# Total valence electrons in a rhodium complex

How many valence electrons does the following metal complex have? 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?

• 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. – Edison Hua Sep 1 '16 at 16:22
• 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 $6 (\ce{Rh^3+}) + 2 (\ce{H-}) + 2 (\ce{Ph-}) + 2 (\ce{Cl-}) + 2 \times 2 (2 \times \ce{PPh3}) = 16$. – Jan Sep 1 '16 at 21:24

This is not a benzene ligand ($\ce{C6H6}$), but rather a phenyl ligand ($\ce{C6H5}$). Therefore the total electron count is $9 + 1 + 1 + 2 +2 + 1 = 16$.