Question
Iron has three main naturally occurring isotopes which can be investigated using a mass spectrometer.
State the full electronic configurations of a Cu atom and a \({\text{C}}{{\text{u}}^ + }\) ion.
Cu:
\({\text{C}}{{\text{u}}^ + }\):[2]
Explain the origin of colour in transition metal complexes and use your explanation to suggest why copper(II) sulfate, CuSO4(aq), is blue, but zinc sulfate, ZnSO4(aq), is colourless.[4]
\({\text{C}}{{\text{u}}^{2 + }}{\text{(aq)}}\) reacts with ammonia to form the complex ion \({{\text{[Cu(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}{\text{]}}^{2 + }}\). Explain this reaction in terms of an acid-base theory, and outline how the bond is formed between \({\text{C}}{{\text{u}}^{2 + }}\) and \({\text{N}}{{\text{H}}_{\text{3}}}\).[3]
Answer/Explanation
Markscheme
Cu:
\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{1}}}\);
\({\text{C}}{{\text{u}}^ + }\):
\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}\);
Ignore relative order of 3d and 4s.
Penalize only once if noble gas core is given.
d orbitals are split (into two sets of different energies);
frequencies of (visible) light absorbed by electrons moving from lower to higher d levels;
colour due to remaining frequencies/complementary colour transmitted;
\({\text{C}}{{\text{u}}^{2 + }}\) has unpaired electrons/partially filled d sub-level;
\({\text{Z}}{{\text{n}}^{2 + }}\) has filled d sub-shell;
electronic transitions/d-d transitions possible for \({\text{C}}{{\text{u}}^{2 + }}\) / no electronic/d-d transitions possible for \({\text{Z}}{{\text{n}}^{2 + }}\);
Allow wavelength as well as frequency.
\({\text{N}}{{\text{H}}_{\text{3}}}\): Lewis base / \({\text{C}}{{\text{u}}^{2 + }}\): Lewis acid;
each \({\text{N}}{{\text{H}}_{\text{3}}}\)/ligand donates an electron pair (to \({\text{C}}{{\text{u}}^{2 + }}\));
\({\text{N}}{{\text{H}}_{\text{3}}}\) replace \({{\text{H}}_2}{\text{O}}\) ligands around \({\text{C}}{{\text{u}}^{2 + }}\) ion/around central ion;
forming coordinate (covalent)/dative covalent bond;
Examiners report
Many candidates identified the electronic configuration of Cu as an exception but the 3d electron was often removed in forming the ion instead of the 4s.
Precision of language proved to be an issue in (e) with some candidates referring to Cu and Zn and not their ions and some students explained the colour as a result of “reflection” or “emission”.
In (f), many candidates mentioned proton donors and proton acceptors and made no reference to Lewis theory.
Question
Ammonia, \({\text{N}}{{\text{H}}_{\text{3}}}\), is a weak base. It has a \({\text{p}}{K_{\text{b}}}\) value of 4.75.
Salts may form neutral, acidic or alkaline solutions when dissolved in water.
Another weak base is nitrogen trifluoride, \({\text{N}}{{\text{F}}_{\text{3}}}\). Explain how \({\text{N}}{{\text{F}}_{\text{3}}}\) is able to function as a Lewis base.[1]
Calculate the pH of a \({\text{1.00}} \times {\text{1}}{{\text{0}}^{ – 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) aqueous solution of ammonia at 298 K.[4]
\({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \(1.00 \times {10^{ – 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrochloric acid solution is added to \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00}} \times {\text{1}}{{\text{0}}^{ – 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) aqueous ammonia solution. Calculate the concentrations of both ammonia and ammonium ions in the resulting solution and hence determine the pH of the solution.[5]
State what is meant by a buffer solution and explain how the solution in (v), which contains ammonium chloride dissolved in aqueous ammonia, can function as a buffer solution.[3]
State the equations for the reactions of sodium oxide, \({\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}\), and phosphorus(V)oxide, \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\), with water.[2]
Answer/Explanation
Markscheme
it can donate the lone/non-bonding pair of electrons (on the N atom);
\({K_{\text{b}}} = \frac{{{{{\text{[O}}{{\text{H}}^ – }{\text{]}}}^{\text{2}}}}}{{{\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}}} = {10^{ – 4.75}}/1.78 \times {10^{ – 5}}\);
\({\text{[O}}{{\text{H}}^ – }{\text{]}} = \sqrt {(1.00 \times {{10}^{ – 2}} \times {{10}^{ – 4.75}})} = 4.22 \times {10^{ – 4}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);
\({\text{pOH}} = – {\log _{10}}(4.22 \times {10^{ – 4}}) = 3.37/[{{\text{H}}^ + }] = \frac{{1.00 \times {{10}^{ – 14}}}}{{4.22 \times {{10}^{ – 4}}}} = 2.37 \times {10^{ – 11}}\);
\({\text{pH}} = 14 – 3.37 = 10.6\);
Award [2 max] for correct final answer if no working shown.
initial amount of \({\text{HCl}} = \frac{{25.0}}{{1000}} \times 1.00 \times {10^{ – 2}} = 2.50 \times {10^{ – 4}}{\text{ mol}}\) and initial amount of \({\text{N}}{{\text{H}}_3}{\text{ = }}\frac{{50.0}}{{1000}} \times 1.00 \times {10^{ – 2}}{\text{ = }}5.00 \times {10^{ – 4}}{\text{ mol}}\);
final amount of \({\text{NH}}_4^ + \) and \({\text{N}}{{\text{H}}_3}\) both \( = 2.50 \times {10^{ – 4}}{\text{ mol}}\);
final \({\text{[NH}}_4^ + {\text{]}}\) and \({\text{[N}}{{\text{H}}_3}{\text{]}}\) both \( = \frac{{2.50 \times {{10}^{ – 4}}}}{{75.0 \times {{10}^{ – 3}}}} = 3.33 \times {10^{ – 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);
\({\text{[O}}{{\text{H}}^ – }{\text{]}} = {K_{\text{b}}} \times \frac{{{\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}}}{{{\text{[NH}}_4^ + {\text{]}}}} = {K_{\text{b}}} = {10^{ – 4.75}}/1.78 \times {10^{ – 5}}\);
\({\text{pOH}} = 4.75\) hence \({\text{pH}} = 9.25\);
Award final two marking points if half-equivalence method used.
a buffer solution resists a change in pH when small amounts of acid or base are added to it;
Do not accept description in terms of composition of buffer.
when \({{\text{H}}^ + }\) is added it reacts with \({\text{N}}{{\text{H}}_3}\) to form \({\text{NH}}_4^ + \);
when \({\text{O}}{{\text{H}}^ – }\) is added it reacts with \({\text{NH}}_4^ + \) to form \({\text{N}}{{\text{H}}_3}\) and \({{\text{H}}_2}{\text{O}}\);
Accept equations for last two marking points.
\({\text{N}}{{\text{a}}_2}{\text{O}} + {{\text{H}}_2}{\text{O}} \to {\text{2N}}{{\text{a}}^ + } + {\text{2O}}{{\text{H}}^ – }/{\text{N}}{{\text{a}}_2}{\text{O}} + {{\text{H}}_2}{\text{O}} \to {\text{2NaOH}}\);
\({{\text{P}}_4}{{\text{O}}_{10}} + {\text{6}}{{\text{H}}_2}{\text{O}} \to {\text{4}}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\);
Ignore state symbols.
Examiners report
In (a) (iii) some candidate did not mention the need for a lone pair even though they had an understanding of the need for a pair of electrons when explaining the basic properties of nitrogen trifluoride.
Answers to (a) (iv) were encouraging with many candidates able to calculate the pH from the \({\text{p}}{K_{\text{b}}}\) value for ammonia.
The more difficult (a) (v) was only answered correctly by the strongest candidates and a significant number left it blank.
Some candidates lost marks in (a) (vi) as they did not explicitly state that buffers are resistant to changes of pH when small amounts of acid or base are added. Many also did not respond directly to the requirements of the question and explain the action of the specific buffer mixture of ammonia and ammonium chloride. Salt hydrolysis was poorly understood.
Most could give an equation for the reaction of sodium oxide with water but the formation of phosphoric (V) acid from phosphorus (V) oxide proved more problematic.