[bu88le]

Can anyone tell me how to figure out the oxidation state of carbon and nitrogen in CNO-? thanks.

[Borek]

I doubt. In covalent compounds oxidation states loose their meaning. CNO^- is charged, but the bonds between C, N and O are covalent.

Do you need it for anything particular, or to balance some redox reaction?

[bu88le]

yea, i need to balance the oxidation-reduction reaction that occur in this basic solution:

CN-(aq) + MnO4-(aq) --> CNO- (aq) + MnO2 (s)

[Borek]

In this case there is a better way of balancing the equation, although I don't remember its name, and my explanation will be very brief. You start with two half-reactions:

MnO₄^2- + H₂O +e^- -> MnO₂ + OH^-

CN^- + OH^- -> CNO^- + e^- + H₂O

They are not balanced - DIY; I have added H₂O and OH^- on both sides so that they should be possible to balance, but I could be wrong. Note, that you have to balance charge too - there must be the same charge on both sides of equation, that's the most important thing here.

Now, what you have to do is to add both half reactions, multiplying them by such a numbers, that electrons cancel out.

This way you don't have to assign oxidation numbers to C and N in CNO^- nor CN^-. To say the truth, this method is closer to the real thing - in many cases, especially when there are organic compounds being reduced or oxidized,  it is not possible to use oxidation numbers, as there is no way of assigning them. However, half reactions exist always and they can be always combined to cancel out electrons.

Well... almost always. But that's another story

[bu88le]

thanks Borek!