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.
Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.
Deduce any change in the colour of the electrolyte during electrolysis.
Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.
Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.
Markscheme
IE «= ΔE = hν = 6.63 × 10–34 J s × 3.28 × 1015 s–1» = 2.17 × 10–18 «J»
[1 mark]
«\(\lambda = \frac{C}{{\text{v}}} = \frac{{3.00 \times {{10}^8}{\text{ m}}{{\text{s}}^{ - 1}}}}{{3.28 \times {{10}^{15}}{\text{ }}{{\text{s}}^{ - 1}}}} = \)» 9.15 × 10–8 «m»
[1 mark]
no change «in colour»
Do not accept “solution around cathode will become paler and solution around the anode will become darker”.
[1 mark]
oxygen/O2
Accept “carbon dioxide/CO2”.
[1 mark]
Transition metals:
«contain» d and s orbitals «which are close in energy»
OR
«successive» ionization energies increase gradually
Alkali metals:
second electron removed from «much» lower energy level
OR
removal of second electron requires large increase in ionization energy
[2 marks]