IChO 2024 · Theory problem 5

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Metal containing salt A can be prepared in a simple exchange reaction when the cold, saturated aqueous solutions of the two distinctly colored compounds, B and C are mixed in stoichiometric ratio. 10.00 g B in solution mixed with the solution of 12.86 g C, and immediately cooled to \(2\,^\circ\mathrm{C}\) will give 4.90 g solid A. The yield of A is 32.6 %.

To determine the composition of A, first iodometry is performed. A known mass of A is added to a titration flask, acidified with sulfuric acid, then \(\ce{KI}\) is added in excess and there is a precipitate formation. After a few minutes, sodium citrate solution is added until the solution is free of precipitate. The citrate ions form a strong complex with a metal ion present in the mixture, reversing completely the reaction leading to the precipitate. The resulting mixture is titrated with sodium thiosulfate solution (Titration I via reaction 1). To the blue titrated solution another portion of sulfuric acid (significantly more than the initial amount) is added to protonate the citrate. The previous solid precipitate is formed once again [reaction 2]. The mixture is titrated with the same thiosulfate solution (titration II via reaction 1).

The average volumes for 100.0 mg A with \(5.000\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\) thiosulfate are \(54.12\ \mathrm{cm^3}\) in Titration I and \(5.41\ \mathrm{cm^3}\) in Titration II.

When the aqueous solution of A is heated, the blue precipitate D can be observed [reaction 3]. Compound E can be crystallized from the liquid above D. The low temperature during the synthesis of A is important to avoid contamination by E.

5.1 Write balanced equations for reactions [1] and [2]. 4.0 pt

The thermal decomposition of A was studied in detail. When pure A is slowly heated, already at around \(75\,^\circ\mathrm{C}\) it explodes. When the compound is dispersed in aluminium oxide and the mixture is heated, then the matrix absorbs the excess heat, and the explosion can be avoided. Two decomposition steps can be observed. In the first step (at \(65\,^\circ\mathrm{C}\)) in addition to the 14.1 % mass decrease, a two-component solid residue forms [reaction 4]. The components of this residue can be easily separated, as F is well soluble in water, while G does not dissolve at all. The F:G mass ratio is 1.00 : 2.97. On further heating, F decomposes without a solid residue [reaction 5]. That means pure G is the final solid decomposition product of A. G contains two other elements in addition to 27.0 wt% oxygen. One of the two components of the gas mixture forming in the first decomposition step can be easily quantitated if it is absorbed in acid solution.

Some hints

  • Compound A decomposes without releasing oxygen.
  • Compound A contains two different metals.
  • Compounds A and B both contain a complex ion.
  • C is a well-known compound for any chemistry student.
  • Compound F does not contain any metal.

5.2 Give chemical formulae of A—G. You don’t have to show your calculations, but if your compounds are incorrect you may gain partial marks from correct calculations. 28.0 pt

5.3 Write balanced equations for reactions [3]—[5]. 12.0 pt

Periodic table

Figure 1

Official English text of the problem.

Answers

  • 5.1 — Reactions [1] and [2]

\[ \boxed{\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}} \tag{1} \]

\[ \boxed{\ce{2Cu^2+ + 4I- -> 2CuI(s) + I2}} \tag{2} \]

  • 5.2 — Formulae of A–G
Compound Formula Name SMILES or formal formula-unit representation
A \(\ce{[Cu(NH3)4](MnO4)2}\) Tetraamminecopper(II) permanganate [Cu+2](<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[O-][Mn](=O)(=O)=O.[O-][Mn](=O)(=O)=O
B \(\ce{[Cu(NH3)4]SO4 * H2O}\) Tetraamminecopper(II) sulfate monohydrate [Cu+2](<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[O-]S(=O)(=O)[O-].O
C \(\ce{KMnO4}\) Potassium permanganate [K+].[O-][Mn](=O)(=O)=O
D \(\ce{Cu(OH)2}\) Copper(II) hydroxide [Cu+2].[OH-].[OH-]
E \(\ce{NH4MnO4}\) Ammonium permanganate [NH4+].[O-][Mn](=O)(=O)=O
F \(\ce{NH4NO3}\) Ammonium nitrate [NH4+].[O-][N+](=O)[O-]
G \(\ce{CuMn2O4}\) Copper manganese oxide, or copper manganite Formal ionic inventory: [Cu+2].[Mn+3].[Mn+3].[O-2].[O-2].[O-2].[O-2]

Dative-bond SMILES use <- for donation from ammonia to copper. The four nitrogen donors form an approximately square-planar \(\ce{CuN4}\) unit; the strings do not specify axial crystal contacts. Dot-separated representations give ionic composition. D and G are extended solids, so their entries specify formula-unit inventories rather than discrete molecular structures.

  • 5.3 — Reactions [3]–[5]

\[ \boxed{\ce{ [Cu(NH3)4](MnO4)2 + 2H2O ->[\Delta] Cu(OH)2(s) + 2NH4MnO4 + 2NH3 }} \tag{3} \]

\[ \boxed{\ce{ [Cu(NH3)4](MnO4)2 ->[\Delta] CuMn2O4(s) + NH4NO3(s) + 2NH3(g) + H2O(g) }} \tag{4} \]

\[ \boxed{\ce{NH4NO3 ->[\Delta] N2O(g) + 2H2O(g)}} \tag{5} \]

Solution

5.1 — Identifying the iodometric reactions

The blue solution and the formation of an insoluble iodide identify copper(II). Iodide reduces copper(II) to copper(I), which precipitates as copper(I) iodide:

\[ \ce{2Cu^2+ + 4I- -> 2CuI(s) + I2}. \]

The iodine is titrated with thiosulfate, producing tetrathionate:

\[ \ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}. \]

The molecular and ionic species in these equations are:

Species Name SMILES
\(\ce{Cu^2+}\) Copper(II) ion [Cu+2]
\(\ce{CuI}\) Copper(I) iodide [Cu+].[I-]
\(\ce{I-}\) Iodide ion [I-]
\(\ce{I2}\) Iodine II
\(\ce{S2O3^2-}\) Thiosulfate ion [O-]S(=O)(=O)[S-]
\(\ce{S4O6^2-}\) Tetrathionate ion [O-]S(=O)(=O)SSS(=O)(=O)[O-]

Citrate, \(\ce{C6H5O7^3-}\), SMILES [O-]C(=O)CC(O)(CC(=O)[O-])C(=O)[O-], strongly complexes copper(II). Removing free copper(II) shifts reaction [2] backwards: copper(I) iodide dissolves, and the iodine originally generated by copper(II) is consumed.

Consequently, Titration I measures the other oxidizing component of A. Subsequent acidification protonates citrate, releases copper(II), and permits reaction [2] to occur again. Titration II therefore measures copper alone.

Combining reactions [1] and [2] shows that one mole of copper(II) corresponds to one mole of thiosulfate in Titration II.

5.2 — Identifying A–G

The calculations below use the supplied atomic masses; intermediate results retain guard digits.

Copper content and the composition of G.

For the \(100.0\ \mathrm{mg}\) sample,

\[ \begin{aligned} n_{\mathrm{I}}(\ce{S2O3^2-}) &=(5.000\times10^{-2})(54.12\times10^{-3})\\ &=2.706\times10^{-3}\ \mathrm{mol},\\[2mm] n_{\mathrm{II}}(\ce{S2O3^2-}) &=(5.000\times10^{-2})(5.41\times10^{-3})\\ &=2.705\times10^{-4}\ \mathrm{mol}. \end{aligned} \]

Thus,

\[ n(\ce{Cu})=2.705\times10^{-4}\ \mathrm{mol}, \qquad m(\ce{Cu})=17.19\ \mathrm{mg}. \]

The first decomposition step leaves \(85.9\%\) of the original mass. Using the stated F:G mass ratio, the amount of G obtained from \(100.0\ \mathrm{mg}\) of A is

\[ m(\mathrm G) =(100.0\ \mathrm{mg})(0.859)\frac{2.97}{1.00+2.97} =64.26\ \mathrm{mg}. \]

Its oxygen content is

\[ m(\ce{O})=(0.270)(64.26)=17.35\ \mathrm{mg}. \]

Therefore,

\[ \frac{n(\ce{O})}{n(\ce{Cu})} = \frac{0.01735/16.00}{2.705\times10^{-4}} =4.009\approx4. \]

The remaining mass belongs to the second metal:

\[ m(\text{second metal}) =64.26-17.35-17.19 =29.72\ \mathrm{mg}. \]

Per mole of copper, this is

\[ \frac{0.02972}{2.705\times10^{-4}} =109.88\ \mathrm{g\,mol^{-1}} =2(54.94)\ \mathrm{g\,mol^{-1}}. \]

This identifies two manganese atoms per copper atom, giving

\[ \boxed{\mathrm G=\ce{CuMn2O4}}. \]

The calculated oxygen percentage confirms the assignment:

\[ M(\mathrm G)=63.55+2(54.94)+4(16.00) =237.43\ \mathrm{g\,mol^{-1}}, \]

\[ w_{\ce{O}}(\mathrm G) =\frac{64.00}{237.43}\times100\% =26.955\%\approx27.0\%. \]

The oxidizing anion and the complex in A.

The titration-volume ratio is

\[ \frac{n_{\mathrm I}}{n_{\mathrm{II}}} =\frac{54.12}{5.41} =10.00. \]

Thus, the non-copper oxidant accepts ten electrons per copper atom. Since there are two manganese atoms per copper atom, each manganese accepts five electrons. This identifies the acidic reduction of permanganate to manganese(II):

\[ \ce{2MnO4- + 16H+ + 10I- -> 2Mn^2+ + 5I2 + 8H2O}. \]

Here the permanganate ion is \(\ce{MnO4-}\), SMILES [O-][Mn](=O)(=O)=O; manganese(II) is [Mn+2], and a proton is [H+]. Permanganate has tetrahedral coordination around manganese; its SMILES gives one localized resonance representation.

One permanganate ion generates iodine requiring five thiosulfate ions. Independently,

\[ n(\ce{MnO4-}) =\frac{2.706\times10^{-3}}{5} =5.412\times10^{-4}\ \mathrm{mol}, \]

\[ \frac{n(\ce{MnO4-})}{n(\ce{Cu})} =\frac{5.412\times10^{-4}}{2.705\times10^{-4}} =2.001\approx2. \]

The mass of A per mole of copper is

\[ \frac{0.1000\ \mathrm g}{2.705\times10^{-4}\ \mathrm{mol}} =369.7\ \mathrm{g\,mol^{-1}}. \]

Subtracting one copper and two permanganates leaves approximately

\[ 369.7-63.55-2(54.94+64.00) =68.3\ \mathrm{g\,mol^{-1}}. \]

The acid-absorbable gas suggests ammonia, \(\ce{NH3}\), SMILES N. Four ammonia ligands contribute

\[ 4M(\ce{NH3}) =4[14.01+3(1.008)] =68.136\ \mathrm{g\,mol^{-1}}. \]

Hence the complex ion is tetraamminecopper(II),

\[ \ce{[Cu(NH3)4]^2+}, \]

SMILES [Cu+2](<-[NH3])(<-[NH3])(<-[NH3])<-[NH3], and

\[ \boxed{\mathrm A=\ce{[Cu(NH3)4](MnO4)2}}. \]

Its calculated molar mass is

\[ M(\mathrm A)=369.566\ \mathrm{g\,mol^{-1}}. \]

This formula predicts

\[ V_{\mathrm I} =\frac{10(0.1000/369.566)}{0.05000} =0.0541175\ \mathrm{dm^3} =54.12\ \mathrm{cm^3}, \]

\[ V_{\mathrm{II}} =\frac{0.1000/369.566}{0.05000} =0.00541175\ \mathrm{dm^3} =5.41\ \mathrm{cm^3}, \]

matching both titrations.

Identifying B and C from the preparation.

The theoretical mass of A is

\[ m_{\mathrm{theor}}(\mathrm A) =\frac{4.90}{0.326} =15.0307\ \mathrm g, \]

corresponding to

\[ n_{\mathrm{theor}}(\mathrm A) =\frac{15.0307}{369.566} =0.0406712\ \mathrm{mol}. \]

In the exchange reaction, B supplies one tetraamminecopper(II) ion per formula unit of A, while two formula units of the monovalent permanganate salt C are required. The inferred molar masses are therefore

\[ M(\mathrm B)\approx\frac{10.00}{0.0406712} =245.9\ \mathrm{g\,mol^{-1}}, \]

\[ M(\mathrm C)\approx\frac{12.86}{2(0.0406712)} =158.1\ \mathrm{g\,mol^{-1}}. \]

The familiar purple compound C is potassium permanganate:

\[ M(\ce{KMnO4})=39.10+54.94+64.00 =158.04\ \mathrm{g\,mol^{-1}}. \]

The deep-blue copper ammine salt matching B is tetraamminecopper(II) sulfate monohydrate:

\[ \begin{aligned} M(\ce{[Cu(NH3)4]SO4 * H2O}) &=63.55+4(17.034)+96.06+18.016\\ &=245.762\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

Thus,

\[ \boxed{\mathrm B=\ce{[Cu(NH3)4]SO4 * H2O}}, \qquad \boxed{\mathrm C=\ce{KMnO4}}. \]

The preparation is

\[ \ce{ [Cu(NH3)4]SO4 * H2O + 2KMnO4 -> [Cu(NH3)4](MnO4)2(s) + K2SO4 + H2O }. \]

The other salt is potassium sulfate, \(\ce{K2SO4}\), SMILES [K+].[K+].[O-]S(=O)(=O)[O-]. Water has SMILES O.

Checking the given quantities:

\[ m(\mathrm C) =\frac{10.00}{245.762}(2)(158.04) =12.861\ \mathrm g, \]

\[ m_{\mathrm{isolated}}(\mathrm A) =\frac{10.00}{245.762}(369.566)(0.326) =4.902\ \mathrm g, \]

which round to \(12.86\ \mathrm g\) and \(4.90\ \mathrm g\).

Identifying D and E.

Heating the aqueous ammine complex promotes ammonia loss. The blue copper-containing precipitate is copper(II) hydroxide:

\[ \boxed{\mathrm D=\ce{Cu(OH)2}}. \]

Formation of one formula unit of hydroxide uses two water molecules. Two ammonia ligands become ammonium ions, and the other two are released as ammonia. The two permanganate ions consequently remain with ammonium:

\[ \boxed{\mathrm E=\ce{NH4MnO4}}. \]

The ammonium ion, \(\ce{NH4+}\), has SMILES [NH4+]. This explains why warming the synthesis mixture can introduce E as a contaminant.

Identifying F.

One formula unit of A gives one formula unit of G. From the residue mass ratio, the mass of F accompanying one mole of G is

\[ \frac{237.43}{2.97} =79.94\ \mathrm g. \]

This matches one mole of ammonium nitrate:

\[ M(\ce{NH4NO3}) =2(14.01)+4(1.008)+3(16.00) =80.052\ \mathrm{g\,mol^{-1}}. \]

Ammonium nitrate is water-soluble, contains no metal, and decomposes into gaseous products. Therefore,

\[ \boxed{\mathrm F=\ce{NH4NO3}}. \]

Its nitrate ion, \(\ce{NO3-}\), has SMILES [O-][N+](=O)[O-]; the ion is trigonal planar with delocalized bonding.

5.3 — Aqueous and thermal decomposition equations

For aqueous decomposition, the assignments of D and E give

\[ \boxed{\ce{ [Cu(NH3)4](MnO4)2 + 2H2O ->[\Delta] Cu(OH)2(s) + 2NH4MnO4 + 2NH3 }}. \tag{3} \]

For the first thermal decomposition, subtracting \(\ce{CuMn2O4}\) and \(\ce{NH4NO3}\) from the composition of A leaves exactly two ammonia molecules and one water molecule:

\[ \boxed{\ce{ [Cu(NH3)4](MnO4)2 ->[\Delta] CuMn2O4(s) + NH4NO3(s) + 2NH3(g) + H2O(g) }}. \tag{4} \]

The predicted mass decrease is

\[ \frac{2M(\ce{NH3})+M(\ce{H2O})}{M(\mathrm A)} \times100\% = \frac{2(17.034)+18.016}{369.566}\times100\% =14.093\%\approx14.1\%. \]

The predicted solid mass ratio is

\[ m(\mathrm F):m(\mathrm G) =80.052:237.43 =1.00:2.966 \approx1.00:2.97. \]

The gas readily quantified by absorption in acid is ammonia:

\[ \ce{NH3 + H+ -> NH4+}. \]

Reaction [4] is also consistent with internal electron transfer: two manganese atoms are reduced from \(+7\) to the formal \(+3\) state, accepting eight electrons in total, while one ammine nitrogen is oxidized from \(-3\) to \(+5\) in nitrate, releasing eight electrons.

Finally, ammonium nitrate decomposes without a solid residue or oxygen evolution:

\[ \boxed{\ce{NH4NO3 ->[\Delta] N2O(g) + 2H2O(g)}}. \tag{5} \]

The nitrogen-containing gas is nitrous oxide, \(\ce{N2O}\), SMILES N#[N+][O-]. It has a linear N–N–O arrangement; the SMILES represents one resonance form.

The solution was checked against the official solution.