IChO 2005 · Theory problem 4

Gold Capital of Asia

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A

Gold Capital of Asia

Chiufen, the old mining town located within the hills in the northeast Taiwan, is a place where you can really experience Taiwan's historical legacy. It was the site of one of the largest gold mines In Asia. Accordingly, Chiufen is often referred to as the Gold Capital of Asia. The compound \(\ce{KCN}\) is traditionally used to extract gold from ore. Gold dissolves in cyanide (\(\ce{CN^-}\)) solutions in the presence of air to form \(\ce{Au(CN)2^-}\), which is stable in aqueous solution.

Figure 1

4.A-1 Draw a structure for \(\ce{Au(CN)2^-}\) showing the spatial arrangements of the atoms.

4.A-2 How many grams of \(\ce{KCN}\) are needed to extract 20 g of gold from ore? Show your work.

Aqua regia, a 3 : 1 mixture (by volume) of concentrated hydrochloric acid and nitric acid, was developed by the alchemists as a means to “dissolve” gold. The process is actually a redox reaction with the following simplified chemical equation:

Figure 2

4.A-3 Write down the half reactions, and use them to obtain a balanced redox reaction for this process.

4.A-4 What are the oxidizing and reducing agents for 4.A-3 process?

Gold is too noble to react with nitric acid. However, gold does react with aqua regia because the complex ion \(\ce{AuCl4^-}\) forms. Consider the following half-reactions:

Figure 3

An electrochemical cell can be formed from these two redox couples.

4.A-5 Calculate the formation constant for \(\ce{AuCl4^-}\) at \(25\,^\circ\mathrm{C}\):

\[ K=\frac{[\ce{AuCl4^-}]}{[\ce{Au^3+}][\ce{Cl^-}]^4} \]

4.A-6 The function of \(\ce{HCl}\) is to provide \(\ce{Cl^-}\). What is the purpose of the \(\ce{Cl^-}\) for the above reaction. Select your answer from the following choices.

(a) \(\ce{Cl^-}\) is an oxidizing agent
(b) \(\ce{Cl^-}\) is a reducing agent
(c) \(\ce{Cl^-}\) is a complexing agent
(d) \(\ce{Cl^-}\) is a catalyst

B

Gold Nanoparticles

The synthesis and characterization of gold nanoparticles is currently an active research area. The Brust-Schiffrin method for the synthesis of gold nanoparticle (AuNP) allows the facile preparation of thermally stable and air-stable AuNPs of reduced polydispersity with a controlled size distribution ranging in diameter between 1.5 and 5.2 nm. The preparative procedure is briefly described as follows. An aqueous solution of \(\ce{HAuCl4}\) is mixed with a toluene solution of tetra-n-octylammonium bromide. The solution is mixed with dodecanethiol and is treated with an excess of \(\ce{NaBH4}\). Formation of the AuNPs is evidenced by the immediate, pronounced darkening of the toluene phase. After ca. 24 h, the toluene solvent is removed with a rotary evaporator and the resulting solid washed on a frit with ethanol and hexane to remove excess thiol. These AuNPs can be repeatedly isolated and re-dissolved in common organic solvents without irreversible aggregation or decomposition.

4.B-1 Is the methodology for this fabrication referred to a top-down or a bottom-up approach? Select your answer from the following choices.

(a) top-down approach, which entails reducing the size of the smallest structures to the nanoscale
(b) bottom-up approach, which involves manipulating individual atoms and molecules into nanostructures

4.B-2 The trimethyl-n-octylammonium bromide can also be used as a phase-transfer reagent. It can carry \(\ce{AlCl4^-}\) from an aqueous phase to an organic phase. Which property does trimethyl-n-octylammonium bromide possess to function as an efficient phase-transfer reagent? Select your answer from the following choices.

(a) one side of the molecule is electropositive, the other side is electronegative.
(b) one side of the molecule is hydrophilic, the other side is hydrophobic.
(c) one side of the molecule is acidic, the other side is basic.

4.B-3 What is the function of \(\ce{NaBH4}\) in this preparation? Select your answer from the following choices.

(a) reducing agent
(b) oxidizing agent
(c) neutralization agent
(d) complexing agent

4.B-4 If the average diameter of a gold nanoparticle is 3 nm, what is the estimated number of \(\ce{Au}\) atoms in each nanoparticle? (the atomic radius of \(\ce{Au}\) is 0.144 nm). Select your answer from the following choices and show your work.

(a) \(10^2\)
(b) \(10^3\)
(c) \(10^4\)
(d) \(10^5\)

4.B-5 What is the estimated percentage of \(\ce{Au}\) atoms on the surface of a nanoparticle? Select your answer from the following choices and show your work.

(a) 20 – 30 %
(b) 40 – 50 %
(c) 60 – 70 %
(d) 80 – 90 %

Official English text of the problem.

Answers

  • 4.A-1: Dicyanidoaurate(I), \(\ce{Au(CN)2^-}\), SMILES N#C[Au-]C#N. All five atoms lie on a straight line in the order nitrogen–carbon–gold–carbon–nitrogen; the carbon–gold–carbon angle is \(180^\circ\).

  • 4.A-2: \(13.2\ \mathrm{g}\) of potassium cyanide, \(\ce{KCN}\), [K+].[C-]#N; approximately \(13\ \mathrm{g}\) to two significant figures.

  • 4.A-3: The half-reactions are \[ \ce{Au + 4Cl^- -> AuCl4^- + 3e^-} \] \[ \ce{NO3^- + 2H+ + e^- -> NO2 + H2O}. \] Their balanced sum is \[ \boxed{\ce{Au + 3NO3^- + 4Cl^- + 6H+ -> AuCl4^- + 3NO2 + 3H2O}}. \]

  • 4.A-4: Oxidizing agent: nitrate ion, \(\ce{NO3^-}\), [O-][N+](=O)[O-]. Reducing agent: elemental gold, \(\ce{Au(s)}\); [Au] represents an individual gold atom.

  • 4.A-5: \(\boxed{K\approx2.3\times10^{25}}\) for formation of tetrachloridoaurate(III), \(\ce{AuCl4^-}\), [Au-](Cl)(Cl)(Cl)Cl.

  • 4.A-6: (c) — chloride ion, \(\ce{Cl^-}\), [Cl-], is a complexing agent.

  • 4.B-1: (b) — bottom-up approach.

  • 4.B-2: (b) — a hydrophilic charged head and a hydrophobic hydrocarbon tail.

  • 4.B-3: (a) — sodium borohydride, \(\ce{NaBH4}\), [Na+].[BH4-], is a reducing agent.

  • 4.B-4: (b) — approximately \(10^3\) gold atoms per nanoparticle.

  • 4.B-5: (b) — \(40\text{–}50\%\); the outer-layer estimate gives approximately \(47\%\).

Solution

The SMILES below specify connectivity and formal charge. Formal charges assigned to metals in these covalent SMILES representations are not their oxidation states; coordination geometries are stated separately.

4.A-1

Each cyanide ion, \(\ce{CN^-}\), [C-]#N, has charge \(-1\). Therefore the oxidation state \(x\) of gold satisfies

\[ x+2(-1)=-1, \qquad x=+1. \]

The two-coordinate gold(I) center has a \(d^{10}\) electron configuration and linear coordination. Both cyanide ligands bind through their carbon atoms.

The drawing should therefore show the dicyanidoaurate(I) ion, N#C[Au-]C#N, as

\[ \boxed{\ce{[N#C-Au-C#N]^-}}. \]

Draw all five atoms on one straight line, with two carbon–nitrogen triple bonds and two gold–carbon bonds. Enclose the complete structure in brackets with an overall \(-1\) charge.

4.A-2

The given dissolution equation is

\[ \ce{4Au(s) + 8CN^-(aq) + O2(g) + 2H2O(l) <=> 4Au(CN)2^-(aq) + 4OH^-(aq)}. \]

Thus each mole of gold requires two moles of cyanide, supplied by two moles of potassium cyanide, [K+].[C-]#N.

Using standard molar masses,

\[ M(\ce{Au})=196.97\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{KCN})=39.10+12.01+14.01 =65.12\ \mathrm{g\,mol^{-1}}. \]

Hence

\[ n(\ce{Au}) =\frac{20\ \mathrm{g}}{196.97\ \mathrm{g\,mol^{-1}}} =0.10154\ \mathrm{mol}, \]

\[ n(\ce{KCN})=2(0.10154)=0.20308\ \mathrm{mol}, \]

and

\[ m(\ce{KCN}) =(0.20308\ \mathrm{mol})(65.12\ \mathrm{g\,mol^{-1}}) =\boxed{13.2\ \mathrm{g}}. \]

This is the theoretical stoichiometric requirement, approximately \(13\ \mathrm{g}\) to two significant figures.

The other molecular or ionic species in the dissolution equation are dioxygen O=O, water O, and hydroxide [OH-].

4.A-3

Gold changes from oxidation state \(0\) to \(+3\), losing three electrons. Including coordination by chloride, the oxidation half-reaction is

\[ \boxed{\ce{Au(s) + 4Cl^-(aq) -> AuCl4^-(aq) + 3e^-}}. \]

Nitrogen changes from \(+5\) in nitrate to \(+4\) in nitrogen dioxide, gaining one electron. Balancing oxygen with water and hydrogen with hydrogen ions gives

\[ \boxed{\ce{NO3^-(aq) + 2H+(aq) + e^- -> NO2(g) + H2O(l)}}. \]

Multiplying the reduction half-reaction by three and adding cancels all electrons:

\[ \boxed{ \ce{Au(s) + 3NO3^-(aq) + 4Cl^-(aq) + 6H+(aq) -> AuCl4^-(aq) + 3NO2(g) + 3H2O(l)} }. \]

The total charge is \(-1\) on each side, and all atoms balance.

The species are tetrachloridoaurate(III), \(\ce{AuCl4^-}\), [Au-](Cl)(Cl)(Cl)Cl; nitrate, [O-][N+](=O)[O-]; chloride, [Cl-]; hydrogen ion, [H+]; nitrogen dioxide, [O-][N+]=O; and water, O. The gold complex is square planar. Nitrogen dioxide is a bent radical molecule.

4.A-4

The oxidizing agent is nitrate, \(\ce{NO3^-}\), [O-][N+](=O)[O-], because it accepts electrons and is reduced: nitrogen changes from \(+5\) to \(+4\).

The reducing agent is elemental gold, \(\ce{Au(s)}\), because it supplies electrons and is oxidized from \(0\) to \(+3\). Its atomic SMILES is [Au]; this does not represent the extended metallic lattice.

4.A-5

The required formation equilibrium is

\[ \ce{Au^3+(aq) + 4Cl^-(aq) <=> AuCl4^-(aq)}. \]

Here the gold(III) ion is [Au+3], and tetrachloridoaurate(III) is [Au-](Cl)(Cl)(Cl)Cl.

Use the first given half-reaction in the reduction direction:

\[ \ce{Au^3+(aq) + 3e^- -> Au(s)}, \qquad E^\circ_{\mathrm{red}}=+1.50\ \mathrm{V}. \]

Reverse the second half-reaction:

\[ \ce{Au(s) + 4Cl^-(aq) -> AuCl4^-(aq) + 3e^-}, \qquad E^\circ_{\mathrm{ox}}=-1.00\ \mathrm{V}. \]

Their sum is the desired formation reaction, with

\[ E^\circ_{\mathrm{cell}}=1.50-1.00=0.50\ \mathrm{V}. \]

Three electrons are transferred in the electrochemical cycle. Therefore,

\[ \Delta_rG^\circ=-3FE^\circ_{\mathrm{cell}} =-RT\ln K, \]

so

\[ \ln K=\frac{3FE^\circ_{\mathrm{cell}}}{RT}. \]

Using \(F=96485\ \mathrm{C\,mol^{-1}}\), \(R=8.31446\ \mathrm{J\,mol^{-1}\,K^{-1}}\), and \(T=298.15\ \mathrm{K}\),

\[ \log_{10}K =\frac{3(96485)(0.50)} {(8.31446)(298.15)\ln 10} =25.355. \]

Thus

\[ \boxed{K=10^{25.355}\approx2.3\times10^{25}}. \]

The thermodynamic equilibrium constant is dimensionless; the printed concentration expression implicitly refers concentrations to the standard concentration.

4.A-6

Chloride, [Cl-], binds gold(III) to form the strongly stabilized tetrachloridoaurate(III) complex:

\[ \ce{Au^3+ + 4Cl^- <=> AuCl4^-}. \]

This greatly decreases the concentration of free gold(III) ions and favors dissolution of gold into the complexed form. Chloride remains in oxidation state \(-1\).

Therefore the answer is (c), complexing agent.

4.B-1

The nanoparticles form by reduction of ionic gold precursors, followed by nucleation and growth into particles containing many gold atoms. The nanostructures are assembled from atomic or molecular starting species.

Therefore this is (b), a bottom-up approach.

4.B-2

Trimethyl-n-octylammonium bromide has SMILES CCCCCCCC[N+](C)(C)C.[Br-].

Its quaternary ammonium head is charged and hydrophilic, whereas its eight-carbon alkyl chain is hydrophobic and compatible with the organic phase. The ammonium cation can associate with an anion, allowing the resulting ion pair to enter the organic phase.

The transferred anion printed in the question is tetrachloridoaluminate(III), \(\ce{AlCl4^-}\), [Al-](Cl)(Cl)(Cl)Cl, which has tetrahedral coordination.

The required property is therefore (b): one side is hydrophilic and the other is hydrophobic.

4.B-3

Sodium borohydride, \(\ce{NaBH4}\), [Na+].[BH4-], supplies reducing equivalents that convert ionic gold into metallic gold. Its borohydride ion, [BH4-], has tetrahedral geometry.

For the gold(III) precursor, the net gold reduction can be represented as

\[ \ce{AuCl4^- + 3e^- -> Au + 4Cl^-}. \]

Therefore sodium borohydride acts as (a), a reducing agent.

4.B-4

Approximate both the nanoparticle and an individual gold atom as spheres. Their radii are

\[ R=\frac{3\ \mathrm{nm}}{2}=1.5\ \mathrm{nm}, \qquad r=0.144\ \mathrm{nm}. \]

Ignoring the spaces between packed atoms for this order-of-magnitude estimate,

\[ N\approx \frac{V_{\mathrm{particle}}}{V_{\mathrm{atom}}} = \frac{\frac43\pi R^3}{\frac43\pi r^3} = \left(\frac{R}{r}\right)^3. \]

Thus

\[ N\approx \left(\frac{1.5}{0.144}\right)^3 =1.13\times10^3. \]

Accounting for packing voids changes the numerical prefactor but not the order of magnitude. Hence the answer is

\[ \boxed{\text{(b), }10^3\text{ atoms per nanoparticle}}. \]

4.B-5

Approximate the surface atoms as an outer shell one atomic diameter thick:

\[ t=2r=2(0.144)=0.288\ \mathrm{nm}. \]

The interior radius is then

\[ R_{\mathrm{interior}}=R-t=1.5-0.288=1.212\ \mathrm{nm}. \]

Using the same average atomic number density in the surface shell and the interior, the fraction of atoms in the shell equals its fraction of the particle volume:

\[ f_{\mathrm{surface}} \approx \frac{R^3-(R-t)^3}{R^3} = 1-\left(\frac{R-t}{R}\right)^3. \]

Therefore,

\[ f_{\mathrm{surface}} \approx 1-\left(\frac{1.212}{1.5}\right)^3 =0.4725. \]

Thus approximately \(47\%\) of the atoms are in the outer atomic layer, giving

\[ \boxed{\text{(b), }40\text{–}50\%}. \]

The solution was checked against the official solution.