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PS3 Key 1. A) PPh2 = L = 2e-­‐ X2(per metal) Cl = X = 1e-­‐ X1(per metal) Pd = 10 e-­‐ Pd—Pd bond = 1e-­‐ Total: 16 e-­‐ Pd(I), d9 B) Cp = L2X = 5e-­‐ X2 Cl = X = 1e-­‐ X2 Zr = 4 e-­‐ Total: 16 e-­‐ Zr(IV), d0 C) CpMe = L2X = 5e-­‐ X2 C4H8O = L = 2e-­‐ H = X = 1e-­‐ H = L = 2e-­‐ Y = 3 e-­‐ Total: 18 e-­‐ Y(III), d0 Note :Alternative accepted answer has total count at 16e-­‐ on each Y D) BN3 = L3 = 6e-­‐ H = X = 1e-­‐ CH3 = X = 1e-­‐ Ph = X = 1e-­‐ Formal -­‐1 charge on BN3 Pt = 10 e-­‐ Total: 18 e-­‐ Pt(IV), d6 E) Cp = L2X = 5e-­‐ Chelating NHC ligands = L = 2e-­‐ X2 CO = L = 2e-­‐ Fe = 8 e-­‐ -­‐1 total for positive charge Total: 18 e-­‐ Fe(II), d6 F) PRtBu2 = L = 2e-­‐ X2 C6H3 = X = 1e-­‐ N = L = 2e-­‐ Ir = 9 e-­‐ Total: 16 e-­‐ Ir(I), d8 G) =NR group = L = 2e-­‐ Me2RN group = L = 2e-­‐ Cl = X = 1e-­‐ X2 RO = x = 1e-­‐ +1 for -­‐1 charge on metal Ni = 10 e-­‐ Total: 18 e-­‐ Ni(II), d8 H) Os 1) CO = L = 2e-­‐ X4 Os—Os = 1e-­‐ X2 Os = 8e-­‐ Total: 18 e-­‐; Os(0), d8 Os 2) CO = L = 2e-­‐ X3 C—H = L = 2e-­‐ H = X = 1e-­‐ Os—Os = 1e-­‐ Os = 8e-­‐ Total: 18 e-­‐; Os(I), d7 Os 3) CO = L = 2e-­‐ X3 H = L = 2e-­‐ Os—Os = 1e-­‐ CH3 = X = 1e-­‐ Os = 8e-­‐ Total: 18 e-­‐; Os(I), d7 1 2 3 Note: Alternative accepted answer is 19e-­‐ for both Os 2 and 3. I) Formal -­‐1 charge on BN3 BN3 = L3 = 6e-­‐ PhN = X = 1e-­‐ NpN=C = X = 1e-­‐ NpN≡C = L = 2e-­‐ Rh = 9e-­‐ Total: 18 e-­‐ Rh(III), d6 J) N = LX2 = 4e-­‐ R2C = X2 = 2e-­‐ Me3CO = LX = 3e-­‐ X2 Mo = 6e-­‐ Total: 18 e-­‐ Mo(VI), d0 Alternative accepted answer is Mo(IV), d2. 2. (a) [OsCl6]2– contains an Os(IV), d4, low-­‐spin ion, t2g4eg0, with two unpaired electrons. It could also be high spin, t2g2eg2, (four unpaired electrons) but it isn’t because all compounds of 2nd and 3rd row transition metals are low spin. (b) [CrF6]4– contains a Cr(II), d4, high spin ion, t2g3eg1, with four unpaired electrons. It could also be low spin, t2g4eg0, (two unpaired electrons) but it isn’t. –
Cr(II) complexes are only low spin with very strong field ligands, while F is a weak-­‐field ligand. Thus this complex is high spin. (c) [Fe(CN)6]3-­‐ Fe(III) [called ferricyanide], d5, low spin, t2g5eg0, with one unpaired electron. It could also be high spin, t2g3eg2, (five unpaired electrons) but it isn’t. The benchmark compound near the high spin/low spin crossover point for Fe(III) is low spin Fe(bpy)33+. [Fe(II) is similar, with low spin Fe(bpy)32+ being near the crossover point.] Since cyanide is above bpy on the spectrochemical series, this complex is low spin. (d) [Fe(H2O)6]2+ Fe(II), d6, high spin ion, t2g4eg2, with four unpaired electrons. It could also be low spin, t2g6eg0, (one unpaired electron) but it isn’t. As noted in part (c), The benchmark compound near the high spin/low spin crossover point for Fe(II) is low spin Fe(bpy)32+. Since water is below bpy in the spectrochemical series, this complex is high spin. Note that all aquo complexes of 1st-­‐row transition metals are high spin except [Co(H2O)6]3+. (e) [Ru(H2O)6]2+ Ru(II), d6, low spin, t2g6eg0, with no unpaired electrons. It could also be high spin, t2g4eg2, (no unpaired electrons) but it isn’t. Ruthenium is a second-­‐row transition metal, and all octahedral complexes of second and third row metals are low spin. (f) [Cu(H2O)6]2+ Cu(II), d9, t2g6eg3, one unpaired electron. There is no other choice. In the demonstration I did in class, this is the ion that gives the light blue color. (g) [CrCl6]3-­‐ Cr(III), d3, t2g3eg0, three unpaired electrons. There is no other choice. (h) [Co(NH3)6]3+ Co(III), d6, low spin, t2g6eg0, with no unpaired electrons. It could also be high spin, t2g4eg2, (four unpaired electrons) but it isn’t. This is one of the benchmark compounds we discussed in class (you need to remember this). Note that the complexes in (e) and (h) are isoelectronic – they all have the same structure and the same number and configuration of valence electrons (low spin d6). The complex in (d) has the same number of electrons but is not isoelectronic because it is high spin. The complexes in (a) and (c) are also isoelectronic (low spin d5). Another example of isoelectronic molecules and ions is: N2, CO, NO+, and CN-­‐. Isoelectronic molecules often have similar or related structure and reactivity. 3. (a) Having bond angles a bit off of 90˚ and 180˚ will not strongly affect the 2-­‐above-­‐3 d-­‐orbital splitting pattern. It C N
Mn N C Mn C N
will make the t2g orbitals a little involved in the bonding, but the overlap still seems pretty poor, and it should only n
reduce the overlaps with the eg orbitals by a little bit. This is illustrated by the drawing at right, where the donor orbital above is 9˚ off the vertical, and it still overlaps high-spin Mn(III)
low-spin Mn(III)
pretty well with the s-­‐bonding d orbital. seems to me more with 2 nitrogenwith 2 carbonlikely to slightly change the magnitude of Do rather than bound CN - ligands
bound CN- ligands
the pattern. If only the bond angles change, then t2g orbitals will take on a little antibonding character, as now they don’t have strictly zero overlap with the ligands. And the e2g orbitals will be a little less antibonding, as their overlap with the ligands will be slightly reduced. But it is also possible that the constraints of the chelate ring will push the ligating atoms closer to the metal, increasing Do. (b) (i) [Mn(acac)3] Mn(III), d4, t2g3eg1, high spin four unpaired electrons. It could also be low spin, t2g4eg0, (two unpaired electrons) but it isn’t. Quite strong field ligands like cyanide are needed to make Mn(III) low spin. (ii) [Ru(en)3]2+ Ru(II), d6, low spin, t2g6eg0, with no unpaired electrons. It could also be high spin, t2g4eg2, (no unpaired electrons) but it isn’t. Ruthenium is a second-­‐row transition metal, and all complexes of second and third row metals are low spin. (iii) [Fe(salen)(H2O)2]+ Fe(III), d5, t2g3eg2, high spin, with five unpaired electrons. One could make a case that it might be high spin, t2g5eg0, (one unpaired electron), but it isn’t. The iron(III) ion has four oxygen donors (two waters and two phenoxides from the salen) and two imine donors (from the salen). Fe3+ with six oxygen ligands is always high spin, while iron with six imine donors, as in [Fe(bpy)3]3+ is low spin. 4. The [Mn(salen)CN]n polymer has one manganese(III) ion with two carbon-­‐bound cyanide ligands alternating with a Mn(III) with two nitrogen-­‐bound cyanide ligands, as shown below. The Mn(III) ion with two C-­‐bound cyanides is low spin, t2g4eg0,with S = 1 (two unpaired spins). C-­‐bound cyanide is a very strong-­‐field ligand. N-­‐bound cyanides, however, are weaker field ligands and the Mn(III) ion with these ligands is high spin, t2g3eg1,with S = 2 (four unpaired spins). This work is described in “[{Mn(salen)CN}n]: The First One-­‐Dimensional Chain with Alternating High-­‐Spin and Low-­‐Spin MnIII Centers Exhibits Metamagnetism” by N. Matsumoto, Y. Sunatsuki, H. Miyasaka, Y. Hashimoto, D. Luneau, and J.-­‐P. Tuchagues Angew. Chem. Int. Ed. 1999, 38, 171-­‐173. You can access this paper online if you want from a UW-­‐based computer wither through the UW online catalog, or by clicking the click the Wiley Interscience box at: http://www.wiley-­‐vch.de/publish/en/journals/alphabeticIndex/2002/ 5. C2v (ortho) E C2 ΓC-­‐N 2 0 ΓC-­‐N = A1 + B1; 2 IR bands, 2 Raman bands σ 2 σ 0 C2v (meta) E C2 ΓC-­‐N 2 0 ΓC-­‐N = A1 + B1; 2 IR bands, 2 Raman bands σ 2 σ 0 D2h E C2(z) C2(y) C2(x) (para) ΓC-­‐N 2 0 2 0 ΓC-­‐N = Ag + B2u; 1 IR band (B2u), 1 Raman band (Ag) i σ (z) σ (y) σ (x) 0 2 0 2 Para-­‐dicyano-­‐benzene can be unequivocally identified by vibrational spectroscopy. Extra Credit. Any reasonable answers.