What is the oxidation state of bromine in BrO3?

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Bretherick’s Handbook of Reactive Chemical Hazards lists bromine trioxide $\ce{BrO3}$ structure as the one composed of cationic and anionic parts [1, p. 116]:

structure of BrO3

The solid produced at $\pu{—5 °C}$ by interaction of bromine and ozone is only stable at $\pu{—80 °C}$ or in presence of ozone, and decomposition may be violently explosive in presence of trace impurities [...]. The structure may be the dimeric bromyl perbromate, analogous to $\ce{Cl2O6}$ [2].

so that formula written as $\ce{BrO3}$ is fine being a formula unit, but more correctly bromine trioxide should be written as $\ce{[BrO2]+[BrO4]-}$, implying that bromine is in two oxidation states: $+5$ and $+7$.

References

  1. Bretherick, L.; Urben, P. G.; Pitt, M. J. Bretherick’s Handbook of Reactive Chemical Hazards, 7th ed.; Elsevier: Amsterdam, The Netherlands ; Boston, MA, 2007. ISBN 978-0-12-373945-2.
  2. Pflugmacher, A.; Rabben, H.-J.; Dahmen, H. Zur Kenntnis der Bromoxyde. Über das Bromtrioxyd. Zeitschrift für anorganische und allgemeine Chemie 1955, 279 (5–6), 313–320. https://doi.org/10.1002/zaac.19552790507.
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Eagle
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Eagle

Updated on August 01, 2022

Comments

  • Eagle
    Eagle over 1 year

    I came across this compound $\ce{BrO3}$ in my textbook, and was trying to find the oxidation state of bromine.

    This is how I proceeded:

    • Since this is not a superoxide or a peroxide, or any other case where the oxidation state of oxygen is not −2 according to this, so the oxidation state of oxygen should be −2.
    • For oxidation state of Br as $x$, I can have $x + 3(-2) = 0$, which gives $x = +6$.

    But, according to Wikipedia, the +6 oxidation state of Br is not listed.

    I was confused because my book (NCERT Part I for class XII) says that "The bromine oxides Br2O, BrO2, BrO3 are the least stable halogen oxides (middle row anomaly) and exist only at low temperatures."

    Please tell me where I am going wrong.

    P. S. This is not an ion $(\ce{BrO3-})$ but a neutral compound.