= Oxidation state sum rule
{title2=$\sum_i\mathrm{OS}_i=z$}
Sum <oxidation states> over all <chemical atoms> of the specified <molecule>, <ion> or <formula unit>, including multiplicities:
$$
\sum_i\mathrm{OS}_i=z,\qquad q=ze.
$$
Here $q$ is the net <electric charge> and $e$ the positive <elementary charge>. For a neutral entity the sum is zero; for an <ion> it equals its signed <electric charge> number $z$. The reason is simple: the allocation assigns every <valence electron> exactly once, so the sum is the neutral <Electron> budget minus the actual <Electron> count.
For <water>, $2(+1)+(-2)=0$; for <hydrogen peroxide>, $2(+1)+2(-1)=0$. In <permanganate>, $\mathrm{MnO}_4^-$, assigning each <oxygen> $-2$ gives $x+4(-2)=-1$, so <manganese> has <oxidation state> $+7$. This is not a free $\mathrm{Mn}^{7+}$ <ion>: the entire <permanganate> <anion> has net <electric charge> $-e$. The sum rule supplies a constraint; without additional assignments it cannot determine every individual <oxidation state>.
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