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How many valence electrons does the following metal complex have?

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I have read that benzene donates 6 electrons to a transition metals valence electrons. Rh has 9 from itself, and H and Cl add 1 each; hence in this assignment above I thought one would end up with: 9+1+1+2+2+6=21 valence electrons.

However, the given answer is 16. How is that so?

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  • $\begingroup$ You add the charge of the ligands, H: and Cl: both have -1. I'm getting 17 though, but if I assume the benzene has a charge of -1 it's 16. That might explain why the benzene isn't donating all 6 aromatic electrons, but rather a single bond of 2. So, 9 - 3 + 10 = 16. $\endgroup$ – Edison Hua Sep 1 '16 at 16:22
  • $\begingroup$ What you did is neutral counting. I typically find oxidation state counting easier. In that case, realise that rhodium is rhodium(III) (three monoanions and a neutral complex) so count $6 (\ce{Rh^3+}) + 2 (\ce{H-}) + 2 (\ce{Ph-}) + 2 (\ce{Cl-}) + 2 \times 2 (2 \times \ce{PPh3}) = 16$. $\endgroup$ – Jan Sep 1 '16 at 21:24
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This is not a benzene ligand ($\ce{C6H6}$), but rather a phenyl ligand ($\ce{C6H5}$).

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In the benzene complex, all the carbons have their hydrogens. The pi electrons of the ring act as donors to the metal. On the other hand, in the phenyl complex, the ring carbon bonded to the metal does not have its hydrogen. That carbon is formally negatively charged and the lone pair on carbon acts as a donor to the metal.

Aryl ligands, just like alkyl ligands, count as 1-electron donors within the "covalent" electron-counting system (which you seem to be using, since you counted H and Cl as 1e donors as well).

Therefore the total electron count is $9 + 1 + 1 + 2 +2 + 1 = 16$.

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