMCT in chrome yellow however would be completely allowed since it will be from O-p orbitals to d orbitals and the ion is tetrahedral giving it a much larger extinction coefficient ( > 1000).

8.When ethylenediamine is added to CoCl2 dissolved in isopropanol a dark blue solid precipitates.  This deep blue solid has a magnetic moment measured as 4.2 B.M.  When this blue solid is dissolved in dimethylsulfoxide, it produces a blue solution that absorbs at 3220, 6,000 and 15,100 cm-1 with extinction coefficients of ~600 M-1 cm-1.  However, when the blue precipitate is dissolved in water, a pink solution is produced that absorbs at 8000, 16,000 and 19,4000 cm-1 with extinction coefficients of ~5 M-1 cm-1.

What are the identities of the blue and the pink species?  Explain your answer accounting for all the information provided.

The intense blue color and the Co(2+) ox. state should make you think of the similarly blue CoCl4]2- complex which has that color due to the Td environment. However the blue species in question apparently only formed with ethylene diamine was added. So perhaps it is Coen2Cl2, a neutral complex that might preciptate, or possible [Coen2]2+. In either case, the magnetic data corresponds best to a 3 unpaired electrons which fits a d7 Co2+ (in either Td or Oh environments actually). When water is added ( and you all got this) the paler pink species is octahedral Co(H2O)6]2+. The lack of inversion in the blue Td complex give it a high absorptivity and the inversion of the Oh complex explains its low absorptivity.

9.Three flasks have been mixed up in the lab. (Unfortunately, this really happens in research!) 
The three flasks contain solutions of three different colors:  yellow, pale blue and violet. 
It is also known that:
                     one flask contains [Cr(NH3)6]3+;
                     the second contains [Cr(H2O)6]2+;
                     the third contains [Ni(NH3)6]2+. 

To try to determine which complex is in which flask, you have taken the visible electronic spectrum of all three solutions and your data is below:

                     Flask A:       lmax at energies 10,740 ; 17,500; 28,200 cm-1
                     Flask B:       lmax at energies  20,000 ; 27,500; 38,000 cm-1
                     Flask C:       lmax  at  energy  14,000 cm-1

            Which complexes are in flasks A, B and C?

The first step is to recognize that the three octahedral metal complexes will have a certain number of d-d transitions:
         Cr(II) is d4 and will have 1 d-d transition,
         Cr(III) is d3 and will have 3 d-d transitions,
         Ni(II) is d8 and will have 1 d-d transition.

Therefore, Flask C must contain [Cr(H2O)6]2+, the only complex expected to show one absorption.

Next to distinquish [Cr(NH3)6]3+ from [Ni(NH3)6]2+ it should be recognized that the higher charge on the Cr(III) complex will cause the Do splitting parameter to be larger.  Hence Flask B where the energies (in cm-1) are clearly larger must contain [Cr(NH3)6]3+; by default, Flask A contains [Ni(NH3)6]3+.

10. Many oil and fresco masterpieces have deteriorated due to reaction of the metal-based pigments with hydrogen sulfide, H2S.  The H2S originates from air pollution as well as from the sulfur-rich egg yolks used in preparing tempera.   Most metal sulfides are black; hence the result of H2S reaction with the oil paints is a darkening of the painting. 

Lead and copper-based pigments are particularly notorious for turning dark, even black. Some examples of such pigments are lead white, Pb2CO3(OH)2, and the green verdigris, Cu2(OH)2(CH3COO)2 . 

Here is a balanced equation for the reaction of lead white with hydrogen sulfide:
 
                                  Pb2CO3(OH)2 +  2 H2S --> 2 PbS + 3 H2O + CO2

Acid-base reaction, both Bronsted (H+ + OH-) and Lewis (Pb2+ + S2-)

HSAB explains the facorable formation of PbS since both are soft, where the Pb reagent is mismatched with a soft acid Pb2+ and hard anionic bases, carbonate and hydroxide. HSAB also predicts the favorable formation of H2O from a Pb-OH unit.

Entropy explains a favorable consequence of CO2 gas formation.