Worked examples — Group 16 (Oxygen family) — allotropes of O (O₂, O₃); ozone chemistry, ozone layer

Before we start, two words we will keep using — earn them once here:

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The scenario matrix

Every ozone exam question falls into one of these cells. The worked examples below are each tagged with the cell(s) they cover.

#	Case class	What makes it tricky	Example
C1	Direct 1-step stoichiometry	just read coefficients	Ex 1
C2	Two-step (linked) titration	ratio chains multiply	Ex 2
C3	Bond-order / bond-length comparison	lower B.O. ⇒ longer bond	Ex 3
C4	Magnetism from MOT	Lewis lies; count unpaired e⁻	Ex 4
C5	Zero / degenerate input	"no reaction" answer is valid	Ex 5
C6	Limiting / catalytic behaviour	Cl regenerated → huge factor	Ex 6
C7	Real-world word problem	extract numbers from prose	Ex 7
C8	Exam twist (percentage in a mix)	ozonised O₂, find % O₃	Ex 8

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Example 1 — Direct stoichiometry (cell C1)

Forecast: Guess before reading — is the answer bigger or smaller than 0.50? (The "4" in front of O₃ hints bigger.)

1. Read the ratio. The equation says $1PbS:4O_3$. Why this step? The coefficients are the stoichiometric ratio — nothing else matters for a direct problem.
2. Multiply. $\text{mol }{O}_3=4\times \text{mol }PbS=4\times 0.50=2.0$ mol. Why this step? Ratio 4:1 means four times as much ozone as $PbS$.

Answer: $2.0$ mol O₃.

Verify: $2.0$ mol O₃ oxidises $2.0/4=0.50$ mol $PbS$. ✔ Matches the given amount. Bigger than 0.50, as forecast.

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Example 2 — Two-step linked titration (cell C2)

This is the classic ozone-estimation problem. The trick is a chain of ratios.

Forecast: The thiosulphate number is the biggest. Will the O₃ answer be $0.020$, half of it, or double?

1. Thiosulphate → I₂. The second equation gives $I_2:S_2{O}_3^{2-}=1:2$. So $\text{mol }{I}_2=\frac{\text{mol }{S}_2{O}_3^{2-}}{2}=\frac{0.020}{2}=\mathrm{0.010.}$ Why this step? Each I₂ molecule consumes two thiosulphate ions, so I₂ is half the thiosulphate.
2. I₂ → O₃. The first equation gives $O_3:I_2=1:1$. So $\text{mol }{O}_3=\text{mol }{I}_2=\mathrm{0.010.}$ Why this step? One ozone molecule liberates exactly one I₂, a clean 1:1 link.

Answer: $0.010$ mol O₃.

Verify: Chain it forward: $0.010$ mol O₃ → $0.010$ mol I₂ → $0.020$ mol thiosulphate. ✔ Lands exactly on the given $0.020$.

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Example 3 — Bond order & bond length (cell C3, geometric)

Forecast: Which do you think is a stronger, tighter bond — the double bond of O₂ or the "one-and-a-half" bond of ozone?

1. Get the bond orders. O₂ has bond order $2$ (from MOT, see Molecular Orbital Theory). O₃ has bond order $1.5$ (resonance average, see Resonance and Delocalisation). Why this step? Bond length is controlled by bond order — we need those numbers first.
2. Apply the rule. Higher bond order ⇒ more shared electron "glue" ⇒ atoms pulled closer ⇒ shorter, stronger bond. Why this step? Bond order is proportional to bond strength, and strength is inversely related to length.
3. Order them. $\text{B.O.}(O_2)=2>\text{B.O.}(O_3)=1.5$, so O₂'s bond is shorter. $\underset{O_2,\text{B.O.}=2}{\underbrace{121\text{ pm}}}<\underset{O_3,\text{B.O.}=1.5}{\underbrace{128\text{ pm}}}<\underset{\text{a pure O–O single bond}}{\underbrace{148\text{ pm}}}$

Answer: O₂ (121 pm) < O₃ (128 pm).

Verify: Ozone's 128 pm sits between a pure single bond (148 pm) and a pure double bond (121 pm) — exactly what a bond order between 1 and 2 should give. ✔ Sanity check passes.

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Example 4 — Magnetism from MOT (cell C4)

Forecast: Lewis says all electrons are paired. Do you trust it, or does the magnet tell a different story?

1. Fill the MOT diagram (16 e⁻). ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$ Why this step? Only MOT has antibonding orbitals; Lewis structures cannot show them.
2. Look at the top orbitals. The last two electrons occupy the two degenerate ${\pi }_{2p}^{\ast }$ orbitals — one each, unpaired (Hund's rule: spread out before pairing up). Why this step? Two orbitals of equal energy are filled singly first, leaving 2 unpaired electrons.
3. Conclude. Unpaired electrons ⇒ the molecule is attracted by a magnet ⇒ paramagnetic, with $2$ unpaired electrons.

Answer: 2 unpaired electrons; O₂ is paramagnetic.

Verify (bond order too): $N_b=10$, $N_a=6$, so $\text{B.O.}=(10-6)/2=2$. ✔ A double bond, consistent with the $O=O$ Lewis picture — Lewis got the bond order right but the magnetism wrong.

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Example 5 — Zero / degenerate input (cell C5)

Forecast: Trick question. What can ozone oxidise if there is nothing oxidisable present?

1. Check for a reducing partner. Ozone works by releasing nascent oxygen $[O]$ to oxidise something. Pure water has nothing in a lower oxidation state to give up electrons here. Why this step? An oxidising agent needs a reducing agent to react with — this is the "zero-input" edge case (see Oxidation and Reducing Agents).
2. Result. No iodide ⇒ no iodine can form. mol $I_2=0$. Why this step? The KI reaction $O_3+2KI+H_{2}O\to I_2+\dots$ cannot run without $KI$.

Answer: $0$ mol I₂; no meaningful oxidation of water occurs under these conditions.

Verify: Plug into the titration chain of Example 2 with mol I₂ = 0 ⇒ mol thiosulphate needed $=2\times 0=0$. ✔ A "nothing happens" answer is a valid answer — always check whether a reducing partner exists.

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Example 6 — Limiting / catalytic behaviour (cell C6)

Forecast: Guess the order of magnitude — tens? thousands? Why is Cl so dangerous?

1. Count O₃ per cycle. Each full turn of the cycle consumes exactly one O₃ (in the first step). Why this step? The second step consumes a free O atom, not O₃ — so O₃ loss is 1 per cycle.
2. Multiply by the number of cycles. $8000\text{ cycles}\times 1O_3=8000$ molecules of O₃ destroyed. Why this step? Cl is regenerated every cycle (it appears on both sides), so it is a catalyst — not used up.
3. Add the two steps to get the net reaction (Cl and ClO cancel): $O_3+O\to 2O_2.$ Why this step? Adding the steps, $\dot{Cl}$ and $ClO\cdot$ appear once on each side and cancel, leaving only the net destruction.

Answer: One Cl destroys $8000$ O₃ molecules; net reaction $O_3+O\to 2O_2$.

Verify: Add left sides ($Cl+O_3+ClO+O$) and right sides ($ClO+O_2+Cl+O_2$); cancel $Cl$ and $ClO$ ⇒ $O_3+O\to 2O_2$. ✔ Balanced: 3 O atoms on left of first species + 1 = 4 O on each side. This tiny-amount-does-huge-damage logic is why the Montreal Protocol banned CFCs (see Environmental Chemistry — Air Pollution).

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Example 7 — Real-world word problem (cell C7)

Forecast: Before calculating — is ground-level ozone the "good sunscreen" ozone or the "bad pollutant" ozone?

1. Thiosulphate → I₂. $\text{mol }{I}_2=0.0060/2=0.0030$. Why this step? Same 1:2 link as Example 2 — each I₂ needs 2 thiosulphate.
2. I₂ → O₃. $\text{mol }{O}_3=\text{mol }{I}_2=0.0030$. Why this step? 1:1 ratio between O₃ and the I₂ it liberates.
3. Interpret (b). This is tropospheric (ground-level) ozone — a toxic pollutant/oxidant in smog, not the protective stratospheric layer. Why this step? Location decides the verdict: high up = shield, low down = harm.

Answer: (a) $0.0030$ mol O₃. (b) Harmful — it is a ground-level pollutant.

Verify: Forward chain: $0.0030$ mol O₃ → $0.0030$ mol I₂ → $0.0060$ mol thiosulphate. ✔ Matches the measured $0.0060$.

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Example 8 — Exam twist: percentage of O₃ in an ozonised stream (cell C8)

Forecast: Small number of moles of O₃ in a bigger total — will the percentage be closer to 10% or 50%?

1. Set up the percentage. Mole % of ozone $=\frac{\text{mol }{O}_3}{\text{total mol}}\times 100$. Why this step? "Mole percentage" means the fraction of total moles that is O₃, expressed out of 100.
2. Plug in. $\frac{0.010}{0.100}\times 100=10.0\%$. Why this step? Direct substitution of the two given amounts.

Answer: $10.0\%$ ozone by moles.

Verify: If $10.0\%$ of $0.100$ mol is O₃, then $0.100\times 0.10=0.010$ mol. ✔ Recovers the given O₃ amount. Closer to 10%, as forecast.

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Recap of every cell

Recall Did each case class get covered?

• C1 direct stoichiometry ::: Ex 1 (PbS + O₃)
• C2 linked two-step titration ::: Ex 2 (KI → I₂ → thiosulphate)
• C3 bond order/length ::: Ex 3 (O₂ vs O₃ lengths)
• C4 magnetism from MOT ::: Ex 4 (O₂ paramagnetic, 2 unpaired e⁻)
• C5 zero/degenerate input ::: Ex 5 (no reducing partner → 0 mol I₂)
• C6 catalytic/limiting ::: Ex 6 (one Cl destroys 8000 O₃)
• C7 real-world word problem ::: Ex 7 (smog ozone, harmful)
• C8 exam-twist percentage ::: Ex 8 (10% O₃ in mixture)