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When given a compound like (NH4)2MoO4 and asked to find the oxidation number of Mo, how do I know which oxidation number to sub in for Nitrogen? On the periodic table, Nitrogen has oxidation numbers of $-3, -2, -1, +1, +2, +3, +4, +5$.

I know the process will look like: (2xNitrogen's oxidation number) + (8x1) + x + (4x-2) = 0

Is there a process to figuring out which number to sub in for Nitrogen's oxidation number?

Maurice
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user114992
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1 Answers1

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This problem can be solved by dealing with separate ions. In water, $\ce{(NH4)2MoO4}$ is dissociated in $\ce{2 NH4^+ + MoO4^{2-}}$. Here each ion can be analyzed separately, to get the unknown oxidation numbers.

In the ammonium ($\ce{NH4^+}$} cation, Nitrogen is at oxidation number $x$, and H at $+1$. So that $x + 4ยท(+1) = +1$, so that $x = -3$. In the anion $\ce{MoO4^{2-}}$, the same calculation yields $+6$ for oxidation number of molybdenum.

Maurice
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