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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.

Erisin
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    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. – bobthechemist Apr 18 '14 at 01:50

1 Answers1

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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 ($\ce{BiO3-})$.

Klaus-Dieter Warzecha
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    I guess you meant $\ce{Bi^{(V)}}$? Also relativistic effects may play an important role lowering the energy of the 6s Orbital. – Martin - マーチン Apr 16 '14 at 06:21
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    @Martin Yes, fixed. And yes, possibly. Apparently, I need more coffee :D – Klaus-Dieter Warzecha Apr 16 '14 at 06:28
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    See also, http://chemistry.stackexchange.com/questions/8717/what-is-inert-pair-effect/8774#8774 and http://chemistry.stackexchange.com/questions/2795/why-does-radium-have-a-higher-first-ionisation-energy-than-barium for more on the relativist effects. – Ben Norris Apr 16 '14 at 10:34
  • 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. – bobthechemist Apr 18 '14 at 01:47
  • Bismuth can also form oxocations like $\ce{BiO+}$ which is also a +3 ion. – Nilay Ghosh Feb 05 '16 at 15:20