How do you calculate the formal charge of O3?

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How do you calculate the formal charge of O3?

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Answer 1 · 2015-11-07T19:15:25

The formal charge of the ozone molecule is zero. Its Lewis structures do present charge separation.

Explanation:

With simple VSEPR considerations, there are 18 valence electrons to distribute around the 3 oxygen atoms (24 electrons in total; 6 are inner core).

Typically, a Lewis structure of $O = {{..} O}^{+} - O^{-}$ $O=stackrel(ddot)O^(+)-O^(-)$ , would be depicted. Going from left $O$ $O$ to right $O$ $O$ and including the 2 inner core electrons on each atom, there are 8, 7, and 9 electrons around each oxygen centre, resulting in formal charges of $0$ $0$ , $+ 1$ $+1$ , and $- 1$ $-1$ , respectively. Of course, I can draw the other resonance structure, but the Lewis structure has the same electronic formulation. The $∠ O - O - O$ $/_O-O-O$ $≅ 117^{\circ}$ $~=117^@$ ; this is slightly LESS than the normal $s p_{2}$ $sp_2$ bond angle of $120^{\circ}$ $120^@$ , due to disproportionate influence of the oxygen lone pair, which tends to compress $∠ O - O - O$ $/_O-O-O$ .

Since the central oxygen has 3 regions of electron density, this molecule is bent.

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How do you calculate the formal charge of O3?

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