What is the oxidation state of an individual nitrogen atom in NH2OH?

Nov 10, 2015

Oxidation number is the charge left on the central atom when all the bonding electrons are removed with the charge going to the most electronegative atom.

Explanation:

Electronegativity is a formalism or a concept, it is a contrived parameter; formally it is the tendency of an atom in an element to polarize electron density towards itself. Electronegativity increases across a Period (because nuclear charge, $Z$, increases sequentially). Because oxygen has greater nuclear charge than nitrogen, it formally is more electronegative.

So if we split up hydroxylamine: ${H}_{2} N - O H \rightarrow {H}_{2} {N}^{+} + H {O}^{-}$.

Then split up the nitrogen containing fragment:

${H}_{2} {N}^{+} \rightarrow 2 {H}^{+} + {N}^{-}$

So the oxidation state of nitrogen in hydroxylamine is $- I$. Remember that this is a formalism, that does not have real physical significance.

If I do the same for ammonia, or methane, I get $N \left(- I I I\right)$, and $C \left(- I V\right)$. What about nitric oxide, $N {O}_{2}$?

See this link for related discussion: http://socratic.org/questions/oxygen-has-1-oxidation-state-in-peroxide-but-each-oxygen-forms-two-bonds-why