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Date	November 2019	Marks available	1	Reference code	19N.2.hl.TZ0.1
Level	HL	Paper	2	Time zone	TZ0
Command term	Identify	Question number	1	Adapted from	N/A

Question

The equations show steps in the formation and decomposition of ozone in the stratosphere, some of which absorb ultraviolet light.

Step 1 O₂ → 2O•

Step 2 O• + O₂ → O₃

Step 3 O₃ → O• + O₂

Step 4 O• + O₃ → 2O₂

Draw the Lewis structures of oxygen, O₂, and ozone, O₃.

[2]

Outline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.

Reason:

Length:

[2]

Predict the bond angle in the ozone molecule.

[1]

Discuss how the different bond strengths between the oxygen atoms in O₂ and O₃ in the ozone layer affect radiation reaching the Earth’s surface.

[2]

Identify the steps which absorb ultraviolet light.

[1]

e(i).

Determine, showing your working, the wavelength, in m, of ultraviolet light absorbed by a single molecule in one of these steps. Use sections 1, 2 and 11 of the data booklet.

[2]

e(ii).

Ozone depletion is catalysed by nitrogen monoxide, NO, which is produced in aircraft and motor vehicle engines, and has the following Lewis structure.

Show how nitrogen monoxide catalyses the decomposition of ozone, including equations in your answer.

[2]

Markscheme

NOTES: Coordinate bond may be represented by an arrow.

Do not accept delocalized structure for ozone.

resonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔

NOTE: Accept any length between these two values.

any value from 110°–119° ✔

«bond» in O₂ stronger than in O₃ ✔

ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔

NOTE: Accept ozone «layer» absorbs a range of frequencies.

steps 1 AND 3 ✔

e(i).

ALTERNATIVE 1:
for oxygen:

$E = « \frac{{498000{\text{ Jmo}}{{\text{l}}^{ - 1}}}}{{6.02 \times {{10}^{23}}{\text{mo}}{{\text{l}}^{ - 1}}}} =» 8.27 \times {10^{ - 19}}«{\text{J}}»$ ✔

$\lambda {\text{ = }}«\frac{{6.63 \times {{10}^{ - 34}}{\text{ J s}} \times {\text{3}}{\text{.00}} \times {\text{1}}{{\text{0}}^8}{\text{ m }}{{\text{s}}^{ - 1}}}}{{8.27 \times {{10}^{ - 19}}{\text{J}}}} =» 2.40 \times {10^{ - 7}}«{\text{m}}»$ ✔

ALTERNATIVE 2:
for ozone:

similar calculation using 200 < bond enthalpy < 400 for ozone, such as

$E = « \frac{{300000{\text{ Jmo}}{{\text{l}}^{ - 1}}}}{{6.02 \times {{10}^{23}}{\text{mo}}{{\text{l}}^{ - 1}}}} =» 4.98 \times {10^{ - 19}}«{\text{J}}»$ ✔

$\lambda {\text{ = }}«\frac{{6.63 \times {{10}^{ - 34}}{\text{ J s}} \times {\text{3}}{\text{.00}} \times {\text{1}}{{\text{0}}^8}{\text{ m }}{{\text{s}}^{ - 1}}}}{{4.98 \times {{10}^{ - 19}}{\text{J}}}} =» 3.99 \times {10^{ - 7}}«{\text{m}}»$ ✔