IChO 1994 · Theory problem 3

Sulphur oxohalides: structures and identification

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Sulphur forms many different compounds with oxygen and halogens (sulphur as the central atom). These compounds are mainly molecular, and many are easily hydrolysed in water.

3.1 Write Lewis structures for molecules \(\ce{SCl2}\), \(\ce{SO3}\), \(\ce{SO2ClF}\), \(\ce{SF4}\), and \(\ce{SBrF5}\).

3.2 Carefully draw the geometries of the 5 molecules. (Disregard small deviations from "ideal" angles.)

3.3 A compound, consisting of sulphur (one atom per molecule), oxygen and one or more atoms of the elements \(\ce{F}\), \(\ce{Cl}\), \(\ce{Br}\), and \(\ce{I}\), was examined. A small amount of the substance reacted with water. It was completely hydrolyzed without any oxidation or reduction, and all reaction products dissolved. 0.1 M solutions of a series of test reagents were added to separate small portions of a diluted solution of the substance.

Which ions are being tested for in the following tests?

i) Addition of \(\ce{HNO3}\) and \(\ce{AgNO3}\).

ii) Addition of \(\ce{Ba(NO3)2}\).

iii) Adjustment to \(pH = 7\) with \(\ce{NH3}\) and addition of \(\ce{Ca(NO3)2}\).

Write the equations for the possible reactions in the tests:

iv) Addition of \(\ce{KMnO4}\) followed by \(\ce{Ba(NO3)2}\) to an acid solution of the substance.

v) Addition of \(\ce{Cu(NO3)2}\).

3.4 In practice, the tests in 3.3 gave the following results:

i) A yellowish precipitate.

ii) No precipitate.

iii) No visible reaction.

iv) The main features were that the characteristic colour of permanganate disappeared, and a white precipitate was formed upon addition of \(\ce{Ba(NO3)2}\).

v) No precipitate.

Write the formulas of the possible compounds, taking the results of these tests into account.

3.5 Finally, a simple quantitative analysis was undertaken:

7.190 g of the substance was weighed out and dissolved in water to give 250.0 \(\mathrm{cm^3}\) of a solution. To 25.00 \(\mathrm{cm^3}\) of this solution, nitric acid and enough \(\ce{AgNO3}\) was added to secure complete precipitation. After washing and drying the precipitate weighed 1.452 g. Determine the formula of the compound.

3.6 Write the equation describing the reaction of the substance with water.

If you have not found the formula for the compound, use \(\ce{SOClF}\).

Official English text of the problem.

Answers

  • 3.1
Molecule SMILES Bonds at sulphur Lone pairs on sulphur
Sulphur dichloride, \(\ce{SCl2}\) ClSCl Two single bonds 2
Sulphur trioxide, \(\ce{SO3}\) O=S(=O)=O Three double bonds 0
Sulphuryl chloride fluoride, \(\ce{SO2ClF}\) O=S(=O)(F)Cl Two oxygen double bonds and two halogen single bonds 0
Sulphur tetrafluoride, \(\ce{SF4}\) FS(F)(F)F Four single bonds 1
Bromopentafluorosulphur, \(\ce{SBrF5}\) FS(F)(F)(F)(F)Br Six single bonds 0

Each terminal halogen has three lone pairs; each oxygen in these structures has two lone pairs. All formal charges are zero.

  • 3.2 \(\ce{SCl2}\): bent; \(\ce{SO3}\): trigonal planar; \(\ce{SO2ClF}\): tetrahedral; \(\ce{SF4}\): seesaw, with an equatorial lone pair; \(\ce{SBrF5}\): octahedral.

  • 3.3

  • i) Chloride [Cl-], bromide [Br-], and iodide [I-].
  • ii) Sulphate [O-]S(=O)(=O)[O-].
  • iii) Fluoride [F-].
  • iv) Sulphite [O-]S(=O)[O-] is oxidized to sulphate, which then precipitates: \[ \ce{5SO3^{2-} + 2MnO4^- + 6H+ -> 5SO4^{2-} + 2Mn^{2+} + 3H2O} \] \[ \ce{Ba^{2+} + SO4^{2-} -> BaSO4(s)} \]

  • v) Iodide gives copper(I) iodide and iodine: \[ \ce{2Cu^{2+} + 4I^- -> 2CuI(s) + I2} \]

  • 3.4 The possible compounds are thionyl bromide, \(\ce{SOBr2}\), O=S(Br)Br, and thionyl bromide chloride, \(\ce{SOBrCl}\), O=S(Cl)Br.

  • 3.5 The intended compound is thionyl bromide chloride, \(\boxed{\ce{SOBrCl}}\). The printed masses have a small numerical inconsistency, discussed below.

  • 3.6 \[ \boxed{\ce{SOBrCl + H2O -> SO2(aq) + HBr(aq) + HCl(aq)}} \]

Solution

3.1 — Lewis structures

The total valence-electron counts are:

\[ \begin{array}{c|c} \text{Molecule}&\text{Valence electrons}\\ \hline \ce{SCl2}&6+2(7)=20\\ \ce{SO3}&6+3(6)=24\\ \ce{SO2ClF}&6+2(6)+2(7)=32\\ \ce{SF4}&6+4(7)=34\\ \ce{SBrF5}&6+6(7)=48 \end{array} \]

Using the bonds and lone pairs listed in the Answers accounts for every electron. For example, in \(\ce{SF4}\), the four bonds use eight electrons, the fluorine lone pairs use twenty-four, and the remaining two form one sulphur lone pair.

These are conventional expanded-octet Lewis structures for sulphur where necessary. SMILES records connectivity and formal bond orders; it does not explicitly show the lone pairs or the geometries described below. Ionic SMILES used later likewise do not describe a solid’s crystal lattice.

3.2 — Molecular geometries

In VSEPR counting, each single or multiple bond constitutes one electron domain.

Molecule VSEPR type Precise description of the requested drawing
\(\ce{SCl2}\) \(\mathrm{AX_2E_2}\) Four electron domains have a tetrahedral arrangement. Two positions contain chlorine atoms and two contain lone pairs, giving a bent molecule. The ideal tetrahedral reference angle is \(109.5^\circ\).
\(\ce{SO3}\) \(\mathrm{AX_3}\) Put sulphur at the centre of an equilateral triangle, with the three oxygen atoms at its vertices. All four atoms are coplanar; the angles are \(120^\circ\).
\(\ce{SO2ClF}\) \(\mathrm{AX_4}\) Put sulphur at the centre of a tetrahedron and the two oxygen atoms, chlorine, and fluorine at its four vertices. Ideal angles are \(109.5^\circ\).
\(\ce{SF4}\) \(\mathrm{AX_4E}\) Start from a trigonal bipyramid. Put the lone pair in one equatorial position, two fluorines in the other equatorial positions, and two fluorines in the axial positions. This gives a seesaw: axial–axial \(180^\circ\), equatorial–equatorial \(120^\circ\), and axial–equatorial \(90^\circ\).
\(\ce{SBrF5}\) \(\mathrm{AX_6}\) Put sulphur at the centre of an octahedron. Place bromine above sulphur, one fluorine directly below, and the other four fluorines at the corners of a square in the perpendicular plane. Adjacent bonds make \(90^\circ\), and opposite bonds make \(180^\circ\).

The lone pair in \(\ce{SF4}\) occupies an equatorial position because that position has only two \(90^\circ\) interactions with neighbouring domains, whereas an axial position would have three.

3.3 — Analytical tests

i) Nitric acid and silver nitrate

Silver(I) ions [Ag+] detect chloride, bromide, and iodide by precipitation:

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

\[ \ce{Ag+ + Br^- -> AgBr(s)} \]

\[ \ce{Ag+ + I^- -> AgI(s)} \]

The products are:

  • Silver chloride [Ag+].[Cl-]: white.
  • Silver bromide [Ag+].[Br-]: cream or pale yellow.
  • Silver iodide [Ag+].[I-]: yellow.

Silver fluoride [Ag+].[F-] is soluble, so this is not a fluoride test. Nitric acid O[N+](=O)[O-] suppresses interference from sulphite precipitation without introducing a halide.

ii) Barium nitrate

Barium ions [Ba+2] detect sulphate through formation of white, acid-insoluble barium sulphate [Ba+2].[O-]S(=O)(=O)[O-]:

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

The acidity of the hydrolysed sample matters: sulphite does not give the same persistent, acid-insoluble precipitate.

iii) Neutralization and calcium nitrate

Calcium ions [Ca+2] detect fluoride by precipitation of calcium fluoride [Ca+2].[F-].[F-]:

\[ \ce{Ca^{2+} + 2F^- -> CaF2(s)} \]

Neutralization prevents fluoride from remaining predominantly as hydrogen fluoride, F. Ammonia N accepts a proton, forming ammonium [NH4+]:

\[ \ce{HF + NH3 -> NH4+ + F^-} \]

iv) Acidic permanganate, followed by barium nitrate

Permanganate [O-][Mn](=O)(=O)=O oxidizes sulphur(IV) to sulphur(VI) and is reduced to manganese(II), [Mn+2].

Using sulphite notation, the half-reactions are:

\[ \ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-} \]

\[ \ce{MnO4^- + 8H+ + 5e^- -> Mn^{2+} + 4H2O} \]

Multiplying the first by five and the second by two gives:

\[ \boxed{\ce{5SO3^{2-} + 2MnO4^- + 6H+ -> 5SO4^{2-} + 2Mn^{2+} + 3H2O}} \]

Here water is O, and hydrogen ions are represented by [H+].

In acidic solution, sulphur(IV) occurs mainly as dissolved sulphur dioxide O=S=O and hydrogen sulphite O=S(O)[O-], rather than free sulphite. An equivalent equation using hydrogen sulphite is:

\[ \ce{5HSO3^- + 2MnO4^- + H+ -> 5SO4^{2-} + 2Mn^{2+} + 3H2O} \]

The subsequent precipitation is:

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

Permanganate can also oxidize bromide and iodide:

\[ \ce{2MnO4^- + 10X^- + 16H+ -> 2Mn^{2+} + 5X2 + 8H2O} \]

For \(X=\ce{Br}\) or \(\ce{I}\), the products are bromine BrBr or iodine II. Under sufficiently strongly acidic conditions, chloride can also be oxidized by this equation to chlorine ClCl.

Thus, disappearance of the permanganate colour alone is not specific to sulphur(IV). Formation of sulphate, demonstrated by the subsequent barium precipitate, supplies the decisive additional evidence.

v) Copper(II) nitrate

Copper(II) ions [Cu+2] oxidize iodide to iodine. The reduced copper precipitates as white copper(I) iodide [Cu+].[I-]:

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

This is the diagnostic iodide reaction relevant here.

3.4 — Possible compounds

Under the qualitative-test assumptions of the problem:

  1. The yellowish silver precipitate indicates bromide and/or iodide. Chloride may also be present in a mixed precipitate.
  2. The negative copper(II) test excludes iodide. Therefore, bromide is present.
  3. The negative calcium test excludes fluoride.
  4. No initial barium precipitate means sulphate is absent. Formation of barium sulphate after oxidation shows that the original hydrolysate contains sulphur(IV).

Because hydrolysis involved no oxidation or reduction, the sulphur in the original compound also had oxidation state \(+4\).

Let its composition be \(\ce{SO_{a}X_{b}}\), where \(X\) represents monovalent halogens. With oxygen at \(-2\) and halogens at \(-1\),

\[ 2a+b=4. \]

Both oxygen and halogen must be present, so \(a\geq1\) and \(b\geq1\). The only integer solution is

\[ a=1,\qquad b=2. \]

The compound is consequently a thionyl dihalide. Since bromine must occur, chlorine may occur, and fluorine and iodine are excluded, the possibilities are:

\[ \boxed{\ce{SOBr2}\quad\text{or}\quad\ce{SOBrCl}}. \]

Their names and SMILES are thionyl bromide O=S(Br)Br and thionyl bromide chloride O=S(Cl)Br. The analysis does not determine an absolute configuration at sulphur in the mixed halide.

3.5 — Quantitative identification

The mass of original substance represented by the aliquot is

\[ m_{\text{aliquot}} = 7.190\ \mathrm{g} \left(\frac{25.00\ \mathrm{cm^3}}{250.0\ \mathrm{cm^3}}\right) = 0.7190\ \mathrm{g}. \]

The problem gives no atomic masses. Using standard rounded atomic masses in \(\mathrm{g\,mol^{-1}}\),

\[ \begin{aligned} M(\ce{S})&=32.06,&M(\ce{O})&=16.00,\\ M(\ce{Cl})&=35.45,&M(\ce{Br})&=79.90,&M(\ce{Ag})&=107.87, \end{aligned} \]

gives

\[ \begin{aligned} M(\ce{AgCl})&=143.32\ \mathrm{g\,mol^{-1}},\\ M(\ce{AgBr})&=187.77\ \mathrm{g\,mol^{-1}}. \end{aligned} \]

For thionyl bromide, each mole gives two moles of silver bromide:

\[ m_{\text{precipitate}} = \frac{0.7190\ \mathrm{g}}{207.86\ \mathrm{g\,mol^{-1}}} \left(2\times187.77\ \mathrm{g\,mol^{-1}}\right) = 1.299\ \mathrm{g}. \]

For thionyl bromide chloride, each mole gives one mole each of silver bromide and silver chloride:

\[ m_{\text{precipitate}} = \frac{0.7190\ \mathrm{g}}{163.41\ \mathrm{g\,mol^{-1}}} \left(187.77+143.32\right)\ \mathrm{g\,mol^{-1}} = 1.457\ \mathrm{g}. \]

Candidate Predicted precipitate mass
\(\ce{SOBr2}\) \(1.299\ \mathrm{g}\)
\(\ce{SOBrCl}\) \(1.457\ \mathrm{g}\)
Experimental value \(1.452\ \mathrm{g}\)

The reported value clearly selects the mixed halide as the intended answer:

\[ \boxed{\ce{SOBrCl}}. \]

Nevertheless, its predicted mass differs from the printed value by approximately \(0.0048\ \mathrm{g}\), or \(0.33\%\). Ordinary atomic-mass rounding does not explain that difference.

3.6 — Hydrolysis

Thionyl bromide chloride hydrolyses to sulphur dioxide, hydrogen bromide Br, and hydrogen chloride Cl:

\[ \boxed{\ce{SOBrCl + H2O -> SO2(aq) + HBr(aq) + HCl(aq)}} \]

The aqueous designation respects the statement that all products dissolved.

Using the conventional sulphurous-acid representation, sulphurous acid O=S(O)O, the same hydrolysis is commonly written:

\[ \ce{SOBrCl + 2H2O -> H2SO3 + HBr + HCl}. \]

Here \(\ce{H2SO3}\) denotes the aqueous sulphur(IV) system. Sulphur remains at oxidation state \(+4\), so the hydrolysis involves no redox change.

The solution was checked against the official solution.