Question 18M.2.hl.TZ2.3
| Date | May 2018 | Marks available | [Maximum mark: 6] | Reference code | 18M.2.hl.TZ2.3 |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Deduce, Explain | Question number | 3 | Adapted from | N/A |
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]
IE «= ΔE = hν = 6.63 × 10–34 J s × 3.28 × 1015 s–1» = 2.17 × 10–18 «J»
[1 mark]
Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.
[1]
«» 9.15 × 10–8 «m»
[1 mark]
Deduce any change in the colour of the electrolyte during electrolysis.
[1]
no change «in colour»
Do not accept “solution around cathode will become paler and solution around the anode will become darker”.
[1 mark]
Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.
[1]
oxygen/O2
Accept “carbon dioxide/CO2”.
[1 mark]
Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.
[2]
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]