This work is Created by Burke Scott Williams (swilliams@isd claremont.edu) and p

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This work is Created by Burke Scott Williams (swilliams@isd claremont.edu) and posted on VIPEr on 01/10/2009. licensed under the Creative Commons Attribution Non-commercial Share Alike License in 2009. To view a copy of this visit hitp sratixecommons.ong about license. Modified by Gerard Rowe 03/2017 on ligand is the allyl anion, CaHs. Figure out the totals for yourself! I suggest counting about whether vou'd expect these to be 16 e or 18 e metals. For 9. A final comm the metals together, and think the purposes of geometry, allyl takes up as much space as two mon odentate ligands Ni H-OH TOTAL n/a 10. Most of the time, closed-shell and neutral-ligand counting will give you the same answer because the assumptions they make about the nature of the M-L bond ultimately work the same. There is a situation where one of the methods can fail, though. Consider the acid-base reaction below. ?? H20 Ir NH3 4-pyridinecarboxylate n/a Ir NH3 4-pyridinecarboxylic n/a acid (N-donor) 3+ charge TOTAL N-donor) 2+ charge TOTAL n/a 18 18 n/a Which method gave you a different electron count for the two species? Does it make sense for the metal's electron count to change if you modified a completely different part of the molecule? What assumption has led to this counting failure?

Explanation / Answer

9. Total electron count,

in CS method = 8 + (2 * 1) + 4 = 8 + 2 + 4 = 14

in NL method = 10 + ( 2 * 1/2 ) + 3 = 10 + 1 + 3 = 14

10. CS and NL method can give rise two different electron count for the product molecule. This can be explained by the resonance effect of the ligand.

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