Exercises — Oxidation number rules — assigning, change

$P_4$ is a free element — only phosphorus atoms, bonded to each other. When an atom bonds to a copy of itself, neither is greedier, so no pretend electron transfer happens. By rule 1, oxidation number $=0$. ✅ P is $0$.

$C{l}^{-}$ is a monatomic ion — one atom, real charge $-1$. There are no bonds to pretend about; the pretend charge equals the real charge. By rule 2, oxidation number $=$ its charge $=-1$. ✅ Cl is $-1$.

• In $H_{2}O$: oxygen is bonded to hydrogen, the less-greedy partner, so O takes the shared electrons → rule 4 default, $-2$.
• In $O_2^{2-}$: there is an O–O bond. Between two identical oxygens neither wins the tug-of-war, so that bond contributes $0$ to each. The result is the peroxide exception: $-1$. ✅ $-2$ in water, $-1$ in peroxide.

Let N $=x$. Hydrogen bonds to a nonmetal here, so H $=+1$ (rule 5). Molecule is neutral, $Q=0$. $x+3(+1)=0\Rightarrow x+3=0\Rightarrow x=-3$ ✅ N is $-3$.

Let Mn $=x$. Oxygen is default $-2$ (no O–O bond, no F). This is an ion, so $Q=-1$ (this is the trap — see below). $x+4(-2)=-1\Rightarrow x-8=-1\Rightarrow x=+7$ ✅ Mn is $+7$.

Let N $=x$, oxygen $-2$, ion charge $Q=-1$. $x+3(-2)=-1\Rightarrow x-6=-1\Rightarrow x=+5$ ✅ N is $+5$.

Fluorine is the most electronegative element in the universe — it beats even oxygen. So here fluorine grabs the electrons, F $=-1$ (rule 3, which outranks the oxygen rule). Molecule neutral, $Q=0$. Let O $=x$. $x+2(-1)=0\Rightarrow x=+2$ ✅ O is $+2$ — one of the rare cases where oxygen is positive, because it lost the tug-of-war.

Oxygen $-2$, neutral solid so $Q=0$. Let the average Fe $=x$. $3x+4(-2)=0\Rightarrow 3x=8\Rightarrow x=+\frac{8}{3}\approx +2.67$ ✅ Average Fe is $+\frac{8}{3}$. This is fractional because $F{e}_3{O}_4$ is really $FeO\cdot F{e}_2{O}_3$ — two of the irons are $+3$ and one is $+2$. The bookkeeping average $\frac{8}{3}$ is $\frac{2(+3)+1(+2)}{3}$.

Oxygen $-2$, ion charge $Q=-2$. Let average S $=x$. $2x+3(-2)=-2\Rightarrow 2x-6=-2\Rightarrow 2x=4\Rightarrow x=+2$ ✅ Average S is $+2$. (In reality the two sulfurs differ — one central, one terminal — but the sum rule only gives the average.)

• Na: starts as free element ($0$) → in $NaCl$ it is Group 1, so $+1$. Change $=+1-0=+1$ per atom. Increase ⇒ oxidised.
• Cl: starts as free element $C{l}_2$ ($0$) → in $NaCl$ it is $-1$. Change $=-1-0=-1$ per atom. Decrease ⇒ reduced. Electron bookkeeping: 2 Na atoms lose $2\times 1=2$ electrons total; 2 Cl atoms gain $2\times 1=2$ electrons total. They cancel — electrons are conserved. ✅ Na oxidised ($0\to +1$), Cl reduced ($0\to -1$); it is the reducing agent Na, oxidising agent $C{l}_2$.

From L2.2, Mn in $Mn{O}_4^{-}$ is $+7$. In $M{n}^{2+}$ it is $+2$ (monatomic ion, rule 2). Change $=+2-(+7)=-5$. The oxidation number drops by 5, so Mn is reduced, gaining $5$ electrons per atom. ✅ Mn: $+7\to +2$, gains $5e^{-}$, reduced — this is why $Mn{O}_4^{-}$ is a powerful oxidising agent (see Oxidising and reducing agents).

Sodium is Group 1, $+1$ (rule 6). Oxygen $-2$. Neutral, $Q=0$. Let average C $=x$. $2(+1)+2x+4(-2)=0\Rightarrow 2+2x-8=0\Rightarrow 2x=6\Rightarrow x=+3$ ✅ Average C is $+3$.

Oxide: oxygen $-2$, neutral, $Q=0$. Let M $=n$. $2n+3(-2)=0\Rightarrow 2n=6\Rightarrow n=+3$ So M is $+3$. Chloride: chlorine is $-1$ here (monatomic-like, it's the greedy partner). Neutral compound, and M must again be $+3$: $(+3)+x(-1)=0\Rightarrow x=3$ ✅ M is $+3$; the chloride is $MC{l}_3$.

This is the mastery test: not every oxygen is $-2$. Hydrogen $+1$ (×2). Of the five oxygens: 2 are peroxide ($-1$ each) and 3 are normal ($-2$ each). Neutral, $Q=0$. Let S $=x$. $2(+1)+x+2(-1)+3(-2)=0$ $2+x-2-6=0\Rightarrow x=+6$ ✅ S is $+6$. Contrast: if you had blindly used all five oxygens at $-2$, you'd get $2+x-10=0\Rightarrow x=+8$, which is impossible for sulfur (max $+6$). The absurd answer flags the missed peroxide.