USNCO 2004 Question 45 states:

Which element exhibits the greatest number of oxidation states in its compounds?

(A) Ca (B) V (C) Cu (D) As

I ruled out Ca, as I know it only exists as +2. I then drew electron configurations for V, Cu and As. I ruled out Cu as it had half filled 4s and full 3d orbitals, also because I've only ever seen copper in a +2 oxidation state.

However, I wasn't able to differentiate between V and As. They both have 3 unpaired electrons, albeit As has an extra full 3d orbital.

I was tempted to go with As, as being less metallic than V, it is more likely to have negative oxidation states, but the answer states Vanadium has the most oxidation states.

Is there any way I can work this question out by looking at electron configurations, or will I have to memorize the oxidation states?

  • $\begingroup$ V and As are evenly matched - both on all states between -3 and 5. BTW I'd expect you heard about Cu +1 compounds, it goes up to +4, actually and there's also exotic -2. $\endgroup$ – Mithoron Jan 10 at 0:15
  • $\begingroup$ @Mithoron I saw on a Wiki page that V and As do have the same number of oxidation states, but is there any way I can work out which have more 'common' states which I think USNCO is asking for? $\endgroup$ – George Tian Jan 10 at 0:21
  • $\begingroup$ Well, I wager V has more "common" ones, just like tr. met. do - unpaired electrons aren't a problem for them. $\endgroup$ – Mithoron Jan 10 at 1:55

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