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

Before we begin, three tiny reminders that every problem below leans on:

Level 1 — Recognition

L1.1 State whether liquid $O_{2}$ is attracted to a magnet or not, and give the one-word name for that property.

L1.2 Name the allotrope of oxygen that is (a) colourless and odourless, (b) pale blue with a pungent smell.

L1.3 Write the balanced equation for the lab preparation of ozone and state the sign of $Δ H$ .

Recall Solutions L1

L1.1 Liquid $O_{2}$ is attracted to a magnet. The property is paramagnetism. Why: Molecular Orbital Theory places the last two electrons unpaired, one each in $π_{2 p_{x}}^{*}$ and $π_{2 p_{y}}^{*}$ . Unpaired electrons act like tiny magnets, so the molecule responds to a magnetic field.

L1.2 (a) $O_{2}$ (dioxygen). (b) $O_{3}$ (ozone).

L1.3 $3 O_{2} silent electric discharge 2 O_{3}, Δ H = + 142 kJ mol^{- 1} (positive, endothermic) .$ Reading $Δ H$ carefully: this $+ 142 kJ$ is per mole of the reaction as written — i.e. per $2$ mol of $O_{3}$ formed. So per mole of $O_{3}$ the value is $142/2 = 71 kJ mol^{- 1}$ . Both are positive; just be clear which you are quoting. Why positive: you must push energy in (the electric discharge) to force stable $O_{2}$ into higher-energy $O_{3}$ . Energy stored = why ozone is later so reactive.

Level 2 — Application

L2.1 Compute the bond order of $O_{2}$ from its MOT configuration and state how many unpaired electrons it has.

L2.2 Ozone's O–O bond length is 128 pm. $O_{2}$ is 121 pm and a plain O–O single bond is 148 pm. Using bond order, explain why 128 pm sits where it does.

L2.3 Balance: ozone oxidising lead sulphide, $P b S + O_{3} \to P b S O_{4} + O_{2}$ .

Recall Solutions L2

L2.1 Filling the 16 electrons of $O_{2}$ : $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 1}$ Count: bonding electrons $N_{b} = 2 + 2 + 2 + 2 + 2 = 10$ ; antibonding $N_{a} = 2 + 2 + 1 + 1 = 6$ . $B.O. = \frac{N _{b} - N _{a}}{2} = \frac{10 - 6}{2} = 2.$ Two singly-filled $π^{*}$ orbitals ⇒ 2 unpaired electrons ⇒ paramagnetic.

The figure below is the ladder you just filled. Read it from bottom to top: the green rungs are the bonding orbitals (count 10 electrons on them, that is $N_{b}$ ), the coral rungs are the antibonding orbitals (count 6, that is $N_{a}$ ). Look especially at the top coral rung labelled $π_{2 p_{x}}^{*} π_{2 p_{y}}^{*}$ : it holds two arrows both pointing up in separate boxes — those are the two unpaired electrons that make $O_{2}$ stick to a magnet.

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

L2.2 Bond order tells us bond strength, and stronger bonds pull atoms closer (shorter). Ranking bond orders: $O_{2}$ has B.O. $= 2$ (shortest, 121 pm); a single bond has B.O. $= 1$ (longest, 148 pm). Ozone's B.O. $= 1.5$ sits between these, so its length (128 pm) must sit between too. It does. ✔

L2.3 Sulphur goes from $- 2$ (in $P b S$ ) to $+ 6$ (in $P b S O_{4}$ ): a jump of 8 electrons lost. Each $O_{3} \to O_{2} + [O]$ supplies one $[O]$ that accepts 2 electrons, so we need $8/2 = 4$ molecules of $O_{3}$ : $P b S + 4 O_{3} \to P b S O_{4} + 4 O_{2} .$ Check atoms: left O $= 12$ , right O $= 4 + 8 = 12$ . ✔

Level 3 — Analysis

L3.1 Ozone is drawn as $O = O - O \leftrightarrow O - O = O$ . A student insists one bond is 121 pm and the other 148 pm. Explain in words why the real molecule has two equal bonds, and give their length relative to those two values.

L3.2 Ozone's bond angle is $\approx 11 7^{\circ}$ , not the $12 0^{\circ}$ of a perfect $s p^{2}$ shape. Explain why it is squeezed below $12 0^{\circ}$ .

L3.3 Why is ozone a powerful oxidising agent while ordinary $O_{2}$ is a mild one? Answer using energy, not just "it reacts."

Recall Solutions L3

L3.1 The two Lewis drawings are not two real states the molecule flickers between — they are two incomplete sketches of one real thing, the resonance hybrid. The extra $π$ electrons are delocalised (smeared) equally over both O–O regions. So both bonds are identical, each with bond order $\frac{2 + 1}{2} = 1.5$ , and each measures 128 pm — a value between 121 pm and 148 pm, never one of them.

In the figure, notice that the pale lavender cloud lies symmetrically over both O–O bonds — it is not fatter on one side. That symmetry is the whole point: the extra $π$ electrons belong equally to both bonds, so both bonds are labelled the same 128 pm and the bent angle is marked $\approx 11 7^{\circ}$ . There is no "short bond" and "long bond" to find.

L3.2 The central O carries 2 bonding pairs + 1 lone pair ( $s p^{2}$ -like, 3 electron domains → base angle $12 0^{\circ}$ ). A lone pair spreads out more than a bonding pair, so it pushes the two bonding pairs together, squeezing the angle from $12 0^{\circ}$ down to $\approx 11 7^{\circ}$ .

L3.3 Formation of ozone is endothermic ( $Δ H = + 142 kJ mol^{- 1}$ ): it is a loaded spring storing energy. Releasing that energy is favourable, so ozone readily decomposes $O_{3} \to O_{2} + [O]$ . The freed nascent oxygen $[O]$ is a lone, electron-hungry atom that grabs electrons off other species — that is oxidising. $O_{2}$ has no such stored energy to shed and no easy $[O]$ to release, so it oxidises only sluggishly.

Level 4 — Synthesis

L4.1 A $25.0 cm^{3}$ ozonised gas stream is bubbled through excess acidic $K I$ . The liberated iodine is titrated and needs 0.020 mol of $N a_{2} S_{2} O_{3}$ . Find the moles of $O_{3}$ in the stream. Reactions: $O_{3} + 2 K I + H_{2} O \to I_{2} + 2 K O H + O_{2}$ and $I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}$ .

L4.2 Suppose the same experiment used 0.050 mol thiosulphate. How many moles of $O_{3}$ now, and how many litres of $O_{3}$ at STP (1 mol gas $= 22.4 L$ ) does that represent?

Recall Solutions L4

L4.1 Walk the chain of ratios, one step at a time.

Step 1 (WHAT): thiosulphate → iodine. $I_{2} : S_{2} O_{3}^{2 -} = 1 : 2$ , so $mol I_{2} = \frac{0.020}{2} = 0.010$ .
Step 2 (WHAT): iodine → ozone. From the first equation $O_{3} : I_{2} = 1 : 1$ , so $mol O_{3} = mol I_{2} = 0.010$ .
Answer: $0.010 mol O_{3}$ . Why divide first, then keep equal: two thiosulphates mop up each $I_{2}$ (so halve), but each ozone made exactly one $I_{2}$ (so no change).

L4.2

$mol I_{2} = \frac{0.050}{2} = 0.025$ .
$mol O_{3} = 0.025$ (1:1 with $I_{2}$ ).
Volume at STP $= 0.025 \times 22.4 = 0.560 L = 560 cm^{3}$ .
Answer: $0.025 mol$ , i.e. $0.560 L at STP$ .

Level 5 — Mastery

L5.1 (mechanism reasoning) Write the two-step CFC catalytic cycle that destroys ozone, add the steps to get the net reaction, and explain in one sentence why one chlorine atom can destroy thousands of ozone molecules.

L5.2 (quantitative catalysis) If a single Cl atom completes the cycle once every $2 s$ and survives, on average, $1, 000, 000 s$ before being removed, how many $O_{3}$ molecules does that one Cl atom destroy? (Each full cycle removes one $O_{3}$ .)

L5.3 (concept trap) A newspaper says "we should make more ground-level ozone to fix the ozone hole." Give the two independent reasons this is wrong.

Recall Solutions L5

L5.1 From Free Radical Chain Reactions: $\dot{C l} + O_{3} \to C l O \cdot + O_{2} (step 1)$ $C l O \cdot + O \to \dot{C l} + O_{2} (step 2)$ Add them; the $C l O \cdot$ produced in step 1 is consumed in step 2, and the $\dot{C l}$ used in step 1 is re-made in step 2, so both cancel: $Net: O_{3} + O \to 2 O_{2} .$ Why thousands: Cl is a catalyst — it is regenerated at the end of each cycle, so it walks away unharmed to attack the next $O_{3}$ . It is never "used up," so its damage multiplies. This chain logic is exactly the radical chain idea, and links to Environmental Chemistry — Air Pollution.

L5.2 Cycles completed $= \frac{1 , 000 , 000 s}{2 s/cycle} = 500, 000$ cycles. Each cycle removes 1 $O_{3}$ , so 500,000 ozone molecules destroyed by one Cl atom. That is the "little cause, huge effect" of catalysis. (Real estimates run to $\sim 1 0^{5}$ , matching this order of magnitude.)

L5.3 Two independent reasons:

Wrong place. The protective ozone is in the stratosphere, tens of km up. Ozone made at the ground stays in the troposphere and cannot rise to patch the hole.
Wrong effect there. Ground-level ozone is a toxic pollutant/oxidant (a smog component) — it harms lungs and plants (see Environmental Chemistry — Air Pollution). Making more of it damages, not protects.

Active Recall

Recall One-line answers (cover them)

B.O. and magnetism of $O_{2}$ ? ::: Bond order 2; paramagnetic (2 unpaired e⁻ in $π_{2 p}^{*}$ ). Why both O–O bonds in ozone are equal? ::: Delocalised π makes it one resonance hybrid, each bond order 1.5. Ozone bond length vs $O_{2}$ ? ::: 128 pm, between 121 pm ( $O_{2}$ ) and 148 pm (single) because B.O. 1.5 is between 2 and 1. Moles of $O_{3}$ if 0.020 mol thiosulphate used? ::: 0.010 mol. Net CFC catalytic equation? ::: $O_{3} + O \to 2 O_{2}$ , with Cl regenerated.