# How do you find oxidation numbers of coordination compounds?

Aug 15, 2016

By the usual means. You assign charges to the metal ions in the complexes, to the ligands that bind to the metal centre, and note the charge on the complex itself.

#### Explanation:

Consider $\left[C o {\left(N {H}_{3}\right)}_{6}\right] C {l}_{3}$. You know (or should know) that the chloride counterions each have $- 1$ charge. Thus we can separate the charges.

${\left[C o {\left(N {H}_{3}\right)}_{6}\right]}^{3 +} + 3 \times C {l}^{-}$. I have preserved charged here: I started with a neutral complex, and it's still neutral as required. Then I dismember the metal complex:

$C {o}^{3 +} + 6 \times N {H}_{3}$ You have to recognize the ammine ligand as a neutral entity; the same as I would for a phosphine, $P {R}_{3}$, or carbonyl. $C \equiv O$, ligand.

Another example, $P d {\left(P {R}_{3}\right)}_{2} C {l}_{2}$. Splitting this up I get $P {d}^{2 +} + 2 \times P {R}_{3} + 2 \times C {l}^{-}$. Reasonably we have a $P {d}^{2 +}$ centre.

If you have got specific questions post them here, but the important thing is to keep practising, and don't be scared of being wrong - anyone can be.