serunato Posted October 9, 2004 Share Posted October 9, 2004 Can someone explain why this is? In [Fe(en)3][picrate]2 this means it is FeII. So using valence bond theory they should be sp3d2 hybrid orbs. Crystal field theory is not an option in this problem, which would have made more sense to me. So in [Fe(1,10-phenanthroline)3][ClO4]2 it is also FeII but it must be in the low spin state meaning that all the e- are paired in the remaining 3d orbs not used for the hybrid bonds. While in the [Fe(en)3][picrate]2 it must be using the 4d orbs and leaving the 3d orbs alone to have the 6e- spread out over the five. Both ligand types are bidentate and both bond by the nitro. This explains the difference in magnetic moment, but why in the hell does one go high spin and the other go low? Link to comment Share on other sites More sharing options...
budullewraagh Posted October 9, 2004 Share Posted October 9, 2004 the latter goes low because it's an ionic compound Link to comment Share on other sites More sharing options...
serunato Posted October 9, 2004 Author Share Posted October 9, 2004 Are you saying that [Fe(en)3][picrate]2 is covalent? It seems that both are ionic to me. How else would Fe form coordination complexes? Or are you saying that Fe(en)3 is somehow covalent? That makes no sense at all. Link to comment Share on other sites More sharing options...
Skye Posted October 10, 2004 Share Posted October 10, 2004 Why can't you use the spectrochemical series? Link to comment Share on other sites More sharing options...
serunato Posted October 10, 2004 Author Share Posted October 10, 2004 I finally found one that listed phen. Both have N bidentates. I still have a little trouble understanding why, when they are so close in the series, phen is low spin and en is high. It doesn't matter I have more than the answer I need, I was just curious. Thanks. Now on to crystal field Link to comment Share on other sites More sharing options...
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