Is the nitrogen oxidation number in $N H_{4}^{+}$ $NH_4^+$ $- I I I$ $-III$ ? Why?

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Is the nitrogen oxidation number in $N H_{4}^{+}$ $NH_4^+$ $- I I I$ $-III$ ? Why?

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Answer 1 · 2016-06-13T20:00:55

The answer is indeed $- I I I$ $-III$ . How?

Explanation:

Oxidation number is formally the charge left on the central atom when all the bonds are broken, and the electrons distributed to the most electronegative atom.

We start with ammonium ion, $N H_{4}^{+}$ $NH_4^+$ , i.e.

$N H_{4}^{+} \to N^{3 -} + 4 \times H^{+}$ $NH_4^(+) rarr N^(3-) + 4xxH^+$

Both charge and mass have been conserved, as they must be. The formal charge on $N$ $N$ (and this is very much a formalism, a convention with marginal significance) is $- 3$ $-3$ , hence an oxidation state of $- I I I$ $-III$ is specified.

Using the same process,can you demonstrate a $- I I$ $-II$ oxidation state for nitrogen in hydrazine, $H_{2} N - N H_{2}$ $H_2N-NH_2$ ?

And what about the oxidation state of sulfur in the counterion, $S O_{4}^{2 -} ?$ $SO_4^(2-)?$

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Is the nitrogen oxidation number in $N H_{4}^{+}$ $NH_4^+$ $- I I I$ $-III$ ? Why?

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Is the nitrogen oxidation number in $N H_{4}^{+}$ $NH_4^+$ $- I I I$ $-III$ ? Why?