psi * psi Posted December 7, 2011 Share Posted December 7, 2011 I'm struggling majorly with a problem. The directions are to draw all possible isomers for [Ru(bpy)(NH3)2(H2O)Cl]+. I'm struggling with drawing the structure of the compound itself, let alone the isomers. I know bpy is a bidentate ligand, but I'm not sure I'm clear on how that effects the structure. Thanks for any guidance! Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 Let's try to work through this. Alright so I count six coordination bonds. Ruthenium(I) is a d^5 cation so what is the likely coordination geometry? Next, consider that bpy is a bidentate ligand. Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 Let's try to work through this. Alright so I count six coordination bonds. Ruthenium(I) is a d^5 cation so what is the likely coordination geometry? Next, consider that bpy is a bidentate ligand. It's d5, so it's either trigonal bipyramidal or square pyramidal. I'm going to go with trigonal bipyramidal because of the charge on the complex? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 7, 2011 Share Posted December 7, 2011 The are 6 coordinate bonds. Might be a touch hard for it to adopt trigonal bipyrimidal or square pyrimidal geometry given that both of those molecular geometries contain 5 ligand bonds. Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 (edited) It's d5, so it's either trigonal bipyramidal or square pyramidal. I'm going to go with trigonal bipyramidal because of the charge on the complex? Guess again. By the way. Earlier I said Ru(I) when the complex actually has a Ru(II) metal center. Edited December 7, 2011 by mississippichem Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 My bad, I saw d5 and thought 5 bonds, even though I know there are 6. So it's octahedral. Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 My bad, I saw d5 and thought 5 bonds, even though I know there are 6. So it's octahedral. Right. Now tell me how many different ways bpy can bind to the ruthenium center (assuming for a moment there are only two types of ligand, 1 bpy and four generic "other"). Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 (edited) Bidentate ligands can only bind in cis positions. I have trouble picturing 3d molecules in my head, but I'm counting 8 possible ways? I could be way off Edited December 7, 2011 by psi * psi Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 Bidentate ligands can only bind in cis positions. I have trouble picturing 3d molecules in my head, but I'm counting 8 possible ways? I could be way off Try two. Equitorial-equitorial and axial-equitorial. Remember we are assuming all other ligands are generic and identical for now. Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 Ok, I see that better now. I wasn't accounting for the other ligands Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 (edited) Ok, I see that better now. I wasn't accounting for the other ligands Bringing the other ligands back in: Now try to draw all the possible isomers for bpy being eq-eq and then all the possibilities for bpy being eq-ax. Remember that there are two amines and they can be axial or equatorial. Edited December 7, 2011 by mississippichem Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 for bpy being eq-eq, I get one configuration with the amines being trans (ax-ax) and the others being cis, and one with the amines cis (eq-eq) and the others trans. Can I also then put an amine axial with chlorine and another amine being equitorial and the H20 as the other axial, or is that repetitive? I've always had trouble telling what's redundant and what's not... Link to comment Share on other sites More sharing options...
mississippichem Posted December 7, 2011 Share Posted December 7, 2011 for bpy being eq-eq, I get one configuration with the amines being trans (ax-ax) and the others being cis, and one with the amines cis (eq-eq) and the others trans. Can I also then put an amine axial with chlorine and another amine being equitorial and the H20 as the other axial, or is that repetitive? I've always had trouble telling what's redundant and what's not... I know what you mean. Sometimes the best way is to write it out. For example here's one isomer. Bpy: eq-eq H2O: ax Amine: eq Amine:eq Cl: ax Now it's easier to compare and see what is redundant. Link to comment Share on other sites More sharing options...
psi * psi Posted December 7, 2011 Author Share Posted December 7, 2011 I know what you mean. Sometimes the best way is to write it out. For example here's one isomer. Bpy: eq-eq H2O: ax Amine: eq Amine:eq Cl: ax Now it's easier to compare and see what is redundant. The example you just gave is one of the isomers I have. Which means I'm not completely off . Can the example you just gave have an enantiomer? Because I see one... I also have: bpy: eq amine: axial amine: axial H2O: eq Cl: eq ^^ which also seems like it'd have an enantiomer...? Then I also have: bpy: eq amine: axial H2O: axial Cl: eq amine: eq which I don't THINK is repetitive since I haven't had amine and water trans to each other yet. same with the next one since I haven't had chlorine and amine trans to each other either. Am I completely off base? bpy: eq amine: ax H2O: eq Cl: ax amine: eq Link to comment Share on other sites More sharing options...
mississippichem Posted December 8, 2011 Share Posted December 8, 2011 (edited) The example I gave doesn't have an enantiomer. Your first example does have an enantiomer and so does your second. If you think this is painful. Try one with a tougher geometry like the seven-coordinate capped-trigonal-anti-prismatic . These exercises are not fun but they really build your 3D visualization chops which are crucial in chemistry. Edited December 8, 2011 by mississippichem Link to comment Share on other sites More sharing options...
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