| Date | May 2010 | Marks available | 2 | Reference code | 10M.2.hl.TZ1.1 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Explain | Question number | 1 | Adapted from | N/A |
Question
Brass is a copper containing alloy with many uses. An analysis is carried out to determine the percentage of copper present in three identical samples of brass. The reactions involved in this analysis are shown below.
\[\begin{array}{*{20}{l}} {{\text{Step 1: Cu(s)}} + {\text{2HN}}{{\text{O}}_3}{\text{(aq)}} + {\text{2}}{{\text{H}}^ + }{\text{(aq)}} \to {\text{C}}{{\text{u}}^{2 + }}{\text{(aq)}} + {\text{2N}}{{\text{O}}_2}{\text{(g)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}}} \\ {{\text{Step 2: 4}}{{\text{I}}^ - }{\text{(aq)}} + {\text{2C}}{{\text{u}}^{2 + }}{\text{(aq)}} \to {\text{2CuI(s)}} + {{\text{I}}_2}{\text{(aq)}}} \\ {{\text{Step 3: }}{{\text{I}}_2}{\text{(aq)}} + {\text{2}}{{\text{S}}_2}{\text{O}}_3^{2 - }{\text{(aq)}} \to {\text{2}}{{\text{I}}^ - }{\text{(aq)}} + {{\text{S}}_4}{\text{O}}_6^{2 - }{\text{(aq)}}} \end{array}\]
In step 1 the copper reacts to form a blue solution.
State the full electronic configuration of \({\text{C}}{{\text{u}}^{2 + }}\).
Explain why the copper solution is coloured.
Markscheme
\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{\text{9}}}\);
Do not allow [Ar]3d9.
d orbitals are split;
(3d) electrons move between orbitals and absorb light/energy / complementary colour is transmitted when energy absorbed by d electrons moving / unpaired d electrons move between the different orbitals;
Accept levels instead of orbitals.
Examiners report
Several errors were seen in the electron configuration, the commonest of which was to give that of elemental copper.
Few attempts at the explanation of colour referred to the splitting of the d orbitals and electron transitions, and in several instances candidates referred to emission instead of absorption. This proved to be the most difficult part of question 1.
