Worked examples — Variable oxidation states — reasons

This page is the practice ground for Variable oxidation states — reasons. The parent gave you the reasons; here we run those reasons through every kind of case you can be asked. Before we start, one promise: nothing below uses a symbol you have not met. If a new number or word appears, it is defined the moment it appears.

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The scenario matrix

Every exam question about variable OS is really one of these case classes. We will hit each cell.

#	Case class	What makes it tricky	Example that covers it
C1	Lowest possible OS ($+2$, lose only $ns$)	why it's so common	Ex 1 (Fe → +2)
C2	One step up ($+2\to +3$)	is the extra $d$-electron cheap?	Ex 2 (Fe → +3)
C3	Maximum OS = all valence e⁻	uses every $s+d$ electron	Ex 3 (Mn → +7)
C4	Degenerate / dead end (only ONE OS)	filled shell, no variation	Ex 4 (Zn)
C5	The "group-number ceiling" breaks	theoretical vs realised max	Ex 5 (Fe vs group 8)
C6	Sign of $\Delta H$ decides existence (Born–Haber balance)	must compute, not guess	Ex 6 (numeric energy check)
C7	Partner-atom limit (O/F allow high OS, I⁻ forbids it)	redox self-destruct	Ex 7 (MnI₇ does not exist)
C8	Real-world word problem	assign OS from a formula	Ex 8 (rust / KMnO₄ titration)
C9	Exam twist (find OS in a weird ion)	mixed anions, peroxide trap	Ex 9 (${\text{Cr}}_2{\text{O}}_7^{2-}$, ${\text{CrO}}_5$)

Read the matrix as a checklist. When you can do all nine, you can do the whole topic.

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The one tool we reuse: the OS-balance rule

Standard partner values we will lean on: oxygen is $-2$, fluorine is $-1$, hydrogen is $+1$ (with normal partners). We flag the exceptions (peroxide) when they bite in Ex 9.

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Worked examples

Ex 1 — Case C1: the lowest, most common state

Forecast: guess which two electrons leave first — the $4s$ pair or two of the $3d$? Write your guess before reading.

1. Identify the outermost electrons. Fe has two $4s$ electrons sitting outside the $3d^6$. Why this step? The electrons removed first are the ones highest in energy / furthest out. Once we know which they are, we know the cheapest ionisation.
2. Remove the two $4s$ electrons. This gives ${\text{Fe}}^{2+}$, configuration $[\text{Ar}]3d^6$. Why this step? Removing the $ns$ pair only needs the first two ionisation enthalpies ($I{E}_1+I{E}_2$), which are small for a d-metal (parent's "gradual rise"). See Ionisation Enthalpy of Transition Metals.
3. Check the compound $\text{FeO}$. Oxygen is $-2$. Rule: $\text{OS}(\text{Fe})+(-2)=0\Rightarrow \text{OS}(\text{Fe})=+2$. ✓

Verify: $+2+(-2)=0$, the neutral molecule. Units: a charge count, dimensionless. The state matches the electrons we removed. ✔ (See Stability of +2 and +3 states (Fe, Mn).)

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Ex 2 — Case C2: one cheap step up

Forecast: what does the $3d^6\to 3d^5$ change give the ion? (Hint: think "half-filled".)

1. Remove one more electron from ${\text{Fe}}^{2+}$. $[\text{Ar}]3d^6\to [\text{Ar}]3d^5$, i.e. ${\text{Fe}}^{3+}$. Why this step? Reaching $+3$ costs the third ionisation enthalpy $I{E}_3$. We must check it is still "repayable".
2. Notice the reward: $3d^5$ is half-filled. A half-filled $d$ shell is extra stable, so $I{E}_3$ for Fe is not as brutal as it would otherwise be. Why this step? The parent's rule says an OS exists when the ionisation cost is repaid by lattice/hydration energy. Landing on a stable $3d^5$ lowers the effective cost — that is the "payback".
3. Confirm from ${\text{Fe}}_2{\text{O}}_3$. Two Fe, three O$(-2)$: $2x+3(-2)=0\Rightarrow 2x=6\Rightarrow x=+3$. ✓

Verify: $2(+3)+3(-2)=6-6=0$. Neutral. ✔ Both $+2$ and $+3$ are real → iron varies its OS, exactly the topic's claim.

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Ex 3 — Case C3: maximum OS uses every valence electron

Forecast: how many electrons can Mn possibly give away? Count before reading.

1. Count valence electrons. $4s^2$ gives 2, $3d^5$ gives 5, total $=7$. Why this step? The theoretical maximum OS equals the number of $(ns+(n-1)d)$ electrons — every one that can be engaged in bonding.
2. Why all 7 are reachable for Mn. All five $3d$ electrons are unpaired (one per orbital), so each is accessible without first breaking a pair. Hence max OS $=+7$. Why this step? Unpaired, accessible electrons are the ones a strong partner like O can pull into bonding — this is why Mn is the peak of the series.
3. Confirm from ${\text{MnO}}_4^{-}$. Charge $q=-1$; four O at $-2$: $x+4(-2)=-1\Rightarrow x-8=-1\Rightarrow x=+7$. ✓

Verify: $+7+4(-2)=7-8=-1=q$. ✔ Matches the ion's charge. This is the single most-asked value in the chapter.

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Ex 4 — Case C4: the degenerate case (only one OS)

Forecast: how many electrons are easy to remove from Zn?

1. Remove the $4s^2$ pair. $\text{Zn}\to {\text{Zn}}^{2+}$, configuration $[\text{Ar}]3d^{10}$. Why this step? Same first move as every d-metal: the loose $ns$ electrons go first.
2. Try to go further — you can't cheaply. The next electron would come from the completely filled, very stable $3d^{10}$. That costs a huge $I{E}_3$ that no lattice/hydration energy can repay. Why this step? This is the degenerate/dead-end cell: a filled shell behaves like the s-block's locked basement — the jump is too big, so the OS is frozen at $+2$.
3. Conclusion. Max OS $=+2$, and there is no lower positive stable state, so Zn has a single OS.

Verify: In $\text{ZnO}$: $x+(-2)=0\Rightarrow x=+2$. ✔ Consistent, and no ${\text{ZnO}}_{?}$ with higher Zn OS exists — matching "$3d^{10}$ filled → no variation".

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Ex 5 — Case C5: the group-number ceiling breaks

Forecast: will the rule that worked for Sc–Mn keep working? Guess yes/no.

1. State the naive prediction. Group $8\Rightarrow$ theoretical max $+8$ (give away all 8 valence electrons of $3d^{6}4s^2$). Why this step? We must write the tempting wrong answer to see where it fails.
2. Add the physics that Sc–Mn didn't need. Past Mn the effective nuclear charge rises (more protons, similar shielding), pulling the $3d$ electrons in tightly and forcing pairing. See Effective Nuclear Charge & Shielding. Why this step? Tightly held, paired $d$-electrons have very high ionisation enthalpies — the cost of the top states can no longer be repaid.
3. Result. Fe reaches $+6$ only in special oxo-anions (ferrate ${\text{FeO}}_4^{2-}$) and commonly stops at $+3$. The realised max is below the group number.

Verify (ferrate ${\text{FeO}}_4^{2-}$): $x+4(-2)=-2\Rightarrow x=+6$. ✔ So $+6$ is the true ceiling, not $+8$ — the group number is only an upper bound. (This is the exact "mistake" the parent flagged.)

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Ex 6 — Case C6: sign of $\Delta H$ decides existence (numeric)

Forecast: just eyeballing, is the big negative $U$ enough to beat the ionisation cost? Guess sign.

1. Write the Born–Haber sum (parent's formula, all terms explicit): $\Delta H_{\text{form}}=\Delta H_{\text{atom}}+(I{E}_1+I{E}_2)+\Delta H_{\text{anion}}+U.$ Why this step? An OS is "stable/occurs" precisely when the whole cycle releases energy — this equation is the existence test. See Born–Haber Cycle & Lattice Energy.
2. Plug numbers. $\Delta H_{\text{form}}=150+(590+1145)+(-456)+(-2526).$ Why this step? Each term is a physical step; adding them tells us the net energy cost/release.
3. Add. $150+1735-456-2526=-1097{\text{kJ mol}}^{-1}$.

Verify: $\Delta H_{\text{form}}=-1097<0$ → the $+2$ state is stable: the lattice energy overpays the ionisation. Units consistent (${\text{kJ mol}}^{-1}$ throughout). ✔ This is the quantitative version of "does this OS exist?".

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Ex 7 — Case C7: the partner-atom limit (why some compounds can't exist)

Forecast: which atom can survive being bonded to a very electron-hungry $+7$ metal — O or I? Guess.

1. Confirm the OS both would demand. In ${\text{Mn}}_2{\text{O}}_7$: $2x+7(-2)=0\Rightarrow x=+7$. In a would-be ${\text{MnI}}_7$: $x+7(-1)=0\Rightarrow x=+7$. Same OS. Why this step? Shows the difference is not the number — it's the partner.
2. Ask who donates the electrons. A $+7$ metal is fiercely electron-greedy. Iodide ${\text{I}}^{-}$ is large and holds its electron loosely, so it is easily oxidised to ${\text{I}}_2$. Oxide ${\text{O}}^{2-}$ is small and holds tightly, so it is not. Why this step? If the anion is oxidised, the metal is reduced back to a lower OS — the high-OS iodide destroys itself.
3. The self-destruct reaction (schematic). Any "${\text{MnI}}_7$" would instantly become a lower-OS iodide plus ${\text{I}}_2$. So only low-OS iodides (e.g. ${\text{MnI}}_2$) are seen, while high OS lives in oxides/fluorides.

Verify: OS arithmetic checks: ${\text{Mn}}_2{\text{O}}_7\to +7$; ${\text{MnI}}_2\to x+2(-1)=0\Rightarrow x=+2$. ✔ High OS with O ✔, only low OS with I ✔ — exactly the parent's "O/F push high, I⁻ keeps it low". (See Acidic vs Basic character of Oxides across OS — ${\text{Mn}}_2{\text{O}}_7$ is acidic.)

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Ex 8 — Case C8: real-world word problem

Forecast: you already met both numbers in earlier examples — predict them, then confirm here in a "real" context.

1. Rust. ${\text{Fe}}_2{\text{O}}_3$: $2x+3(-2)=0\Rightarrow x=+3$. Why this step? Rust is iron that has been oxidised by air — its everyday appearance is literally the $+3$ state from Ex 2.
2. Permanganate — strip the spectator. In ${\text{KMnO}}_4$, K is $+1$, so the rest ${\text{MnO}}_4$ must carry $-1$: it is the permanganate ion of Ex 3. Why this step? Splitting off the counter-ion turns a real reagent into the ion whose OS we already solved.
3. Mn OS. $x+4(-2)=-1\Rightarrow x=+7$.

Verify: (a) $2(+3)+3(-2)=0$ ✔. (b) $(+1)+[(+7)+4(-2)]=1+(7-8)=1-1=0$ for neutral ${\text{KMnO}}_4$ ✔. The strong purple oxidiser owes its power to Mn being at its maximum $+7$ — real-world variable OS at work.

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Ex 9 — Case C9: the exam twist (peroxide trap)

Forecast: if you naively use O $=-2$ everywhere in ${\text{CrO}}_5$, you get an impossible OS. Guess what nonsense value that gives, then see the fix.

1. (a) Dichromate. $2x+7(-2)=-2\Rightarrow 2x-14=-2\Rightarrow 2x=12\Rightarrow x=+6$. Why this step? The clean case first — sets your expectation that Cr's high state is $+6$.
2. (b) The naive (wrong) attempt on ${\text{CrO}}_5$. If all 5 O were $-2$: $x+5(-2)=0\Rightarrow x=+10$. Impossible (Cr has only 6 valence electrons). Why this step? Seeing the absurd $+10$ tells you an assumption is broken — some oxygens are not ordinary oxide.
3. (b) The fix — peroxide oxygens. ${\text{CrO}}_5$ has one ordinary O ($-2$) and two peroxide units ${\text{O}}_2^{2-}$ (each $-2$, i.e. two O at $-1$). Total negative $=(-2)+2\times (-2)=-6$. Then $x+(-6)=0\Rightarrow x=+6$. Why this step? Reassigning the peroxide oxygens recovers a sensible $+6$, consistent with dichromate — the twist tests whether you spot the oxidation-state exception.

Verify: (a) $2(+6)+7(-2)=12-14=-2$ ✔ (matches the ion charge). (b) $(+6)+1(-2)+4(-1)=6-2-4=0$ ✔ (neutral, with two peroxide ${\text{O}}_2^{2-}$). Both give Cr $=+6$, its stable maximum.

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Recall Quick self-test across all nine cells

Assign the metal OS, then name the case class: $\text{FeO}$ ::: $+2$ (C1) ${\text{Fe}}_2{\text{O}}_3$ ::: $+3$ (C2) ${\text{MnO}}_4^{-}$ ::: $+7$ (C3) $\text{ZnO}$ ::: $+2$, only state (C4) ${\text{FeO}}_4^{2-}$ ::: $+6$, ceiling below group 8 (C5) ${\text{MnI}}_2$ vs ${\text{MnI}}_7$ ::: $+2$ exists, $+7$ iodide self-destructs (C7) ${\text{CrO}}_5$ ::: $+6$ using peroxide $-1$ oxygens (C9)

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Connections

• Variable oxidation states — reasons (parent)
• d-Block Overview
• Ionisation Enthalpy of Transition Metals
• Born–Haber Cycle & Lattice Energy
• Stability of +2 and +3 states (Fe, Mn)
• Acidic vs Basic character of Oxides across OS
• Effective Nuclear Charge & Shielding

Case map