Name KEY D# _______ Chemistry 350 Fall 2005 CCM Exam #4, November 18, 2005 50 minutes

100 points on 4 pages + a useful page 5 1. Consider the molecular orbital diagram shown for MLN. (16 pts) a) Indicate the following: Number of Ligands, N, = L = π acceptor σ only donor π donor (circle one) b) On the diagram at right, label the bonding orbitals as more metal (M) or ligand (L) in character. c) Briefly explain what it means to be a: i) “π acceptor” ligand π acceptor has empty π∗ orbitals on ligand in to which d e- from M can be donated ii) “σ only donor” ligand σ donor has only one lone pair facing the central metal atom and no extra lone pairs or empty π∗ orbitals on ligand iii) “π donor” ligand. π donor – extra e- (i.e. lone pairs) on L after σ bonding “donates” to M d) Describe how ∆ compares in the case shown to that in ML6 with only σ bonding.

∆shown > < = ∆O σ only (circle one)

e) Give an example of a molecule for which this MO diagram is applicable. many, e.g. [Cr(CO)6]

6 L

M

L L L

Chem 350, Fall 2005 Exam 4 CCM, Page 2 2. a)Complete the following table. (24 pts)

[Fe(CN)6]4- [Cr(NH3)6]3+ [Ni(OH2)5F]

correct name

hexacyano-ferrate(II)

hexaammine-chromium(III)

pentaaquafluoro-

nickel(I)

oxidation state of

central metal atom

+2 +3 +1

number of valence electrons on central

metal atom

neutral: 8 + 1(6) + 4 =18 ionic: 6 + 2(6) = 18

18

n: 6 + 2(6) - 3 =15 i: 3 + 2(6) = 15

15

n: 10 + 2(5) + 1 =21 i: 9 + 2(5) + 2 = 21

21

number of d electrons

6 3 9

electron configuration of central metal ion

[Ar]3d6 [Ar]3d3 [Ar]3d9

likely molecular geometry around

central atom

octahedral octahedral octahedral

spin: high, low, or not applicable? (circle) HIGH LOW N/A HIGH LOW N/A HIGH LOW N/A

spin, S = 0 3/2 1/2 3. A little potpourri of transition metal questions like those on the problem set… (9 pts) a) Determine the number of valence electrons on the central metal atom in the following complex using either of the methods we learned about in class. You must state which method you are using and show your work.

[Co(PR3)(CN)3]4- Ionic: Co(-1) 8 Neutral: Co 7 PR3 2 PR3 2 3CN 6 3 CN 6 4- 4 16 16 b) Assuming the 18 electron rule applies, identify the second-row transition metal: [MCl(CO)(PPh3)2] neutral (n): 18-2-2*2-1=11=M, Ag ionic (i) : 18-2-2*2-2=10=M+, Ag 18-CO-P-Cl

c) What charge, z, would be necessary for the following complex to obey the 18-electron rule? [Ru(CO)3(NO)]z (linear NO)

z =[8+3*2+3]-18= 1-

Chem 350, Fall 2005 Exam 4 CCM, Page 3 4. Complexes of Co3+ having more than one type of ligand display a variety of colors. For example, CoCl3·6NH3 is yellow, whereas CoCl3·5NH3 is purple, whereas CoCl3·4NH3 is green or purple. Suggest a source for this variability. It is probably helpful to discuss (at least) the likely structures of these complexes and the d orbital splitting of the metal atom. (10 pts) Different ligands affect ∆o, therefore, the color of the light absorbed by the complexes. More NH3 and less Cl- makes for a larger ∆o as Cl- is a better pi donor ligand. As far as structures go, these complexes are probably more accurately written as octahedral complexes with Co3+ and, therefore, d6. They are likely all high-spin complexes as there are no pi acceptor ligands. The structures are probably [Co(NH3)6]Cl3, [Co(NH3)5(Cl)]Cl2, and [Co(NH3)5(Cl)2]Cl, respectively. [Co(NH3)5(Cl)2]Cl exists as purple and green complexes because it can be either cis or trans. 5. The complex [Os(2,2’-bipyridine)3]2+ is a complex with a λmax (in the MLCT region of the spectrum) of 480 nm. (bipy = 2,2’-bipyridine) (15 pts) a) Geometry-wise, [Os(bipy)3]2+ is likely to exist as one of two geometries. Give the coordination number of the central Os and, therefore, the two likely geometries. Bipy is bidentate, so CN =6 and likely geometries are octahedral and trigonal prismatic b) [Os(bipy)3]2+ exists as two enantiomers so only one geometry in a) is possible. Draw each of these enantiomers and determine the point group of one of the enantiomers of [Os(bipy)3]2+. Be sure to label the enantiomers appropriately.

c) Indicate the hybridization of central metal atom.

sp3d2 d) What would you expect to happen to λmax if each bipy was replaced by 2 NH3’s?

replace a π acceptor with a σ donor, so decrease ∆o, ∴decrease E = hc/λ, so λ increases. ∴Longer λmax

Point Group: D3

Chem 350, Fall 2005 Exam 4 CCM, Page 4 6. Three of the five structural fragments below occur in the borane ion B3H8

- of C2v symmetry. i) Provide a projection formula of B3H8

- including these fragments. Indicate the total number of electrons. ii) Using Wade’s rules, determine the number of “skeletal electrons” and predict whether you expect the structure to be best described as closo, nido, or arachno. (12 pts) i) 18 total electrons ii) 3 B-H = 6 e-

5 H = 5 1- = 1 Skeletal e = 12

So, 6 pairs n = 3, so n+ 3 … arachno

It should be noted that this is not the actual structure according to theoretical calculations and experimental data. There, the 3 B’s should all be connected, not just 2,but this violates our “rules”. See J. Phys. Chem. 1991, 95, 10623-10630 for more detail 7. Shown is a portion of an engineering diagram for a common industrial process. (10 pts) a) Name the process Chlora-alkali process b) Write (several) balanced chemical equations for the …

i) sub-processes/half-ractions occurring

2NaCl(aq) 2 Na+(aq) + Cl2(g) + 2e-

2 e- + 2 H2O (l) 2 OH-(aq) + H2 (g) ii) and the overall process.

2NaCl(aq) + 2 H2O (l) 2 NaCl(aq) + Cl2(g) + H2 (g) c) Comment on why the diaphragm must be present to prevent mixing of the two gases being produced. Without the diaphragm the H2 and Cl2 gases would recombine to form HCl in a very exothermic reaction. This HCl could then readily react with the NaOH and get us back to the starting materials of NaCl and H2O!

B

H

BB H

B

B

BB B

B B

B

bridging fragments

B

H B

B

H

H H

H

HH

H

-1

Chem 350, Fall 2005 Exam 4 CCM, Page 5

Brief Spectrochemical Series:

I- < H2O < NH3 < CN-,CO

Ligand Ionic Method Neutral Method

NO (bent) 2 (NO–) 1 (O=N•) NO (linear) 2 (NO+) 3 (O=N• + lp) =O, =S, =NR 4 (O2–, S2–, RN2–) 2 ≡CR 3 (neutral) 3 ≡N 6 (N3–) 3 η5-C5H5 (Cp) 6 ( C5H5 –) 5 η6-C6H6 (benzene) 6 6 η7-C7H7 (cycloheptatrienyl)

6 ( C7H7+) 7

η3-C3H5 (allyl) 4 (C3H5–) 3

RC≡CR 2 or 4 2 or 4

N N

= bipy