IChO 2001 · Theory problem 5

Halogen Chemistry

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Halogens in their reactions among themselves and with a variety of other elements give rise to a large number of compounds with diverse structure, bonding and chemical behaviour. Metal halides, halogen derivatives and interhalogens represent major types of halogen compounds.

(A) Photography

A "black and white" photographic film contains a coating of silver bromide on a support such as cellulose acetate.

5.1 Write the photochemical reaction that occurs when light falls on \(\ce{AgBr(s)}\) coated on a film.

5.2 During the developing process, unexposed \(\ce{AgBr}\) is washed away by complexation of Ag(I) by sodium thiosulphate solution. Write down this chemical reaction.

5.3 These washings are often disposed of as waste. However, metallic silver can be recovered from them by adding cyanide, followed by zinc. Write down the reactions involved.

(B) Shapes, spectra and reactivity

The most reactive halogen, fluorine, reacts with other halogens \(\ce{Cl2}\), \(\ce{Br2}\) and \(\ce{I2}\) under controlled conditions giving a tetra-atomic, hexa-atomic and an octa-atomic molecule, respectively.

5.4 Write the formulae and 3-dimensional structures of these interhalogen molecules on the basis of VSEPR theory. Show the disposition of the lone pairs on the central atom, where appropriate.

A mixture of iodine vapour and chlorine gas when fed into a mass spectrometer gave two sets (A and B) of mass spectral peaks corresponding to molecular ions of two chemical species at \(m/z\):

A: 162, 164

B: 464, 466, 468, 470, 472, 474, 476

5.5 Identify the molecular species corresponding to \(m/z = 162,\ 164,\ 466\) and \(476\). Draw the structure of the heaviest species (\(m/z = 476\)) indicating clearly the lone pairs on atom(s) of I (iodine). Show the isotopic composition of each species.

In aqueous medium chlorine gas oxidises sodium thiosulphate to an ion containing the highest oxidation state of sulphur.

5.6 Write down the chemical equation for this reaction.

5.7 Write down the Lewis dot structure of the thiosulphate ion. Circle the sulphur atom that has the lower oxidation state.

5.8 Chlorine dioxide reacts with sodium hydroxide as shown below. Identify the products X and Y (both containing chlorine) and balance the equation.

5.9 Reaction of chlorine an alkali is used by manufacturing bleach. Write the chemical reaction for its formation.

5.10 Write the oxidation state(s) of chlorine in bleach.

(C) Alkali metal halides and X-ray crystallography

X-ray crystallography reveals many aspects of the structure of metal halides. The radius ratio \((r_+/r_-)\) is a useful parameter to rationalise their structure and stability. A table of radius ratio \((r_+/r_-)\) for some alkali halides with radius ratio (\(r_-\) kept constant) is shown schematically for \(\ce{NaCl}\)-type and \(\ce{CsCl}\)-type crystal structures.

Figure 1
\(\ce{Li+}\) \(\ce{Na+}\) \(\ce{K+}\) \(\ce{Rb+}\) \(\ce{Cs+}\)
\(\ce{Cl-}\) 0.33 0.52 0.74 0.82 0.93
\(\ce{Br-}\) 0.31 0.49 0.68 0.76 0.87
\(\ce{I-}\) 0.28 0.44 0.62 0.69 0.78

5.11 For a given anion, the graph for \(\ce{NaCl}\)-type structure levels off at low \(r_+/r_-\) values because of

(a) cation-cation contact along the face diagonal. \(\square\)

(b) anion-anion contact along the face diagonal. \(\square\)

(c) cation-anion contact along the cell edge. \(\square\)

(Mark X in the correct box.)

5.12 Which among the halides \(\ce{LiBr}\), \(\ce{NaBr}\) and \(\ce{RbBr}\) is likely to undergo phase transition from \(\ce{NaCl}\)-type to \(\ce{CsCl}\)-type structure with change of temperature and / or pressure?

5.13 Show by calculation the radius ratio \((r_+/r_-)\) at which the energy of \(\ce{CsCl}\)-type structure levels off.

5.14 Using CuKα X-rays (\(\lambda = 154\ \mathrm{nm}\)), diffraction by a \(\ce{KCl}\) crystal (fcc structure) is observed at an angle (\(\theta\)) of \(14.2^\circ\). Given that (i) diffraction takes place from the planes with \(h^2+k^2+l^2=4\), (ii) in a cubic crystal \(d_{hkl}=a/(h^2+k^2+l^2)^{1/2}\), where "d" is the distance between adjacent hkl planes and "a" is a lattice parameter, and (iii) reflections in an fcc structure can occur only from planes with "all odd" or "all even" hkl (Miller) indices, calculate the lattice parameter "a" for \(\ce{KCl}\).

5.15 Indicate in the table given below the required information for the \(2^{\mathrm{nd}}\) and \(3^{\mathrm{rd}}\) nearest neighbours of a \(\ce{K+}\) ion in the \(\ce{KCl}\) lattice.

\(2^{\mathrm{nd}}\) nearest neighbours \(3^{\mathrm{rd}}\) nearest neighbours
number sign of the charge distance (pm) number sign of the charge distance (pm)

5.16 Determine the lowest value of diffraction angle \(\theta\) possible for the \(\ce{KCl}\) structure.

Official English text of the problem.

Answers

Names, SMILES, structural descriptions and derivations are supplied in the corresponding solution parts below.

  • 5.1 \[ \ce{2AgBr(s) ->[light] 2Ag(s) + Br2} \]

  • 5.2 \[ \ce{AgBr(s) + 2S2O3^2- -> [Ag(S2O3)2]^3- + Br-} \]

  • 5.3 \[ \ce{[Ag(S2O3)2]^3- + 2CN- -> [Ag(CN)2]- + 2S2O3^2-} \] \[ \ce{2[Ag(CN)2]- + Zn(s) -> 2Ag(s) + [Zn(CN)4]^2-} \]

  • 5.4 \(\ce{ClF3}\): T-shaped, two equatorial lone pairs; \(\ce{BrF5}\): square pyramidal, one lone pair opposite the apical fluorine; \(\ce{IF7}\): pentagonal bipyramidal, no iodine lone pairs.

  • 5.5 The molecular-ion isotopic compositions are \[ \begin{array}{c|l} m/z & \text{Composition}\\ \hline 162 & \ce{[^{127}I ^{35}Cl]^{+\bullet}}\\ 164 & \ce{[^{127}I ^{37}Cl]^{+\bullet}}\\ 466 & \ce{[^{127}I2 ^{35}Cl5 ^{37}Cl]^{+\bullet}}\\ 476 & \ce{[^{127}I2 ^{37}Cl6]^{+\bullet}} \end{array} \] The heavier species derives from the planar iodine trichloride dimer, \(\ce{I2Cl6}\), with two bridging chlorines. Each iodine in the neutral parent has two lone pairs perpendicular to the molecular plane.

  • 5.6 \[ \ce{S2O3^2- + 4Cl2 + 5H2O -> 2SO4^2- + 8Cl- + 10H+} \]

  • 5.7 A Lewis contributor is [O-]S(=O)(=O)[S-]. Circle the terminal sulphur: its oxidation state is \(-1\); the central sulphur has oxidation state \(+5\).

  • 5.8 The products are sodium chlorite and sodium chlorate: \[ \ce{2ClO2 + 2NaOH -> NaClO2 + NaClO3 + H2O} \]

  • 5.9 \[ \ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O} \] using cold, dilute alkali.

  • 5.10 Chlorine is \(+1\) in hypochlorite and \(-1\) in the accompanying chloride. In the active bleaching compound \(\ce{NaClO}\) itself, chlorine is \(+1\).

  • 5.11 (b) Anion-anion contact along the face diagonal.

  • 5.12 \(\ce{RbBr}\).

  • 5.13 \[ \boxed{\frac{r_+}{r_-}=\sqrt3-1=0.732} \]

  • 5.14 Literally using the printed \(\lambda=154\ \mathrm{nm}\): \[ \boxed{a=628\ \mathrm{nm}}. \] With the apparent intended wavelength, \(\lambda=154\ \mathrm{pm}\): \[ \boxed{a=628\ \mathrm{pm}}. \]

  • 5.15 Second neighbours: 12 \(\ce{K+}\) ions, at \(a/\sqrt2\). Third neighbours: 8 \(\ce{Cl-}\) ions, at \(\sqrt3a/2\). The distances are respectively 444 pm and 544 pm if \(\lambda=154\ \mathrm{pm}\), or \(4.44\times10^5\) pm and \(5.44\times10^5\) pm using the printed wavelength.

  • 5.16 \[ \boxed{\theta_{\min}=12.3^\circ} \] from the \((111)\) reflection, using the supplied fcc selection rule.

Solution

SMILES specify connectivity, not the spatial arrangements or lone pairs described below. For ionic solids, a dot-disconnected SMILES represents an ionic formula unit rather than the crystal lattice. Coordination complexes use the dative-bond SMILES extension ->.

5.1

Light causes photochemical decomposition of silver bromide, \(\ce{AgBr}\), ionic formula-unit SMILES [Ag+].[Br-]:

\[ \boxed{\ce{2AgBr(s) ->[light] 2Ag(s) + Br2}} \]

Silver(I) ions ([Ag+]) are reduced to silver atoms, while bromide ions ([Br-]) are oxidised to bromine, \(\ce{Br2}\), BrBr. The resulting metallic silver produces the dark regions of the image.

5.2

Sodium thiosulphate, [Na+].[Na+].[O-]S(=O)(=O)[S-], supplies thiosulphate ions, \(\ce{S2O3^2-}\), [O-]S(=O)(=O)[S-]. Their complexation of silver(I) allows unexposed silver bromide to dissolve:

\[ \boxed{\ce{AgBr(s) + 2S2O3^2- -> [Ag(S2O3)2]^3- + Br-}} \]

The soluble product is bis(thiosulphato)argentate(I):

[O-]S(=O)(=O)[S-]->[Ag+]<-[S-]S(=O)(=O)[O-]

The ligands bind through their terminal sulphur atoms; the idealised two-coordinate arrangement around silver is linear. The equation conserves charge: both sides have total charge \(-4\).

5.3

Cyanide, \(\ce{CN-}\), [C-]#N, replaces thiosulphate to form the very stable dicyanoargentate(I) ion:

\[ \boxed{\ce{[Ag(S2O3)2]^3- + 2CN- -> [Ag(CN)2]- + 2S2O3^2-}} \]

Dicyanoargentate(I) has SMILES [C-](#N)->[Ag+]<-[C-]#N and approximately linear carbon–silver–carbon coordination.

Zinc then reduces silver(I) to metallic silver:

\[ \boxed{\ce{2[Ag(CN)2]- + Zn(s) -> 2Ag(s) + [Zn(CN)4]^2-}} \]

Zinc changes from oxidation state \(0\) to \(+2\), supplying the two electrons required to reduce two silver(I) centres.

The zinc product is tetracyanozincate(II):

[Zn+2](<-[C-]#N)(<-[C-]#N)(<-[C-]#N)<-[C-]#N

Its four carbon donors have approximately tetrahedral coordination around zinc.

5.4

The stated total numbers of atoms give three, five and seven fluorine atoms around chlorine, bromine and iodine, respectively.

Molecule and name SMILES VSEPR description Required three-dimensional structure
\(\ce{ClF3}\), chlorine trifluoride FCl(F)F \(AX_3E_2\); trigonal-bipyramidal electron-domain arrangement T-shaped. Two fluorines occupy axial positions and one occupies an equatorial position. Both lone pairs occupy the other two equatorial positions.
\(\ce{BrF5}\), bromine pentafluoride FBr(F)(F)(F)F \(AX_5E\); octahedral electron-domain arrangement Square pyramidal. Four fluorines form a square base, the fifth is the apex, and the lone pair occupies the position opposite the apex.
\(\ce{IF7}\), iodine heptafluoride FI(F)(F)(F)(F)(F)F \(AX_7\) Pentagonal bipyramidal. Five fluorines form an equatorial pentagon; two lie on the perpendicular axis, on opposite sides. Iodine has no lone pairs.

For \(\ce{ClF3}\), equatorial lone pairs minimise the number of \(90^\circ\) interactions. In idealised \(\ce{IF7}\), adjacent equatorial bonds make \(72^\circ\), and each axial bond is perpendicular to the equatorial plane.

5.5

The relevant isotopes are \(\ce{^{127}I}\), \(\ce{^{35}Cl}\) and \(\ce{^{37}Cl}\). For singly charged molecular ions, replacing one chlorine-35 by chlorine-37 increases \(m/z\) by 2.

For set A:

\[ 127+35=162,\qquad127+37=164. \]

Thus species A is iodine monochloride, \(\ce{ICl}\), ICl.

Set B has seven members, consistent with six chlorine atoms containing \(0,1,\ldots,6\) chlorine-37 atoms. Its lowest mass satisfies

\[ 464-6(35)=254=2(127), \]

so species B is diiodine hexachloride, \(\ce{I2Cl6}\), the iodine trichloride dimer. In general,

\[ m/z=2(127)+(6-k)(35)+k(37)=464+2k. \]

The requested assignments are:

\(m/z\) Isotopic composition of molecular ion One formal, localised radical-cation SMILES
162 \(\ce{[^{127}I ^{35}Cl]^{+\bullet}}\) [127I+][35Cl]
164 \(\ce{[^{127}I ^{37}Cl]^{+\bullet}}\) [127I+][37Cl]
466 \(\ce{[^{127}I2 ^{35}Cl5 ^{37}Cl]^{+\bullet}}\) [127I]1([35Cl])([35Cl])[35Cl+][127I-]([35Cl])([37Cl])[35Cl+]1
476 \(\ce{[^{127}I2 ^{37}Cl6]^{+\bullet}}\) [127I]1([37Cl])([37Cl])[37Cl+][127I-]([37Cl])([37Cl])[37Cl+]1

The \(466\) peak does not establish which chlorine position contains chlorine-37; the SMILES selects one representative placement.

For the requested structural drawing, describe the neutral parent dimer as follows:

  • Draw a planar four-membered ring with alternating iodine and bridging chlorine atoms: iodine–chlorine–iodine–chlorine.
  • Attach two additional terminal chlorines to each iodine, outside the ring.
  • Each iodine therefore has four chlorine neighbours in an approximately square-planar arrangement.
  • Place two lone pairs on each iodine, one above and one below the molecular plane. This is an \(AX_4E_2\) arrangement.
  • Label both iodines \(\ce{^{127}I}\) and all six chlorines \(\ce{^{37}Cl}\).

A charge-separated Lewis SMILES for this neutral heaviest parent is:

[127I-]1([37Cl])([37Cl])[37Cl+][127I-]([37Cl])([37Cl])[37Cl+]1

The formal charges in this representation account for the bridging bonds; they are not oxidation states. Iodine has oxidation state \(+3\).

The observed molecular ion has one fewer electron than its neutral parent. In the localised ion representation above, one iodine retains two lone pairs, while the other has one lone pair and an unpaired electron. The mass data do not determine the actual distribution of the ion’s charge or spin.

5.6

Sulphur’s highest oxidation state is \(+6\), so the product is sulphate, \(\ce{SO4^2-}\), [O-]S(=O)(=O)[O-].

The oxidation half-reaction is

\[ \ce{S2O3^2- + 5H2O -> 2SO4^2- + 10H+ + 8e-}. \]

Chlorine is reduced according to

\[ \ce{4Cl2 + 8e- -> 8Cl-}. \]

Adding gives

\[ \boxed{\ce{S2O3^2- + 4Cl2 + 5H2O -> 2SO4^2- + 8Cl- + 10H+}}. \]

Here chlorine is ClCl, water is O, chloride is [Cl-], and the hydrogen ion is [H+]. Sodium ions are spectators.

5.7

The thiosulphate ion contains

\[ 2(6)+3(6)+2=32 \]

valence electrons. One suitable Lewis contributor has SMILES:

[O-]S(=O)(=O)[S-]

Describe the drawing precisely as follows:

  • Place one sulphur at the centre, bonded to three oxygens and one terminal sulphur.
  • Draw two sulphur–oxygen double bonds, one sulphur–oxygen single bond and one sulphur–sulphur single bond.
  • Put two lone pairs on each doubly bonded oxygen.
  • Put three lone pairs and formal charge \(-1\) on the singly bonded oxygen.
  • Put three lone pairs and formal charge \(-1\) on the terminal sulphur.
  • The central sulphur has no lone pairs. Enclose the structure in brackets with overall charge \(2-\).

There are three equivalent contributors differing in which oxygen bears the single bond and negative charge. The four bond directions around the central sulphur are approximately tetrahedral.

For oxidation states, assign sulphur–oxygen bonding electrons to oxygen and divide the sulphur–sulphur bonding pair equally. This gives

\[ \boxed{\text{central S: }+5,\qquad\text{terminal S: }-1}. \]

As a check,

\[ (+5)+(-1)+3(-2)=-2. \]

Circle the terminal sulphur atom, which is bonded only to the central sulphur.

5.8

Chlorine dioxide, \(\ce{ClO2}\), O=[Cl]=O, is a radical with chlorine in oxidation state \(+4\). In alkali it disproportionates to chlorine in oxidation states \(+3\) and \(+5\):

Product Chlorine oxidation state SMILES
Sodium chlorite, \(\ce{NaClO2}\) \(+3\) [Na+].[O-]Cl=O
Sodium chlorate, \(\ce{NaClO3}\) \(+5\) [Na+].[O-]Cl(=O)=O

Sodium hydroxide has ionic formula-unit SMILES [Na+].[OH-].

One chlorine gains one electron while another loses one electron, giving equal amounts of the two products:

\[ \boxed{\ce{2ClO2 + 2NaOH -> NaClO2 + NaClO3 + H2O}}. \]

Thus X and Y are sodium chlorite and sodium chlorate. Their label order cannot be recovered, because the scheme the problem refers to is not printed.

5.9

With cold, dilute sodium hydroxide, chlorine disproportionates to chloride and hypochlorite:

\[ \boxed{\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}}. \]

The products are sodium chloride, [Na+].[Cl-], and sodium hypochlorite, [Na+].[O-]Cl. Hypochlorite provides the bleaching action.

5.10

In chloride, \(\ce{Cl-}\), chlorine has oxidation state \(-1\).

For the hypochlorite ion, \(\ce{ClO-}\), [O-]Cl, let chlorine’s oxidation state be \(x\):

\[ x+(-2)=-1 \quad\Longrightarrow\quad x=+1. \]

Therefore the mixture prepared in 5.9 contains chlorine in oxidation states

\[ \boxed{-1\text{ and }+1}. \]

The active hypochlorite component itself contains chlorine only in oxidation state \(+1\).

5.11

In the sodium chloride structure, the anions form an fcc array. Their nearest separation is along a face diagonal and equals \(a/\sqrt2\).

When neighbouring anions touch,

\[ \frac{a}{\sqrt2}=2r_-, \qquad a=2\sqrt2\,r_-. \]

Further reduction of the cation radius cannot reduce the lattice parameter in this hard-sphere model because the anions already touch. With \(r_-\) fixed, the lattice dimensions—and hence the schematic energy—level off.

Thus mark X in box (b): anion-anion contact along the face diagonal.

For reference, simultaneous cation–anion contact requires

\[ r_++r_-=\frac a2=\sqrt2\,r_-, \]

giving the familiar limiting ratio \(\sqrt2-1\approx0.414\).

5.12

The three candidates are lithium bromide ([Li+].[Br-]), sodium bromide ([Na+].[Br-]) and rubidium bromide ([Rb+].[Br-]).

Their supplied radius ratios are

\[ 0.31,\qquad0.49,\qquad0.76, \]

respectively. The geometrical limiting ratio for eight-coordinate caesium chloride, [Cs+].[Cl-], is approximately \(0.732\), as derived next.

Rubidium bromide lies near this value, where the two arrangements can compete. It is therefore the most likely candidate in the supplied model:

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

This is a comparative prediction; the radius-ratio model alone does not determine every possible high-pressure phase transition.

5.13

In the caesium chloride structure, take the anions at the cube corners and the cation at its centre.

Cation–anion contact along the body diagonal gives

\[ \sqrt3\,a=2(r_++r_-). \]

The lattice stops shrinking when adjacent corner anions touch along an edge:

\[ a=2r_-. \]

Combining,

\[ \sqrt3(2r_-)=2(r_++r_-), \]

so

\[ \boxed{\frac{r_+}{r_-}=\sqrt3-1=0.732}. \]

Below this ratio, fixed-size anions prevent further contraction of the idealised structure.

5.14

Potassium chloride has ionic formula-unit SMILES [K+].[Cl-].

Since \(h^2+k^2+l^2=4\), the reflection belongs to the \(\{200\}\) family and

\[ d_{200}=\frac a2. \]

First-order Bragg diffraction satisfies

\[ \lambda=2d_{200}\sin\theta=a\sin\theta. \]

Hence

\[ a=\frac{\lambda}{\sin14.2^\circ}. \]

Using the printed wavelength literally:

\[ a=\frac{154\ \mathrm{nm}}{\sin14.2^\circ} =627.784\ \mathrm{nm} \approx\boxed{628\ \mathrm{nm}}. \]

The stated wavelength appears to contain a unit error: CuKα radiation corresponds to approximately \(154\ \mathrm{pm}=0.154\ \mathrm{nm}\), which gives an atomic-scale lattice parameter:

\[ a=\frac{154\ \mathrm{pm}}{\sin14.2^\circ} =627.784\ \mathrm{pm} \approx\boxed{628\ \mathrm{pm}}. \]

Both interpretations are retained below.

5.15

Place a potassium ion, \(\ce{K+}\), [K+], at the origin. In the rock-salt structure:

  • Its first neighbours are six chloride ions at distance \(a/2\).
  • Its second neighbours lie at permutations of \((\pm a/2,\pm a/2,0)\). There are \(3\times4=12\), all potassium ions.
  • Its third neighbours lie at \((\pm a/2,\pm a/2,\pm a/2)\). There are \(2^3=8\), all chloride ions.

Thus

\[ r_2=\sqrt{\left(\frac a2\right)^2+\left(\frac a2\right)^2} =\frac a{\sqrt2}, \]

and

\[ r_3=\sqrt{3\left(\frac a2\right)^2} =\frac{\sqrt3\,a}{2}. \]

Using the apparent intended wavelength, the answer-sheet entries are:

\(2^{\mathrm{nd}}\): number Sign of charge Distance (pm) \(3^{\mathrm{rd}}\): number Sign of charge Distance (pm)
12 \(+\) 444 8 \(-\) 544

The unrounded distances are \(443.910\ \mathrm{pm}\) and \(543.677\ \mathrm{pm}\).

If the printed wavelength \(154\ \mathrm{nm}\) is used literally, the corresponding distances are instead

\[ \boxed{r_2=4.44\times10^5\ \mathrm{pm},\qquad r_3=5.44\times10^5\ \mathrm{pm}}. \]

The neighbour counts and charge signs are unchanged.

5.16

The lowest diffraction angle corresponds to the largest allowed plane spacing, hence the smallest nonzero allowed value of \(h^2+k^2+l^2\).

The \((100)\) and \((110)\) indices have mixed parity and are forbidden by the supplied fcc rule. The first allowed family is \(\{111\}\), for which

\[ h^2+k^2+l^2=3, \qquad d_{111}=\frac a{\sqrt3}. \]

Therefore

\[ \sin\theta_{\min} =\frac{\lambda}{2d_{111}} =\frac{\sqrt3\,\lambda}{2a}. \]

Using \(\lambda/a=\sin14.2^\circ\) from 5.14,

\[ \theta_{\min} =\sin^{-1}\!\left(\frac{\sqrt3}{2}\sin14.2^\circ\right) =12.2655^\circ. \]

Thus

\[ \boxed{\theta_{\min}=12.3^\circ}. \]

The wavelength-unit issue does not affect this angle because it depends on the ratio \(\lambda/a\).

The solution was checked against the official solution.