IB Questionbank

Updates to Questionbank

User interface language: English | Español

Date	November 2019	Marks available	2	Reference code	19N.2.hl.TZ0.1
Level	HL	Paper	2	Time zone	TZ0
Command term	Show and Determine	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]

a.

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]

b.

Predict the bond angle in the ozone molecule.

[1]

c.

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]

d.

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]

f.

Markscheme

NOTES: Coordinate bond may be represented by an arrow.

Do not accept delocalized structure for ozone.

a.

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

NOTE: Accept any length between these two values.

b.

any value from 110°–119° ✔

c.

«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.

d.

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

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

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

NOTE: Award [2] for correct final answer.

e(ii).

•NO + O₃ → •NO₂ + O₂ ✔

•NO₂ + O₃ → •NO + 2O₂ ✔

NOTE: Accept •NO₂ → •NO + •O AND •O + O₃ → 2O₂ for M2.

f.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e(i).

[N/A]

e(ii).

[N/A]

f.

Syllabus sections

Additional higher level (AHL) » Topic 12: Atomic structure » 12.1 Electrons in atoms

Show 29 related questions

17M.1.hl.TZ1.5:
Which statement explains one of the decreases in first ionization energy (I.E.) across period 3?

A. The nuclear charge of element Al is greater than element Mg.

B. The electron-electron repulsion is greater, for the electron with the opposite spin, in element S than in element P.

C. A new sub-level is being filled at element S.

D. The p orbital being filled in element Al is at a lower energy than the s orbital in element Mg.
22M.1.hl.TZ1.5:
The graph shows the first six ionization energies of an element.

^{[Ionization energies of the elements (data page) Available at:}https://en.wikipedia.org/wiki/Ionization_energies_of_the_
^{elements_(data_page) Text is available under the Creative Commons Attribution-ShareAlike License 3.0 (CC BY-SA}
^{3.0) https://creativecommons.org/licenses/by-sa/3.0/deed.en.]}

In which group is the element?

A. 13

B. 14

C. 15

D. 16
18M.1.hl.TZ2.5: The graph shows the first ionization energies of some consecutive elements. Which...
18M.1.hl.TZ1.5: Which transition on the diagram corresponds to the ionization of hydrogen in the ground state?
18M.2.hl.TZ1.2e:
Sketch a graph of the first six ionization energies of calcium.
17M.2.hl.TZ1.2d.iv:
Explain why an aluminium-titanium alloy is harder than pure aluminium.
16N.1.hl.TZ0.6: A period 3 element, M, forms an oxide of the type M2O. Which represents the first four...
18N.1.hl.TZ0.5: The values for the first three successive ionization energies for two elements X and Z are...
18N.2.hl.TZ0.4c: Sketch a graph to show the relative values of the successive ionization energies of boron.
19M.1.hl.TZ1.5: Which element is represented by the first eight successive ionization energies on the...
19M.1.hl.TZ2.5:
Which of the following transitions in the hydrogen atom emits the least energy?

A. n = 2 to n = 1

B. n = 3 to n = 1

C. n = 4 to n = 2

D. n = 4 to n = 3
17N.1.hl.TZ0.6:
The graph represents the first ten ionisation energies (IE) of an element.

What is the element?

A. O

B. S

C. Ne

D. Cl
18M.2.hl.TZ2.3a.iii:
Hydrogen spectral data give the frequency of 3.28 × 10¹⁵ s⁻¹ for its convergence limit.

Calculate the ionization energy, in J, for a single atom of hydrogen using sections 1 and 2 of the data booklet.
16N.2.hl.TZ0.4d:
(i) Explain the convergence of lines in a hydrogen emission spectrum.

(ii) State what can be determined from the frequency of the convergence limit.
17M.2.hl.TZ2.4c:
The first six ionization energies, in kJ mol^–1, of an element are given below.

Explain the large increase in ionization energy from IE₃ to IE₄.
19M.2.hl.TZ1.3a:
Plot the relative values of the first four ionization energies of sodium.
17M.2.hl.TZ1.2d.iii:
Sketch a graph of the first six successive ionization energies of vanadium on the axes provided.
19N.1.hl.TZ0.5: Which shows the first ionization energies of successive elements across period 2, from left...
21M.1.hl.TZ1.6: The diagram shows the first ionisation energies of consecutive elements in the same period of...
21M.2.hl.TZ1.1c: Sketch the first eight successive ionisation energies of sulfur.
21M.1.hl.TZ2.5: The first eight successive ionization energies for an element are shown. In which group is...
19M.2.hl.TZ2.3c:
Explain why the first ionization energy of nitrogen is greater than both carbon and oxygen.

Nitrogen and carbon:

Nitrogen and oxygen:
22M.2.hl.TZ2.5a(ii): Explain why there is a large increase from the 8th to the 9th ionization energy of iron.
20N.1.hl.TZ0.5: Which element is in group 13?
20N.2.hl.TZ0.6a:
The electron configuration of copper makes it a useful metal.

Determine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.
21M.2.hl.TZ2.2a(ii):
Explain why the first ionization energy of sulfur is lower than that of phosphorus.
22M.2.hl.TZ1.1e(iv): Suggest, giving a reason, whether magnesium or nitrogen would have the greater sixth...
21N.2.hl.TZ0.2b:
Sodium emits yellow light with a frequency of 5.09 × 10¹⁴Hz when electrons transition from 3p to 3s orbitals.

Calculate the energy difference, in J, between these two orbitals using sections 1 and 2 of the data booklet.

Darling, D, n.d. D lines (of sodium). [online] Available at <https://www.daviddarling.info/encyclopedia/D/D_lines.html> [Accessed 6 May 2020].
21N.1.hl.TZ0.5: Which statement explains why the second ionization energy of aluminium is higher than the...

Hide related questions

Additional higher level (AHL) » Topic 12: Atomic structure

Additional higher level (AHL)