Bi can reach its 3+ and 5+ oxidation state. Why not 4+? Can you use the inert pair effect here to explain this?

  • $\begingroup$ Well bismuth is in group 15 so, +4 oxidation is probably rare. I don't think that has anything to with inert pair effect. Group 15 generally shows +3 and +5 states because they give closed-shell species. +4 would result in an unpaired electron i.e. a radical $\endgroup$
    – S R Maiti
    May 20, 2021 at 9:28

1 Answer 1


Actually bismuth(IV) is known, and in not so exotic conditions at that. In fact both even oxidation states +2 and +4 appear when acidic bismuth(III) solutions are irradiated:

Bi(IV) has been observed; see A. I. Aleksandrov, I. E. Makarov (1987). "Formation of Bi(II) and Bi(IV) in aqueous hydrochloric solutions of Bi(III)". Bulletin of the Academy of Sciences of the USSR, Division of Chemical Science. 36 (2): 217–220. doi:10.1007/BF00959349.

From the reference cited above:

The formation of bismuth(II) and bismuth(IV) under the action of ionizing radiation on hydrochloric acid solutions of bismuth(III) has been shown by low-temperature and pulse radiolysis. The kinetics of bismuth(IV) disappearance at 300°K have a complex nature and would seem to be caused both by the disproportionation of bismuth (IV) and the reaction of bismuth(IV) with bismuth(II), which leads to bismuth(III).

The properties of bismuth (IV), which is formed from the reaction of Cl2 with bismuth (III) agree with previously established relationships on the variation in optical transition energies and the unpaired electron density at the s atomic orbital of the ion for the series of isoelectronic ions mercury (I), thallium (II), lead (III), and bismuth (IV) in the $^2S_{1/2}$ state.


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