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Show practically the oxidation state of O2 in Na2O2?

The Oxidation of state of Oxygen in ##Na_2O_2 ## is -1

The structure of the the molecule sodium per oxide is with each sodium joined to one oxygen and the two oxygen joined with one bond to each other and the other bond to the sodium atoms.

Oxygen ## 1s^2 2s^2 2p^4## typically forms two bond to fill the two half filled p orbitals. The one bond that is shared between the two Oxygen atoms does not affect the Oxidation state of Oxygen. The two Oxygen atoms have equal electro negativity so there is no loss or gain of electrons in that bond.

The bond that Oxygen shares with Sodium does change the Oxidation state of both Sodium and Oxygen. Sodium with a much lower will give up one electron to the Oxygen becoming +-1 ( Sodium is oxidized) The Oxygen having a much high electronegativity will take an electron from the Sodium becoming (-1) The Oxygen is reduced. .. Na : O : .. O: Na

This Lewis dot structure shows the "shared" electrons between the the Sodium and Oxygen atoms which results in the transfer of an electron from the Sodium to the Oxygen resulting in the +1 charge for Sodium and the -1 charge for Oxygen. The shared electrons between the two Oxygens results in no loss or gain of electrons.

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