Sulfur — allotropes (rhombic, monoclinic); SO₂, SO₃; H₂SO₄ (Contact process); oxoacids of S: Worked examples

This page is the drill floor. The parent note built the ideas; here we push each idea into every corner it can hide in. Before we begin, one promise: you will never meet a case on the exam that is not shown below.

The scenario matrix

The topic can only throw a finite number of question-shapes at you. Here they all are, in one table. Every worked example that follows is tagged with the cell it fills.

#	Case class	What makes it tricky	Filled by
A	Simple oxidation state ( $H_{x} S_{y} O_{z}$ , no S–S/O–O)	mechanical, but sign of O must be watched	Ex 1
B	Peroxo present (–O–O– link)	one O is $- 1$ , not $- 2$ → answer changes	Ex 2
C	S–S bond / non-equivalent S (thiosulfate, dithionous)	"average" hides two different sulfurs	Ex 3
D	Le Chatelier direction (which way does equilibrium shift?)	sign of $Δ H$ , mole change, catalyst trap	Ex 4
E	Yield / conversion numerics (percent of SO₂ → SO₃)	stoichiometry + a real number	Ex 5
F	Allotrope thermodynamics (which form at which $T$ ?)	$G = H - T S$ crossover, both signs of $(H, S)$	Ex 6
G	Redox titration (thiosulfate + iodine, mole ratio)	word problem, real lab number	Ex 7
H	Degenerate / trick oxidation state ( $S_{8}$ ring, elemental)	oxidation state of a lone element = 0	Ex 8
I	Exam twist (SO₂ oxidising vs reducing)	same molecule, opposite behaviour	Ex 9

Nine cells, nine examples. Let's go.

Ex 1 — Cell A: the plain oxidation-state machine

Worked example Find the oxidation state of sulfur in sulfuric acid,

$H_{2} S O_{4}$ . Forecast: guess the number now (you already saw it in the parent). Is it $+ 4$ , $+ 6$ , or $+ 2$ ?

Step 1 — Assign the "known" atoms. Set $H = + 1$ and $O = - 2$ . Why this step? Hydrogen bonded to a non-metal is always $+ 1$ ; oxygen (unless in a peroxide/superoxide) is always $- 2$ . These are the two fixed anchors — see Oxidation states and oxoacids.

Step 2 — Write the neutrality equation. The whole molecule is neutral, so the oxidation states sum to $0$ : $2 (+ 1) + S + 4 (- 2) = 0$ Why this step? Oxidation states are a bookkeeping of charge; a neutral molecule's books must balance to zero.

Step 3 — Solve. $2 + S - 8 = 0 \Rightarrow S = + 6$

Verify: $+ 6$ is sulfur's group-maximum ( $3 s^{2} 3 p^{4}$ → up to $6$ electrons "given up"), so it is chemically possible. Plug back: $2 - 8 + 6 = 0$ . ✓

Ex 2 — Cell B: a peroxo link changes the oxygen count

Worked example Find sulfur's oxidation state in

Caro's acid, $H_{2} S O_{5}$ (peroxomonosulfuric acid). It contains one –O–O– peroxo link. Forecast: The naive machine (treat all O as $- 2$ ) gives $S = + 8$ — impossible! Guess: what is the real value once we fix the peroxo oxygens?

Step 1 — Try the naive machine, watch it fail. $2 (+ 1) + S + 5 (- 2) = 0 \Rightarrow S = + 8$ Why this step? We deliberately break the rule to see the break. $+ 8$ exceeds sulfur's maximum $+ 6$ , so an assumption is wrong.

Step 2 — Fix the peroxo oxygens. In a peroxo link $-O-O-$ each oxygen is $- 1$ , not $- 2$ . Of the 5 oxygens, 2 form the peroxo bridge ( $- 1$ each) and 3 are ordinary ( $- 2$ each). Why this step? Two oxygens sharing a bond split their electrons evenly with each other, so neither can claim the full $- 2$ .

Step 3 — Re-balance. $2 (+ 1) + S + peroxo 2 (- 1) + normal 3 (- 2) = 0$ $2 + S - 2 - 6 = 0 \Rightarrow S = + 6$

Verify: $+ 6$ is now within range. This matches the parent table (Caro's acid, $S = + 6$ ). ✓

Ex 3 — Cell C: two sulfurs that are NOT the same

Worked example In thiosulfate

$S_{2} O_{3}^{2 -}$ , find (a) the average oxidation state of S, and (b) the individual states. Forecast: Will both sulfurs have the same number? Guess yes/no before reading.

Step 1 — Average, via the machine. The ion has charge $- 2$ : $2 S + 3 (- 2) = - 2 \Rightarrow 2 S = 4 \Rightarrow S_{avg} = + 2$ Why this step? For a fast bookkeeping answer, average works.

Step 2 — Ask WHY the two S differ. Thiosulfate = sulfate $S O_{4}^{2 -}$ with one terminal O replaced by an S. The central S keeps its sulfate-like high state; the substituted S is "sulfide-like" and low. Why this step? Structure, not arithmetic, tells us the truth — see Redox titrations - thiosulfate and iodine.

Step 3 — Read the individuals. Central S $\approx + 5$ , terminal (thio) S $\approx - 1$ . Check the average: $\frac{( + 5 ) + ( - 1 )}{2} = + 2$ . ✓

Verify: In $2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$ , only the low ( $- 1$ ) sulfur is oxidised (loses one electron → tetrathionate). Average S in $S_{4} O_{6}^{2 -}$ : $4 S + 6 (- 2) = - 2 \Rightarrow S = + 2.5$ — a fractional value, another fingerprint of non-equivalent sulfurs. ✓

Ex 4 — Cell D: which way does the equilibrium shift?

Worked example For

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}, Δ H = - 196 kJ/mol$ , predict the shift when we (i) raise $T$ , (ii) raise $P$ , (iii) add $V_{2} O_{5}$ . Forecast: Guess the three arrows (→ product, ← reactant, or no shift) before reading.

Step 1 — Temperature up. Forward reaction is exothermic (releases heat). Adding heat is like adding a product; the system relieves it by going backward → less $S O_{3}$ . Why this step? Le Chateliers Principle: a system opposes an imposed change. Raise $T$ ⇒ shift endothermic direction = reverse here.

Step 2 — Pressure up. Left side: $2 + 1 = 3$ moles of gas. Right side: $2$ moles. Higher $P$ pushes toward fewer moles → forward → more $S O_{3}$ . Why this step? Squeezing favours the side that takes less volume.

Step 3 — Add catalyst. $V_{2} O_{5}$ lowers the activation barrier equally for forward and reverse. No shift in position; equilibrium is merely reached faster — see Catalysis and V2O5.

Verify: Answers: (i) ← , (ii) → , (iii) no shift. This is exactly why the plant uses moderate $T$ (720 K, not high) plus a catalyst: temperature would hurt yield, so we buy speed from $V_{2} O_{5}$ instead. ✓

Figure — Sulfur — allotropes (rhombic, monoclinic); SO₂, SO₃; H₂SO₄ (Contact process); oxoacids of S

Ex 5 — Cell E: a conversion-percentage number

Ex 6 — Cell F: the allotrope crossover (both signs of ΔH, ΔS)

Worked example Rhombic sulfur is denser (lower enthalpy) but lower entropy; monoclinic is looser (higher

$H$ , higher $S$ ). The transition $S_{rhombic} ⇌ S_{monoclinic}$ has crossover at $T = 369 K$ . Using $Δ G = Δ H - T Δ S = 0$ at the transition, and taking $Δ H = + 401 J/mol$ for the rhombic→monoclinic change, estimate $Δ S$ . Forecast: Will $Δ S$ be positive or negative for rhombic → monoclinic? (Which one is looser?)

Step 1 — Write the crossover condition. At the transition temperature the two forms are equally stable, so $Δ G = 0$ : $0 = Δ H - T Δ S$ Why this step? Where free energies are equal, neither form is preferred — that is the transition.

Step 2 — Solve for $Δ S$ . $Δ S = \frac{Δ H}{T} = \frac{401}{369} \approx 1.09 J mol^{- 1} K^{- 1}$ Why this step? Rearranging the crossover equation isolates the unknown.

Step 3 — Check the sign logic. $Δ S > 0$ : monoclinic is looser/higher entropy, so rhombic→monoclinic gains disorder. ✓ Consistent with the parent's density argument (2.06 vs 1.98 g/cm³).

Verify: Below $369 K$ : $T Δ S < Δ H$ , so $Δ G > 0$ → forward (to monoclinic) is unfavourable → rhombic stable. Above $369 K$ : $T Δ S > Δ H$ → $Δ G < 0$ → monoclinic stable. Exactly the parent's rule. ✓ See Allotropy - carbon and phosphorus.

Ex 7 — Cell G: the iodine–thiosulfate titration (real lab)

Worked example

$25.0 mL$ of an iodine solution is exactly decolourised by $20.0 mL$ of $0.100 M$ sodium thiosulfate. Find the molarity of the iodine. Forecast: Because two thiosulfates react with one iodine, guess: is the iodine concentration bigger or smaller than $0.100 M$ ?

Step 1 — Write the balanced reaction. $2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}$ Ratio: $S_{2} O_{3}^{2 -} : I_{2} = 2 : 1$ — see Redox titrations - thiosulfate and iodine. Why this step? Mole ratio is the bridge between the two solutions.

Step 2 — Moles of thiosulfate used. $n (S_{2} O_{3}^{2 -}) = 0.100 mol/L \times 0.0200 L = 2.00 \times 1 0^{- 3} mol$ Why this step? Moles $=$ molarity $\times$ volume(L).

Step 3 — Moles of iodine, then its molarity. $n (I_{2}) = \frac{1}{2} n (S_{2} O_{3}^{2 -}) = 1.00 \times 1 0^{- 3} mol$ $[I_{2}] = \frac{1.00 \times 1 0 ^{- 3} mol}{0.0250 L} = 0.0400 M$

Verify: Iodine ( $0.0400 M$ ) is less concentrated than thiosulfate ( $0.100 M$ ) — sensible, since it takes 2 thiosulfates per $I_{2}$ and we used a similar volume. ✓ Units: mol/L throughout. ✓

Ex 8 — Cell H: the degenerate case (elemental sulfur)

Ex 9 — Cell I: the exam twist (same molecule, opposite roles)

Worked example

$S O_{2}$ meets two different partners: (a) $S O_{2} + 2 H_{2} S \to 3 S + 2 H_{2} O$ (b) $S O_{2} + C l_{2} \to S O_{2} C l_{2}$ In which does $S O_{2}$ act as an oxidising agent, and in which as a reducing agent? Forecast: In $S O_{2}$ , S starts at $+ 4$ . Guess: which reaction pulls S down and which pushes it up?

Step 1 — Track S in reaction (a). In $S O_{2}$ , S $= + 4$ ; in product $S$ (element), S $= 0$ . Sulfur went down ( $+ 4 \to 0$ ): it was reduced. Why this step? The species that is reduced is the oxidising agent — it took electrons from $H_{2} S$ . ⇒ Here $S O_{2}$ is the oxidising agent.

Step 2 — Track S in reaction (b). In $S O_{2}$ , S $= + 4$ ; in $S O_{2} C l_{2}$ , compute: $S + 2 (- 2) + 2 (- 1) = 0 \Rightarrow S = + 6$ . Sulfur went up ( $+ 4 \to + 6$ ): it was oxidised. Why this step? The species oxidised is the reducing agent — it gave electrons to $C l_{2}$ . ⇒ Here $S O_{2}$ is the reducing agent.

Verify: This is exactly why the parent stresses that $+ 4$ is intermediate (between $- 2$ and $+ 6$ ): $S O_{2}$ can go both ways. Reaction (a) drives it down, (b) drives it up. ✓

Recall Quick self-test across the matrix

S in $H_{2} S_{2} O_{8}$ (Marshall's, one –O–O– bridge)? ::: $+ 6$ (correct the two peroxo oxygens to $- 1$ ) Raising temperature in Stage 2 shifts equilibrium which way? ::: Backward (exothermic) → less $S O_{3}$ Oxidation state of S in free $S_{8}$ ? ::: $0$ $2 S_{2} O_{3}^{2 -} + I_{2} \to ?$ ::: $S_{4} O_{6}^{2 -} + 2 I^{-}$ Above 369 K, which allotrope is stable and why? ::: Monoclinic — higher entropy, so $- T Δ S$ wins

Return to the parent topic · related: Oxidation states and oxoacids, Le Chateliers Principle, Redox titrations - thiosulfate and iodine.