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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.

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    $\begingroup$ 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

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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 ($\ce{BiO3-})$.

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    $\begingroup$ 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
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    $\begingroup$ @Martin Yes, fixed. And yes, possibly. Apparently, I need more coffee :D $\endgroup$ Apr 16, 2014 at 6:28
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    $\begingroup$ 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

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