IChO 2022 · Theory problem 2

Chromium in Ancient and Modern Times

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Figure 1

(The photo is from the website of Palace Museum)

Black glazed porcelain is a special Chinese porcelain and was popular in the Tang and Song dynasties (ca. 1000 years ago). The ceramic ware contained iron oxides as the main color rendering agent, which was mixed with other transition metal oxides to exhibit different color appearance such as auburn, dark brown or black. Currently, the black glazed porcelain is still popular in China.

The typical black glaze is composed of Fe-containing oxides with a spinel structure. Spinel oxides have the general formula \(\ce{AB2O4}\) and the structure consists of a cubic close packing array of \(\ce{O^{2-}}\) ions in which the A cations occupy one-eighth of the tetrahedral vacancies and the B cations occupy half of the octahedral vacancies, as shown in Figure 2.1(a) for a unit cell.

Figure 2

Figure 2.1 Illustration of spinel structure

The cubic unit cell of a spinel structure can be divided into 8 cubic subunits, and the dashed lines represent the internal edges of the subunits. 4 of the subunits belong to type I, and the other 4 subunits are type II (Figure 2.1(b)). The details of adjacent subunits of type I and type II are shown in Figure 2.1(c).

2.1 How many cations of type A and B are there in a unit cell?

The black ceramic glaze with spinel structure can be produced by roasting \(\ce{Fe2O3}\) and \(\ce{Cr2O3}\) in a certain proportion in a reducing atmosphere (reaction (I)). When \(\ce{Fe2O3}\) and \(\ce{Cr2O3}\) are reacted with the mass ratio of 63.6 : 36.4, they completely transform into a pure stoichiometric compound. The product has a spinel structure, in which the tetrahedral A sites are occupied by iron cations only.

2.2 In reaction (I), which element is reduced?

2.3 Calculate the numbers of \(\ce{Fe^{3+}}\) and \(\ce{Cr^{3+}}\) in the B sites of one unit cell.

Besides doping in black ceramic glaze, chromium-containing pigments are used in painting and printing due to the abundant color originating from chromium with different oxidation states, such as +2, +3 and +6. The pigment, chrome green (\(\ce{Cr2O3}\)) can be converted into a series of other compounds (D-G) in the following process, where E, F, G are yellow, orange and red respectively.

Figure 3

2.4 Write the chemical formula of E.

2.5 Write the reaction equation of F \(\rightarrow\) G.

2.6 Choose the agent which could be H.

(A) \(\ce{FeSO4}\)

(B) \(\ce{FeCl3}\)

(C) \(\ce{ZnSO4}\)

(D) \(\ce{CuSO4}\)

The variable valence of chromium is not only important for the production of pigments, but also useful for catalysis. For example, a typical Phillips catalyst for the polymerization of ethylene consists of chromium oxide species grafted onto a porous support material such as amorphous silica. The tetracoordinated Cr(VI) specie is the core in the pre-catalyst (1), which is proposed to be reduced quickly into a six-coordinated Cr(II) species (2) by the ethylene molecules at the beginning. 2 is proposed to further catalyze polymerization of ethylene molecules.

2.7 The reaction of 1 with ethylene can be traced by UV–vis spectra, which show the light absorption of 1 at \(21500\ \mathrm{cm^{-1}}\) and the light absorption of 2 at \(16700\ \mathrm{cm^{-1}}\) in the visible region, respectively. Choose the appropriate combination of colors of 1 and 2.

(A) orange and white, respectively

(B) orange and blue, respectively

(C) blue and orange, respectively

(D) blue and white, respectively

2.8 Assuming that the Cr(II) ion in 2 is located in a regular octahedral crystal field with a splitting energy \(\Delta_{\mathrm{o}}\) of \(16000\ \mathrm{cm^{-1}}\), draw the configurations of d electrons of the Cr(II) ion in 2, and calculate the crystal field stabilization energy (CFSE) for the Cr(II) ion in 2. (Note: The pairing energy \(P\) for Cr(II) in 2 is \(23500\ \mathrm{cm^{-1}}\).)

2.9 Coordination compounds/ions exhibit paramagnetism when containing unpaired electrons, and the corresponding magnetic moment (\(\mu\)) of transition metal cations is calculated by the function \(\mu=\sqrt{n(n+2)}\mu_B\) where \(n\) is the number of unpaired electrons. Calculate the magnetic moment in term of \(\mu_B\) for the Cr(II) ion in 2.

Official English text of the problem.

Answers

  • 2.1: 8 A cations and 16 B cations per unit cell.
  • 2.2: Iron, reduced from \(\ce{Fe^{3+}}\) to \(\ce{Fe^{2+}}\).
  • 2.3: 7 \(\ce{Fe^{3+}}\) and 9 \(\ce{Cr^{3+}}\) ions in the B sites, using the normal-spinel assignment.
  • 2.4: Sodium chromate, \(\ce{Na2CrO4}\); its yellow ion is chromate, \(\ce{CrO4^{2-}}\).
  • 2.5: \(\boxed{\ce{Na2Cr2O7 + H2SO4 -> 2CrO3(s) + Na2SO4 + H2O}}\), using 98% sulfuric acid. G is chromium(VI) oxide.
  • 2.6: (A) \(\ce{FeSO4}\), iron(II) sulfate.
  • 2.7: (B): 1 is orange; 2 is blue.
  • 2.8: High-spin \(d^4\), \(t_{2g}^{3}e_g^{1}\), with four unpaired electrons. \(\boxed{\mathrm{CFSE}=-9600\ \mathrm{cm^{-1}}}\), corresponding to a stabilization magnitude of \(9600\ \mathrm{cm^{-1}}\).
  • 2.9: \(\boxed{\mu=\sqrt{24}\,\mu_B\approx4.90\,\mu_B}\).

Names, SMILES, and structural qualifications are supplied with the identifications below.

Solution

2.1 — Cation counts

The conventional spinel cell contains eight subunits of the cubic close-packed oxide array. Each accounts for four oxide ions, giving

\[ N_{\mathrm O}=8\times4=32. \]

The oxide ion is \(\ce{O^{2-}}\), SMILES [O-2].

A close-packed array containing \(N\) anions has \(2N\) tetrahedral holes and \(N\) octahedral holes. Therefore,

\[ N_A=\frac18(2\times32)=\boxed{8}, \qquad N_B=\frac12(32)=\boxed{16}. \]

Thus the cell contains eight \(\ce{AB2O4}\) formula units. Spinel is an extended crystal; a molecular SMILES cannot represent its periodic unit-cell geometry.

2.2 — Element reduced

Initially, both metals have oxidation state \(+3\) in \(\ce{Fe2O3}\) and \(\ce{Cr2O3}\).

For the normal-spinel assignment intended here, the B-site ions are trivalent. Charge neutrality of \(\ce{AB2O4}\) then requires

\[ q_A+2(+3)+4(-2)=0, \]

so \(q_A=+2\). Because the A sites contain only iron, iron must undergo reduction:

\[ \boxed{\ce{Fe^{3+} + e- -> Fe^{2+}}}. \]

The species are the iron(III) ion, [Fe+3], and the iron(II) ion, [Fe+2].

2.3 — Iron(III) and chromium(III) in the B sites

Take a \(100.0\ \mathrm g\) mixture. Using approximate molar masses

\[ M(\ce{Fe2O3})=159.7\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{Cr2O3})=152.0\ \mathrm{g\,mol^{-1}}, \]

the amounts of metal atoms are

\[ n(\ce{Fe}) =2\frac{63.6\ \mathrm g}{159.7\ \mathrm{g\,mol^{-1}}} \approx0.796\ \mathrm{mol}, \]

\[ n(\ce{Cr}) =2\frac{36.4\ \mathrm g}{152.0\ \mathrm{g\,mol^{-1}}} \approx0.479\ \mathrm{mol}. \]

Hence,

\[ n(\ce{Fe}):n(\ce{Cr})\approx1.66:1\approx5:3. \]

There are \(8+16=24\) metal ions per unit cell, so the corresponding integer populations are

\[ N_{\mathrm{Fe}}=24\frac58=15, \qquad N_{\mathrm{Cr}}=24\frac38=9. \]

All nine chromium ions must occupy B sites because A sites contain only iron. Chromium remains trivalent: the chromium(III) ion has SMILES [Cr+3].

Under the normal-spinel assignment, the eight A-site ions are \(\ce{Fe^{2+}}\). The remaining seven iron ions occupy B sites as \(\ce{Fe^{3+}}\):

\[ \boxed{N_B(\ce{Fe^{3+}})=15-8=7}, \qquad \boxed{N_B(\ce{Cr^{3+}})=9}. \]

Both the site count and charge balance agree:

\[ 7+9=16, \]

\[ 8(+2)+7(+3)+9(+3)+32(-2)=0. \]

2.4 — Identification of E

Hydrogen peroxide, \(\ce{H2O2}\), SMILES OO, oxidizes chromium(III) to chromium(VI) in alkaline solution. The yellow chromium(VI) species is the chromate ion:

\[ \boxed{\ce{CrO4^{2-}}}, \]

SMILES [O-][Cr](=O)(=O)[O-].

Because the solution contains sodium ions from sodium hydroxide, E is sodium chromate:

\[ \boxed{\ce{Na2CrO4}}, \]

SMILES [Na+].[Na+].[O-][Cr](=O)(=O)[O-].

Chromate has tetrahedral coordination around chromium. The SMILES gives a formal bonding representation; it does not explicitly show the tetrahedral geometry or the delocalization of bonding among the oxygen atoms.

2.5 — Conversion of F into G

Acidification shifts chromate toward the orange dichromate ion, \(\ce{Cr2O7^{2-}}\):

\[ \ce{2CrO4^{2-} + 2H+ <=> Cr2O7^{2-} + H2O}. \]

Dichromate has SMILES [O-][Cr](=O)(=O)O[Cr](=O)(=O)[O-]; structurally, it contains two chromium–oxygen tetrahedra sharing one oxygen atom.

Thus F is sodium dichromate, \(\ce{Na2Cr2O7}\), SMILES [Na+].[Na+].[O-][Cr](=O)(=O)O[Cr](=O)(=O)[O-].

Treatment with 98% sulfuric acid produces red chromium(VI) oxide, also called chromium trioxide, \(\ce{CrO3}\):

\[ \boxed{\ce{Na2Cr2O7 + H2SO4 -> 2CrO3(s) + Na2SO4 + H2O}}. \]

The corresponding net ionic equation is

\[ \boxed{\ce{Cr2O7^{2-} + 2H+ -> 2CrO3(s) + H2O}}. \]

Chromium remains in oxidation state \(+6\). Concentrated sulfuric acid promotes formation of the oxide through acidification and dehydration.

A formula-unit SMILES for chromium trioxide is O=[Cr](=O)=O. This representation does not show the polymeric solid structure, which contains linked chromium–oxygen tetrahedra.

The other species in these equations are:

  • Sulfuric acid, \(\ce{H2SO4}\): OS(=O)(=O)O.
  • Sodium sulfate, \(\ce{Na2SO4}\): [Na+].[Na+].[O-]S(=O)(=O)[O-].
  • Hydrogen ion, \(\ce{H+}\): [H+].
  • Water, \(\ce{H2O}\): O.

2.6 — Identification of H

Dichromate must be reduced from chromium(VI) to chromium(III). Iron(II) supplies electrons by being oxidized to iron(III), so the suitable reagent is iron(II) sulfate, \(\ce{FeSO4}\), SMILES [Fe+2].[O-]S(=O)(=O)[O-].

Each of the two chromium atoms gains three electrons, requiring six iron(II) ions:

\[ \boxed{\ce{Cr2O7^{2-} + 14H+ + 6Fe^{2+} -> 2Cr^{3+} + 7H2O + 6Fe^{3+}}}. \]

Therefore,

\[ \boxed{\text{H is option (A), }\ce{FeSO4}}. \]

2.7 — Colors of 1 and 2

Wavelength and wavenumber are related by

\[ \lambda(\mathrm{nm}) =\frac{10^7}{\widetilde{\nu}(\mathrm{cm^{-1}})}. \]

For 1,

\[ \lambda_1=\frac{10^7}{21500}\ \mathrm{nm} \approx465\ \mathrm{nm}. \]

This lies in the blue region. Absorption of blue light gives the complementary observed color, orange.

For 2,

\[ \lambda_2=\frac{10^7}{16700}\ \mathrm{nm} \approx599\ \mathrm{nm}. \]

This lies in the orange region, giving the complementary observed color, blue.

Thus,

\[ \boxed{\text{Option (B): 1 is orange and 2 is blue.}} \]

2.8 — Electron configuration and CFSE

The chromium(II) ion, \(\ce{Cr^{2+}}\), SMILES [Cr+2], has a \(3d^4\) electron configuration. This SMILES represents the metal ion alone; the stipulated octahedral coordination must be specified separately because the ligands of the supported site are not given.

In an octahedral field:

\[ E(t_{2g})=-0.4\Delta_{\mathrm o}, \qquad E(e_g)=+0.6\Delta_{\mathrm o}. \]

Since

\[ \Delta_{\mathrm o}=16000\ \mathrm{cm^{-1}} < P=23500\ \mathrm{cm^{-1}}, \]

placing the fourth electron in an \(e_g\) orbital costs less energy than pairing it in a \(t_{2g}\) orbital. The configuration is therefore high-spin:

\[ \boxed{t_{2g}^{3}e_g^{1}}. \]

For the requested drawing: draw three degenerate lower \(t_{2g}\) orbitals, each containing one electron with the same spin. Above them, draw two degenerate \(e_g\) orbitals, with one containing a single electron of that same spin and the other empty. The separation between the levels is \(\Delta_{\mathrm o}\).

There are no electron pairs, so

\[ \begin{aligned} \mathrm{CFSE} &=3(-0.4\Delta_{\mathrm o})+1(+0.6\Delta_{\mathrm o})\\ &=-0.6\Delta_{\mathrm o}\\ &=-0.6(16000\ \mathrm{cm^{-1}}). \end{aligned} \]

Thus,

\[ \boxed{\mathrm{CFSE}=-9600\ \mathrm{cm^{-1}}}. \]

The stabilization magnitude is \(9600\ \mathrm{cm^{-1}}\).

As a check, the alternative low-spin configuration \(t_{2g}^{4}\) would have one electron pair and total energy

\[ -1.6\Delta_{\mathrm o}+P =-25600+23500 =-2100\ \mathrm{cm^{-1}}, \]

which is higher than the high-spin energy by \(7500\ \mathrm{cm^{-1}}\).

2.9 — Magnetic moment

The high-spin \(t_{2g}^{3}e_g^{1}\) configuration has four unpaired electrons. Substitution into the supplied expression gives

\[ \mu=\sqrt{n(n+2)}\,\mu_B =\sqrt{4(4+2)}\,\mu_B =\sqrt{24}\,\mu_B. \]

Therefore,

\[ \boxed{\mu\approx4.90\,\mu_B}. \]

The solution was checked against the official solution.