IChO 2015 · Theory problem 4

From one yellow powder to another: A simple inorganic riddle

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The yellow binary compound \(X_1\) was completely dissolved in concentrated nitric acid by heating, the gas evolved is 1.586 times denser than air. Upon adding an excess of barium chloride to the solution formed a white solid \(X_2\) precipitates. It was filtered. The filtrate reacts with an excess of silver sulfate solution forming a precipitate of two solids \(X_2\) and \(X_3\), also separated from solution by filtration. To the new filtrate the solution of sodium hydroxide was being added drop-wise until the solution became nearly neutral (about pH 7). At this time a yellow powder \(X_4\) (77.31 wt.% of \(\ce{Ag}\)) crystallized from the solution. The mass of \(X_4\) is nearly 2.4 times larger than that the mass of the first portion of \(X_2\).

4.1 Determine the chemical formulae of \(X_1\) – \(X_4\).

4.2 Determine the chemical formula of the gas and provide equations for all reactions in ionic or non-ionic form.

4.3 In the structural unit of \(X_1\) all atoms of only one element are in equivalent positions. Draw the structure of \(X_1\).

4.4 Predict the products of \(X_1\) interaction with:

a) excess oxygen;

b) excess of hot concentrated sulfuric acid;

c) solid \(\ce{KClO3}\) with grinding.

Write down the equations of the reactions.

Official English text of the problem.

Answers

  • 4.1
Compound Formula Name SMILES
\(X_1\) \(\ce{P4S3}\) Tetraphosphorus trisulfide, also called phosphorus sesquisulfide P12P3P1SP(S2)S3
\(X_2\) \(\ce{BaSO4}\) Barium sulfate [Ba+2].[O-]S(=O)(=O)[O-]
\(X_3\) \(\ce{AgCl}\) Silver(I) chloride [Ag+].[Cl-]
\(X_4\) \(\ce{Ag3PO4}\) Silver(I) phosphate [Ag+].[Ag+].[Ag+].[O-]P(=O)([O-])[O-]

  • 4.2 The gas is nitrogen dioxide, \(\ce{NO2}\), SMILES [O-][N+]=O. The balanced reaction equations are given below.

  • 4.3 The \(\ce{P4S3}\) molecule has a triangular base of three directly bonded phosphorus atoms. Each basal phosphorus is connected through one sulfur atom to a fourth, apical phosphorus. All three sulfur atoms are equivalent; the phosphorus atoms occupy two distinct environments.

  • 4.4(a) \[ \ce{P4S3 + 8O2 -> P4O10 + 3SO2} \]

  • 4.4(b) \[ \ce{P4S3 + 16H2SO4 -> 4H3PO4 + 19SO2 + 10H2O} \]

  • 4.4(c) \[ \ce{3P4S3 + 16KClO3 -> 3P4O10 + 9SO2 + 16KCl} \]

The disconnected SMILES for salts specify their constituent ions, rather than their crystal structures.

Solution

4.1 — Identification of \(X_1\)–\(X_4\)

The first white precipitate is consistent with barium sulfate, formed from sulfate produced during oxidation of sulfur in \(X_1\). Its precipitation from the strongly acidic solution supports this identification:

\[ X_2=\ce{BaSO4}. \]

The added barium chloride also introduces chloride ions. Subsequent addition of silver sulfate precipitates the remaining barium as more barium sulfate, while silver ions precipitate chloride as silver(I) chloride:

\[ X_3=\ce{AgCl}. \]

The final yellow precipitate is silver(I) phosphate. Using standard atomic molar masses,

\[ M(\ce{Ag3PO4}) =3(107.868)+30.974+4(15.999) =418.57\ \mathrm{g\,mol^{-1}}. \]

Its calculated silver content is

\[ w(\ce{Ag}) = \frac{3(107.868)}{418.57}\times100\% = 77.31\%, \]

matching the stated value. Therefore,

\[ X_4=\ce{Ag3PO4}. \]

The binary starting material consequently contains phosphorus and sulfur. Each mole of phosphorus atoms gives one mole of silver phosphate, while each mole of sulfur atoms gives one mole of barium sulfate in the first precipitation.

With

\[ M(\ce{BaSO4})=233.38\ \mathrm{g\,mol^{-1}}, \]

the mass ratio gives

\[ \frac{m(X_4)}{m(X_2,\text{ first portion})} = \frac{n(\mathrm P)\,M(\ce{Ag3PO4})} {n(\mathrm S)\,M(\ce{BaSO4})} \approx2.4. \]

Thus,

\[ \frac{n(\mathrm P)}{n(\mathrm S)} \approx 2.4\frac{233.38}{418.57} = 1.338 \approx\frac43. \]

The starting compound is therefore

\[ \boxed{X_1=\ce{P4S3}}. \]

As a check, its predicted precipitate mass ratio is

\[ \frac{4M(\ce{Ag3PO4})}{3M(\ce{BaSO4})} = \frac{4(418.57)}{3(233.38)} = 2.391, \]

consistent with “nearly 2.4.”

Only the first portion of barium sulfate is used here: sulfate in the second portion comes from the added silver sulfate.

4.2 — Gas identification and reaction equations

At the same temperature and pressure, gas density is proportional to molar mass. Taking the mean molar mass of air as approximately \(29.0\ \mathrm{g\,mol^{-1}}\),

\[ M_{\text{gas}} = 1.586(29.0) = 46.0\ \mathrm{g\,mol^{-1}}. \]

This identifies nitrogen dioxide, \(\ce{NO2}\), whose molar mass is approximately \(46.0\ \mathrm{g\,mol^{-1}}\). Its SMILES is [O-][N+]=O; it is a bent, odd-electron molecule.

Oxidative dissolution. Hot concentrated nitric acid oxidizes phosphorus to phosphoric acid and sulfur to sulfuric acid:

\[ \boxed{\ce{P4S3 + 38HNO3 -> 4H3PO4 + 3H2SO4 + 38NO2 + 10H2O}}. \]

The molecular species used here are:

Name Formula SMILES
Nitric acid \(\ce{HNO3}\) O=[N+]([O-])O
Phosphoric acid \(\ce{H3PO4}\) OP(=O)(O)O
Sulfuric acid \(\ce{H2SO4}\) OS(=O)(=O)O
Water \(\ce{H2O}\) O

For the electron balance, the three sulfur atoms initially have oxidation state \(-2\), so the total oxidation-state sum for phosphorus is \(+6\). Oxidation of the four phosphorus atoms to \(+5\) releases \(20-6=14\) electrons. Oxidation of three sulfur atoms from \(-2\) to \(+6\) releases \(24\) electrons. The total is \(38\), matching reduction of \(38\) nitrogen atoms from \(+5\) to \(+4\).

First precipitation. Barium chloride, SMILES [Ba+2].[Cl-].[Cl-], supplies barium ions, [Ba+2], which precipitate sulfate ions, [O-]S(=O)(=O)[O-]:

\[ \boxed{\ce{Ba^2+ + SO4^2- -> BaSO4(s)}}. \]

Phosphoric acid remains dissolved under these acidic conditions.

Second precipitation. Silver(I) sulfate, SMILES [Ag+].[Ag+].[O-]S(=O)(=O)[O-], supplies sulfate ions that remove the excess barium by the same barium sulfate precipitation reaction. Its silver(I) ions, [Ag+], also remove chloride ions, [Cl-]:

\[ \boxed{\ce{Ag+ + Cl- -> AgCl(s)}}. \]

Thus, the second precipitate contains both \(\ce{BaSO4}\) and \(\ce{AgCl}\). Excess silver ions remain in the filtrate.

Neutralization and yellow precipitation. Sodium hydroxide, SMILES [Na+].[OH-], supplies hydroxide ions, [OH-], which neutralize hydrogen ions, [H+]:

\[ \boxed{\ce{H+ + OH- -> H2O}}. \]

As the acidity decreases, silver phosphate precipitates. A net ionic equation incorporating phosphate deprotonation is

\[ \boxed{\ce{H3PO4 + 3Ag+ + 3OH- -> Ag3PO4(s) + 3H2O}}. \]

This equation does not require fully deprotonated phosphate to be the predominant dissolved phosphate species at pH 7: precipitation draws the acid–base equilibria toward further deprotonation.

4.3 — Structure of \(X_1\)

Tetraphosphorus trisulfide, \(\ce{P4S3}\), SMILES P12P3P1SP(S2)S3, has a cage structure.

To draw it precisely:

  1. Draw three phosphorus atoms at the vertices of a triangular base, with a single phosphorus–phosphorus bond along each side.
  2. Place a fourth phosphorus atom above the centre of this triangle.
  3. Connect the apical phosphorus to each basal phosphorus through a separate sulfur atom, giving three bent \(\ce{P-S-P}\) bridges.
  4. Draw all bonds as single bonds. Each phosphorus has one lone pair, and each sulfur has two lone pairs.

There are three direct phosphorus–phosphorus bonds and six phosphorus–sulfur bonds, with no sulfur–sulfur bonds.

The molecular symmetry is \(C_{3v}\). The three sulfur atoms are equivalent, as are the three basal phosphorus atoms. The apical phosphorus is distinct: it bonds to three sulfur atoms, whereas each basal phosphorus bonds to two phosphorus atoms and one sulfur atom. This satisfies the stated equivalence condition.

4.4 — Reactions of \(X_1\)

The additional species appearing in these reactions are:

Name Formula SMILES
Dioxygen \(\ce{O2}\) O=O
Tetraphosphorus decaoxide, phosphorus(V) oxide \(\ce{P4O10}\) O=P12OP3(=O)OP(=O)(O1)OP(=O)(O2)O3
Sulfur dioxide \(\ce{SO2}\) O=S=O
Potassium chlorate \(\ce{KClO3}\) [K+].[O-]Cl(=O)=O
Potassium chloride \(\ce{KCl}\) [K+].[Cl-]

The molecular \(\ce{P4O10}\) cage has four phosphorus atoms arranged tetrahedrally, six bridging oxygen atoms—one along each tetrahedral edge—and one terminal oxygen at each phosphorus. The conventional dioxygen SMILES records connectivity; ground-state dioxygen is a triplet.

(a) Excess oxygen

Combustion gives phosphorus(V) oxide and sulfur dioxide:

\[ \boxed{\ce{P4S3 + 8O2 -> P4O10 + 3SO2}}. \]

Phosphorus reaches oxidation state \(+5\), while sulfur reaches \(+4\).

(b) Excess hot concentrated sulfuric acid

Sulfuric acid acts as the oxidant and is reduced to sulfur dioxide. The phosphorus becomes phosphoric acid, and the sulfur originally present in \(\ce{P4S3}\) also becomes sulfur dioxide:

\[ \boxed{\ce{P4S3 + 16H2SO4 -> 4H3PO4 + 19SO2 + 10H2O}}. \]

The phosphorus atoms release \(14\) electrons in total, and the three sulfur atoms release

\[ 3\bigl[4-(-2)\bigr]=18 \]

electrons. Thus, \(32\) electrons are released. Each sulfuric-acid sulfur atom accepts two electrons on changing from \(+6\) to \(+4\), requiring \(16\) sulfuric acid molecules. The \(19\) sulfur dioxide molecules contain \(16\) sulfur atoms from the acid and three from the starting compound.

(c) Solid potassium chlorate with grinding

Chlorate oxidizes the phosphorus and sulfur to phosphorus(V) oxide and sulfur dioxide, while chlorine is reduced from \(+5\) to \(-1\):

\[ \boxed{\ce{3P4S3 + 16KClO3 -> 3P4O10 + 9SO2 + 16KCl}}. \]

Each \(\ce{P4S3}\) molecule releases \(32\) electrons, and each chlorate accepts six. The smallest matching totals are

\[ 3(32)=16(6)=96, \]

giving the coefficients shown.

The solution was checked against the official solution.