Worked examples — Variation of oxidation state across the table

Everything below rests on one master rule you already met in the parent:

Why this tool and not another? Oxidation state is defined as a bookkeeping charge — a made-up number that pretends every bond is fully ionic (see Electronegativity for who takes the electrons). Bookkeeping means the books must balance: total made-up charge = total real charge. That single conservation law solves every example on this page; all we ever change is which numbers are "anchors" (fixed by rule) and which is the unknown $x$.

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The scenario matrix

Every oxidation-state problem falls into one of these cells. Our examples (Ex a–i) are tagged so you can see full coverage.

#	Cell (scenario class)	What makes it tricky	Example
1	Neutral molecule, standard anchors	plain $\sum =0$	Ex a
2	Polyatomic ion, $Q\ne 0$	sum equals the charge, not zero	Ex b
3	Fractional / average OS	atoms in different environments	Ex c
4	Anchor breaks: peroxide O = $-1$	O is not $-2$	Ex d
5	Anchor breaks: O positive (${\text{OF}}_2$)	F beats O in Electronegativity	Ex e
6	Max positive limit reached	check against $+n_v$	Ex f
7	Most-negative floor reached	check against $-(8-n_v)$	Ex g
8	Inert-pair / heavy p-block	lower state preferred	Ex h
9	Transition metal, real-world word problem	many close states	Ex i
—	Degenerate input: free element	OS $=0$ by definition	covered in Ex a note

Keep the number line of possible states in view as we work:

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Example a — Neutral molecule, standard anchors (Cell 1)

1. Write the balance. Neutral molecule $\Rightarrow Q=0$: $x+2(-2)=0.$ Why this step? We use the master rule with $Q=0$; oxygen is the anchor $-2$ (parent rule).
2. Solve for $x$. $x-4=0\Rightarrow x=+4.$ Why this step? Only carbon is unknown, so one equation gives it.

Verify: $(+4)+2(-2)=+4-4=0=Q.✓$ And $+4$ equals carbon's group valence $n_v=4$ — its maximum, which makes sense: oxygen took everything.

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Example b — Polyatomic ion, non-zero charge (Cell 2)

1. Balance to the ion charge. $x+4(-2)=-1.$ Why this step? This is an ion with charge $-1$, so the right-hand side is $-1$, not $0$. This is the one place students slip.
2. Solve. $x-8=-1\Rightarrow x=+7.$

Verify: $(+7)+4(-2)=+7-8=-1=Q.✓$ Manganese's removable electrons are $3d^{5}4s^2=7$, so $+7$ is its ceiling (see Transition Metal Chemistry). Perfectly consistent.

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Example c — Fractional / average oxidation state (Cell 3)

1. Balance. $3x+4(-2)=0.$ Why this step? Neutral solid, $Q=0$; oxygen anchor $-2$.
2. Solve. $3x=8\Rightarrow x=+\frac{8}{3}\approx +\mathrm{2.67.}$ Why this step? Three iron atoms are not all identical — magnetite is really ${\text{FeO⋅Fe}}_2{\text{O}}_3$, i.e. one ${\text{Fe}}^{2+}$ and two ${\text{Fe}}^{3+}$. The formula gives their average.

Verify: $3(\frac{8}{3})+4(-2)=8-8=0=Q.✓$

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Example d — Broken anchor: peroxide oxygen is $-1$ (Cell 4)

1. Try the naive anchor (to see it break). With O $=-2$: $2(+1)+2(-2)=2-4=-2\ne 0$. Contradiction — the books don't balance. Why this step? Demonstrating why the shortcut fails is the whole lesson.
2. Fix: two oxygens share an O–O bond. In $\text{H–O–O–H}$ the central bond is between two identical atoms, so neither wins the electrons — that bond contributes $0$ to each. Each O only fully grabs its H.
3. Balance with O $=x$. $2(+1)+2x=0\Rightarrow 2x=-2\Rightarrow x=-1.$ Why this step? H is still the anchor $+1$ (it's less electronegative than O); solving gives the peroxide value.

Verify: $2(+1)+2(-1)=2-2=0=Q.✓$ Oxygen sits at $-1$, exactly halfway between its usual $-2$ and elemental $0$ — the signature of a peroxide.

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Example e — Broken anchor: oxygen positive in ${\text{OF}}_2$ (Cell 5)

1. Fix the anchors by electronegativity. F is the most electronegative element of all, so F $=-1$ always (parent rule). Oxygen must therefore give electrons to F. Why this step? The negative number always goes to the more electronegative atom (Electronegativity); here that is F, not O.
2. Balance. $x+2(-1)=0\Rightarrow x=+2.$

Verify: $(+2)+2(-1)=+2-2=0=Q.✓$ Oxygen is positive — the rare exception the parent warned about. F beats O, so O plays the "generous" role for once.

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Example f — Confirming the maximum-positive limit (Cell 6)

1. Balance. $(+1)+x+4(-2)=0.$ Why this step? Neutral molecule; H $=+1$, O $=-2$ anchors.
2. Solve. $1+x-8=0\Rightarrow x=+7.$
3. Compare to the ceiling $+n_v$. Chlorine is group 17, so $n_v=7$, giving max $=+7$. We hit it.

Verify: $(+1)+(+7)+4(-2)=1+7-8=0=Q.✓$ And $+7=+n_v$, so chlorine has handed over every valence electron to oxygen. Related driver: high Ionisation Energy means this only happens against a very greedy partner.

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Example g — Confirming the most-negative floor (Cell 7)

1. Balance, but flip who is negative. In ${\text{NH}}_3$, N is more electronegative than H, so H is $+1$ and N goes negative: $x+3(+1)=0.$ Why this step? Anchor H at $+1$ (its partner N wins the electrons).
2. Solve. $x+3=0\Rightarrow x=-3.$
3. Compare to the floor $-(8-n_v)$. Nitrogen has $n_v=5$, so min $=-(8-5)=-3$. We hit it.

Verify: $(-3)+3(+1)=-3+3=0=Q.✓$ Nitrogen filled its octet by grabbing 3 electrons — the most it needs (see Electron Affinity).

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Example h — Inert-pair effect, heavy p-block (Cell 8)

1. Balance each. Chlorine is $-1$ (more electronegative than Pb): ${\text{PbCl}}_2:x+2(-1)=0\Rightarrow x=+2.$ ${\text{PbCl}}_4:x+4(-1)=0\Rightarrow x=+4.$ Why this step? Same master rule, Cl anchored at $-1$.
2. Decide stability by the inert-pair effect. Down group 14 the $6s^2$ pair is held tightly (poor $d/f$ shielding) and resists bonding. So lead prefers to keep that pair and use only its two $6p$ electrons $\Rightarrow +2$ is favoured over $+4$. Why this step? This is the Inert Pair Effect the parent note derives; it explains why ${\text{PbCl}}_4$ decomposes.

Verify: both balances give $0=Q$ ($+2-2=0$ and $+4-4=0$).$✓$ Conclusion: ${\text{PbCl}}_2$ (Pb $=+2$) is the stable one; ${\text{PbCl}}_4$ (Pb $=+4$) is an unstable oxidiser — exactly the group-max $-2$ preference.

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Example i — Transition metal, real-world word problem (Cell 9)

1. Before — balance the ion. $2x+7(-2)=-2.$ Why this step? Dichromate has charge $-2$, so $Q=-2$. $2x-14=-2\Rightarrow 2x=12\Rightarrow x=+6.$
2. After — read the product. ${\text{Cr}}^{3+}$ is a bare ion, so its OS = its charge $=+3$. Why this step? For a monatomic ion, OS equals the charge (parent rule).
3. Electrons gained per Cr. Oxidation state fell $+6\to +3$, a drop of $3$. A drop in OS means electrons were gained (Redox Reactions): $\Delta =(+6)-(+3)=3\text{ electrons gained per Cr}.$

Verify: before-balance $2(+6)+7(-2)=12-14=-2=Q✓$; after: monatomic ${\text{Cr}}^{3+}$ gives $+3✓$; electron count $6-3=3✓$. Because $3d$ and $4s$ lie close in energy, chromium can rest at several states ($+6,+3,+2$) — the many-states behaviour of Transition Metal Chemistry, rooted in Electronic Configuration.

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Recall Quick self-test (reveal after guessing)

S in ${\text{SO}}_3$? ::: $+6$ (from $x+3(-2)=0$) O in ${\text{Na}}_2{\text{O}}_2$ (a peroxide)? ::: $-1$ Cr in ${\text{Cr}}^{3+}$? ::: $+3$ (monatomic ion = charge) Which is more stable, $\text{TlCl}$ (Tl $+1$) or ${\text{TlCl}}_3$ (Tl $+3$)? ::: $\text{TlCl}$, by the inert-pair effect

Connections

• Electronegativity — decides which atom takes the negative number (Ex d, e).
• Ionisation Energy — why high positive states are hard (Ex f).
• Electron Affinity — why negative floors form (Ex g).
• Electronic Configuration — source of $n_v$ and $d$–$s$ spacing (Ex i).
• Inert Pair Effect — heavy p-block preference (Ex h).
• Transition Metal Chemistry — many close states (Ex b, i).
• Redox Reactions — OS change = electron transfer (Ex i).