Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: kriggy on February 02, 2014, 08:54:26 AM
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Why is that that complex M40401 favors folding into octaherdal geometry and M40404 not?
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FQtuUqO2.jpg&hash=50a176fa8bbde58bf84b9e3a053183b719ad2413)
The paper doesnt specify why is that. I suppose it has something to do with the methyl substituents and direction of that bond relative to the fused benzen rings - the bond benzene-nitrogen to be more specific. But I have realy hard time imagining the structure in 3D
Could you give me some hints or help?
Thans
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Humm, I think it's about the equilibrium pentagonal bipyramid ::equil:: mono-capped octhaedra, which, dependng on the conditions, is totally shifted to the left or to the right.
If you look at M40401, you'll notice that methyl and phenyl are eclipsed on each side, which destabilizes the pentagonal bypiramid and shifts the geometry to a possible one where there are no such repulsions. On the other hand, there are no many differences for the two geometries
in terms of repulsions between bulky groups for M40404, so here orbital energetics may play a crucial role. ( sp3d3 likes the pentagonal bipyramidal geometry.)
Do you think that's the answer, what do you say? ( I purely rationalized, I found nothing about NC=7 Mn2+ compounds.)
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Yes this sounds reasonable. I will probably consult it with someone at school when next semester starts.
Thanks for help