π donor L L L π acceptor has empty π orbitals on ligand in to which d e- from M can be donated

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1 Name KEY D# Chemistry 350 Fall 2005 Exam #4, November 18, minutes CCM 100 points on 4 pages + a useful page 5 1. Consider the molecular orbital diagram shown for M N. (16 pts) a) Indicate the following: Number of igands, N, = 6 = π acceptor σ only donor (circle one) π donor M b) On the diagram at right, label the bonding orbitals as more metal (M) or ligand () in character. c) riefly 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 after σ bonding donates to M d) Describe how compares in the case shown to that in M 6 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 ]

2 Chem 350, Fall 2005 Exam 4 CCM, Page 2 2. a)complete the following table. (24 pts) [Fe(CN) 6 ] 4- [Cr(N 3 ) 6 ] 3+ [Ni(O 2 ) 5 F] correct name hexacyanoferrate(ii) hexaamminechromium(iii) pentaaquafluoronickel(i) oxidation state of central metal atom number of valence electrons on central metal atom neutral: 8 + 1(6) + 4 =18 ionic: 6 + 2(6) = n: 6 + 2(6) - 3 =15 i: 3 + 2(6) = n: (5) + 1 =21 i: 9 + 2(5) + 2 = number of d electrons electron configuration of central metal ion [Ar]3d 6 [Ar]3d 3 [Ar]3d 9 likely molecular geometry around central atom octahedral octahedral octahedral spin: high, low, or not IG OW N/A IG OW N/A IG OW N/A applicable? (circle) spin, S = 0 / 2 / 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(PR 3 )(CN) 3 ] 4- Ionic: Co(-1) 8 Neutral: Co 7 PR 3 2 PR 3 2 3CN 6 3 CN b) Assuming the 18 electron rule applies, identify the second-row transition metal: [MCl(CO)(PPh 3 ) 2 ] neutral (n): *2-1=11=M, Ag ionic (i) : *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-

3 Chem 350, Fall 2005 Exam 4 CCM, Page 3 4. Complexes of Co 3+ having more than one type of ligand display a variety of colors. For example, CoCl 3 6N 3 is yellow, whereas CoCl 3 5N 3 is purple, whereas CoCl 3 4N 3 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 N 3 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 Co 3+ and, therefore, d6. They are likely all high-spin complexes as there are no pi acceptor ligands. The structures are probably [Co(N 3 ) 6 ]Cl 3, [Co(N 3 ) 5 (Cl)]Cl 2, and [Co(N 3 ) 5 (Cl) 2 ]Cl, respectively. [Co(N 3 ) 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 λ max (in the MCT 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. ipy 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+. e sure to label the enantiomers appropriately. Point Group: D 3 c) Indicate the hybridization of central metal atom. sp 3 d 2 d) What would you expect to happen to λ max if each bipy was replaced by 2 N 3 s? replace a π acceptor with a σ donor, so decrease o, decrease E = hc/λ, so λ increases. onger λ max

4 Chem 350, Fall 2005 Exam 4 CCM, Page 4 6. Three of the five structural fragments below occur in the borane ion 3-8 of C 2v symmetry. i) Provide - a projection formula of 3 8 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) bridging fragments i) 18 total electrons ii) 3 - = 6 e = 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 s should all be connected, not just 2,but this violates our rules. See J. Phys. Chem. 1991, 95, 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) + Cl 2 (g) + 2e - 2 e O (l) 2 O - (aq) + 2 (g) ii) and the overall process. 2NaCl(aq) O (l) 2 NaCl(aq) + Cl 2 (g) + 2 (g) c) Comment on why the diaphragm must be present to prevent mixing of the two gases being produced. Without the diaphragm the 2 and Cl 2 gases would recombine to form Cl in a very exothermic reaction. This Cl could then readily react with the NaO and get us back to the starting materials of NaCl and 2 O!

5 Chem 350, Fall 2005 Exam 4 CCM, Page 5 igand Ionic Method Neutral Method NO (bent) 2 (NO ) 1 (O=N ) NO (linear) 2 (NO+) 3 (O=N + lp) =O, =S, =NR 4 (O 2, S 2, RN 2 ) 2 CR 3 (neutral) 3 N 6 (N 3 ) 3 η 5 -C 5 5 (Cp) 6 ( C 5 5 ) 5 η 6 -C 6 6 (benzene) 6 6 η 7 -C ( C ) 7 (cycloheptatrienyl) η 3 -C 3 5 (allyl) 4 (C 3 5 ) 3 RC CR 2 or 4 2 or 4 rief Spectrochemical Series: I - < 2 O < N 3 < CN -,CO = bipy N N