DP Chemistry (first assessment 2025)

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Question 21M.2.HL.TZ1.7

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Date May 2021 Marks available [Maximum mark: 7] Reference code 21M.2.HL.TZ1.7
Level HL Paper 2 Time zone TZ1
Command term Discuss, Draw, Explain Question number 7 Adapted from N/A
7.
[Maximum mark: 7]
21M.2.HL.TZ1.7

Oxygen exists as two allotropes, diatomic oxygen, O2, and ozone, O3.

(a(i))

Draw a Lewis (electron dot) structure for ozone.

[1]

Markscheme

Accept any combination of lines, dots or crosses to represent electrons.

Do not accept structures that represent 1.5 bonds.

(a(ii))

Discuss the relative length of the two O−O bonds in ozone.

[2]

Markscheme

both equal ✔

delocalization/resonance ✔


Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.

(b)

Explain why there are frequencies of UV light that will dissociate O3 but not O2.

[2]

Markscheme

bond in O3 is weaker
OR
O3 bond order 1.5/< 2 ✔


Do not accept bond in O3 is longer for M1.


lower frequency/longer wavelength «UV light» has enough energy to break the O–O bond in O3 «but not that in O2» ✔


Accept “lower frequency/longer wavelength «UV light» has lower energy”.

(c)

Explain, using equations, how the presence of CCl2F2 results in a chain reaction that decreases the concentration of ozone in the stratosphere.

[2]

Markscheme

CCl2F2(g) →∙CClF2(g) Cl(g)

Cl•(g)+O3(g)→O2(g)+ClO•(g)
AND
ClO∙(g)+O3(g)→2O2(g)+Cl(g)


Do not penalize missing radical.

Accept:for M2:
Cl∙(g) + O3(g) → O2(g) + ClO(g)
AND
ClO∙(g) + O(g) → O2(g) + Cl(g)