What are the oxidation numbers of each oxygen atom in ozone?

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What are the oxidation numbers of each oxygen atom in ozone?

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Answer 1 · 2015-10-09T22:30:46

What is the oxidation state of the individual $O$ $O$ atoms in ozone: from left to right, $- I, + I, O$ $-I, +I, O$ . ( $O^{-} - O^{+} = O$ $O^(-)-O^(+)=O$ ).

Explanation:

Oxidation state/number is of course a formalism. It is the charge left on the atom when all the bonding pairs are broken, and the charge goes to the most electronegative atom. Here, we have oxygen-oxygen bonds. The bond is cleaved homolytically, with the 2 (or 4) bonding electrons shared between the oxygen atoms.

From left to right, there are $(7 + 2) e^{-}$ $(7+2)e^-$ , $7 e$ $7e$ , and $8 e$ $8e$ . The atomic number of oxygen is $8$ $8$ , i.e. 8 protons; of course this means from left to right the formal charge on each atom is $- 1, + 1,$ $-1, +1,$ and $0$ $0$ ). Of course this a formalism, but it nevertheless means that the formal charge of the bent ozone molecule is ZERO $(- 1 + 1 + 0)$ $(-1 + 1 + 0)$ , even if the individual oxygen atoms have different formal oxidation states, and we can draw the other resonance structure such that the order of charge is reversed.

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What are the oxidation numbers of each oxygen atom in ozone?

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