Question 18M.3.hl.TZ2.18
| Date | May 2018 | Marks available | [Maximum mark: 4] | Reference code | 18M.3.hl.TZ2.18 |
| Level | hl | Paper | 3 | Time zone | TZ2 |
| Command term | Draw, Identify, Outline, State | Question number | 18 | Adapted from | N/A |
The conductivity of a germanium semiconductor can be increased by doping.
Draw the Lewis (electron dot) structure for an appropriate doping element in the box in the centre identifying the type of semiconductor formed.
[2]
ALTERNATIVE 1
B/Ga in circle AND Type of semiconductor: p-type
showing 3 electron pairs AND one lone electron «and hole»
ALTERNATIVE 2
P/As in circle AND Type of semiconductor: n-type
showing 4 electron pairs AND one non-bonded electron
Accept any group 13 element labelled as p-type.
Accept showing 7 electrons.
Accept any group 15 element labelled as n-type.
Accept showing 9 electrons.
Accept dots or crosses for electrons.
[2 marks]
A dye-sensitized solar cell uses a ruthenium(II)–polypyridine complex as the dye. Two ruthenium(II) complexes, A and B, absorb light of wavelengths 665 nm and 675 nm respectively.
State the feature of the molecules responsible for the absorption of light.
[1]
conjugated C=C/carbon–carbon double bonds
OR
«multiple» alternating C=C/carbon–carbon double bonds
OR
«extensive electron» conjugation/delocalization
OR
«many» fused/conjugated aromatic/benzene rings
[1 mark]
Outline why complex B absorbs light of longer wavelength than complex A.
[1]
complex B has greater conjugation/delocalization
[1 mark]