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Date	May 2018	Marks available	1	Reference code	18M.2.hl.TZ2.3
Level	HL	Paper	2	Time zone	TZ2
Command term	Calculate	Question number	3	Adapted from	N/A

Question

The emission spectrum of an element can be used to identify it.

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.

[1]

a.iii.

Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.

[1]

a.iv.

Deduce any change in the colour of the electrolyte during electrolysis.

[1]

c.iv.

Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.

[1]

c.v.

Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.

[2]

Markscheme

IE «= ΔE = hν = 6.63 × 10^–34 J s × 3.28 × 10¹⁵ s^–1» = 2.17 × 10^–18 «J»

[1 mark]

a.iii.

« $\lambda = \frac{C}{{\text{v}}} = \frac{{3.00 \times {{10}^8}{\text{ m}}{{\text{s}}^{ - 1}}}}{{3.28 \times {{10}^{15}}{\text{ }}{{\text{s}}^{ - 1}}}} =$ » 9.15 × 10^–8 «m»

[1 mark]

a.iv.

no change «in colour»

Do not accept “solution around cathode will become paler and solution around the anode will become darker”.

[1 mark]

c.iv.

oxygen/O₂

Accept “carbon dioxide/CO2”.

[1 mark]

c.v.

Transition metals:

«contain» d and s orbitals «which are close in energy»

«successive» ionization energies increase gradually

Alkali metals:

second electron removed from «much» lower energy level

removal of second electron requires large increase in ionization energy

[2 marks]

Examiners report

[N/A]

a.iii.

[N/A]

a.iv.

[N/A]

c.iv.

[N/A]

c.v.

[N/A]

Syllabus sections

Core » Topic 2: Atomic structure » 2.2 Electron configuration

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