It has 5 valence electrons, but only loses 3 of them to make a +3 ion. Why does this occur?

I believe it may have something to do with how losing 3 electrons leaves you with the p sublevel full, but I do not understand why Bismuth is okay with that.

  • 2
    $\begingroup$ losing 3 electrons leaves you with the p sublevel full is incorrect. Removing 3 electrons from Bi results in an electron configuration of [Xe]6s2 4f14 5d10 or an empty 6p subshell. $\endgroup$ Apr 18, 2014 at 1:50

1 Answer 1


Isn't it just that +5 is a lot of (positive) charge for a single cation? Which other (transition) metals do form a $\ce{M^{5+}}$ cation? Just out of my mind, I can't think of any.

Typically, these high oxidation states exist in the form of oxo-anions ($\ce{CrO4-}$, $\ce{MnO4-}$, $\ce{VO4^{3-}}$, etc.) and bismuth is no exception here: $\ce{Bi(V)}$ exists as bismutate ($\ce{BiO3-})$.

  • 3
    $\begingroup$ I guess you meant $\ce{Bi^{(V)}}$? Also relativistic effects may play an important role lowering the energy of the 6s Orbital. $\endgroup$ Apr 16, 2014 at 6:21
  • 1
    $\begingroup$ @Martin Yes, fixed. And yes, possibly. Apparently, I need more coffee :D $\endgroup$ Apr 16, 2014 at 6:28
  • 2
    $\begingroup$ See also, chemistry.stackexchange.com/questions/8717/… and chemistry.stackexchange.com/questions/2795/… for more on the relativist effects. $\endgroup$
    – Ben Norris
    Apr 16, 2014 at 10:34
  • $\begingroup$ I'm not sure why transition metals need to be invoked to answer this question, but a look at the Latimer diagrams of V, Cr, and Mn group transition metals (for example at webelements we see that there are a number of metals with accessible (V) oxidation states. $\endgroup$ Apr 18, 2014 at 1:47
  • $\begingroup$ Bismuth can also form oxocations like $\ce{BiO+}$ which is also a +3 ion. $\endgroup$ Feb 5, 2016 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.