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a paradox on finding an Oxidation number

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amirziv24:
hi everyone.
I apologize in advance because my first language isn't English so you may find it a little not easy to read
So I tried to find the oxidation number of N in the N2O5

we know that for a neutral compound, the sum of the oxidation numbers of all constituent atoms must be equal to zero
the oxidation number of oxygen is usually equal to -2
therefore N+ (5*(-2)) = 0  ==> N = +5

but here's the thing,
there's another way of finding an oxidation number which is based on scoring bonds based on Electronegativity. which goes like this:
in a bond, the more negative element gets -1 and the other one get's +1
then we'll sum them

for example :

CH4:

we have four C-H bonds. C has a more negative number therefore we'll have C = -1 and H = +1
so C is 4 *(-1) = -4
in OF2 there are two O-F bonds
F is more negative so: O = +1 and F = -1
O = 2 * (-1) = -2



now I've tried the previous solution for N in N2O5 but didn't work
can someone please explain to me why?

Hunter2:
Why you cannot find for nitrogen 5 *(-2) / 2 = = -5 so nitrogen has +5.

Babcock_Hall:
I do not have time for a complete answer, but in compounds such as peroxides, the oxidation number of oxygen is -1, not -2.  I do not think that this exception applies in your case.

Hunter2:
Here we have nitrogen pentoxide, the anhydride of nitric acid, no peroxide group.

Babcock_Hall:

--- Quote from: Hunter2 on April 16, 2024, 02:33:39 PM ---Here we have nitrogen pentoxide, the anhydride of nitric acid, no peroxide group.

--- End quote ---
I agree.  On the other hand, I think that the strategy that the more electronegative atom essentially owns the electrons works well here.

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