What is the oxidation number of oxygen in #"sodium peroxide"#, #Na_2O_2#?

2 Answers
Oct 13, 2016

In sodium peroxide the oxidation state of oxygen is #-I#.


When we assign oxidation numbers, we assign conceptual charges to the individual atoms such that when these charges are summed up we get the charge on the molecule or the ion. Thus, elements have an oxidation number of #0#, because they are assumed not to have donated or accepted electrons (the which process defines oxidation and reduction respectively).

For your sodium peroxide we have #Na_2O_2#. Now sodium generally has an oxidation state of #+I# in its salts, and it does so here. Thus #1+1+2xx"Oxidation number of oxygen in peroxide"=0#. Clearly, #"Oxidation number of oxygen in peroxide"=-1#.

Another way of looking at this is to consider the peroxide ion itself, #""^(-)O-O^-#. When we break a bond conceptually, (i.e. in our imagination), we assume that the charge goes to the most electronegative atom. And so we gets #""^(-)O-O^(-)rarr2xxdotO^(-)#; i.e. a formal #-I# charge. And likewise if we do this for #OF_2# we get #2xxF^-# and #O^(2+)#. And so for element-element bonds, the elements have equal electronegativity and the charge, the electrons, should properly be shared, to give again #O^-#, i.e. oxygen with a #-I# oxidation state.

And thus the oxidation state of oxygen in peroxide is conceived to be #-I#.

Oct 13, 2016

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 electronegativity 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.