IChO 2006 · Theory problem 8

Chemistry of Iron Oxides

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The nucleus of iron is the most stable among all elements and, therefore, iron accumulates at the core of massive red giant stars where nucleosynthesis of many elements essential for life (such as \(\ce{C}\), \(\ce{N}\), \(\ce{O}\), \(\ce{P}\), \(\ce{S}\), etc.) takes place. As a result, among heavy elements iron is quite abundant in the universe. Iron is also abundant on Earth.

Development of a technology for reducing iron oxide to iron was a key step in human civilization. Key reactions taking place in the blast furnace are summarized below.

Figure 1

8.1 Indicate the reducing agent in each reaction.

8.2 Balance reaction (3) and calculate the equilibrium constant of reaction (3) at \(1200\,^\circ\mathrm{C}\). \(\Delta H_{\mathrm f}^{\circ}(\ce{Fe2O3(s)} = -824.2\ \mathrm{kJ\,mol^{-1}}\),

\(S^\circ\ (\mathrm{J\,mol^{-1}\,K^{-1}})\): \(\ce{Fe(s)} = 27.28\), \(\ce{Fe2O3(s)} = 87.40\), \(\ce{C(s)} = 5.74\), \(\ce{CO(g)} = 197.674\), \(\ce{CO2(g)} = 213.74\)

Figure 2

In the manufacture of celadon pottery, \(\ce{Fe2O3}\) is partially reduced in a charcoal kiln to mixed oxides of \(\ce{Fe3O4}\) and \(\ce{FeO}\). The amount of the different oxides seems to be related to the “mystic” color of celadon ceramics. \(\ce{Fe3O4}\) (magnetite) itself is a mixed oxide containing \(\ce{Fe^2+}\) and \(\ce{Fe^3+}\) ions and belongs to a group of compounds with a general formula of \(\ce{AB2O4}\). The oxide ions form a face-centered cubic array. The figure shows the array of oxygens (gray circles) and representative sites for divalent A and trivalent B cations. The dark circle represents a tetrahedral site and the white circle an octahedral site.

Figure 3

8.3 How many available octahedral sites for iron ions are there in one \(\ce{AB2O4}\) unit? Certain sites are shared by neighbouring units.

\(\ce{AB2O4}\) can adopt a normal- or an inverse-spinel structure. In normal-spinel structure, two B ions occupy two of the octahedral sites and one A occupies one of the tetrahedral sites. In an inverse-spinel structure, one of the two B ions occupies a tetrahedral site. The other B ion and the one A ion occupy octahedral sites.

8.4 What percentage of available tetrahedral sites is occupied by either \(\ce{Fe^2+}\) or \(\ce{Fe^3+}\) ion in \(\ce{Fe3O4}\)?

8.5 \(\ce{Fe3O4}\) has an inverse-spinel structure. Draw the crystal field splitting pattern of \(\ce{Fe^2+}\) and fill out the electrons. The electron pairing energy is greater than the octahedral field splitting.

Official English text of the problem.

Answers

  • 8.1: (1) Carbon, \(\ce{C(s)}\); (2) carbon, \(\ce{C(s)}\); (3) carbon monoxide, \(\ce{CO}\), [C-]#[O+].
  • 8.2: \(\ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)}\). At \(1200\,^\circ\mathrm{C}\), \(\boxed{K\approx47.9}\), dimensionless, using the supplied thermodynamic data without temperature corrections.
  • 8.3: \(\boxed{4}\) available octahedral sites per \(\ce{AB2O4}\) formula unit.
  • 8.4: \(\boxed{12.5\%}\) of the available tetrahedral sites are occupied.
  • 8.5: Iron(II), \(\ce{Fe^2+}\), [Fe+2], is octahedral and high spin: \(\boxed{t_{2g}^{4}e_g^{2}}\). The lower three orbitals contain one pair and two single electrons; the upper two each contain one electron. There are four unpaired electrons.

Solution

8.1 — Reducing agents

The reducing agent is the species that is oxidized.

  1. Reaction (1) is \[ \ce{C(s) + O2(g) -> CO2(g)}. \] Carbon changes from oxidation state \(0\) to \(+4\), while oxygen changes from \(0\) to \(-2\). Thus carbon is the reducing agent.

The molecular species are dioxygen, \(\ce{O2}\), O=O, and carbon dioxide, \(\ce{CO2}\), O=C=O.

  1. Reaction (2) is \[ \ce{CO2(g) + C(s) -> 2CO(g)}. \] The carbon initially present as \(\ce{C(s)}\) changes from \(0\) to \(+2\). The carbon initially present in carbon dioxide changes from \(+4\) to \(+2\). Thus solid carbon is the reducing agent.

The product is carbon monoxide, \(\ce{CO}\), [C-]#[O+].

  1. In reaction (3), carbon in carbon monoxide changes from \(+2\) to \(+4\), while iron in iron(III) oxide changes from \(+3\) to \(0\). Thus carbon monoxide is the reducing agent.

8.2 — Balanced equation and equilibrium constant

Two iron atoms must be produced per formula unit of iron(III) oxide. Its three oxygen atoms convert three carbon monoxide molecules into carbon dioxide:

\[ \boxed{\ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)}}. \]

First obtain the standard formation enthalpy of carbon monoxide. Adding reactions (1) and (2) gives

\[ \ce{2C(s) + O2(g) -> 2CO(g)}, \]

with enthalpy

\[ -393.51+172.46=-221.05\ \mathrm{kJ\,mol^{-1}}. \]

Therefore,

\[ \Delta_{\mathrm f}H^\circ(\ce{CO(g)}) =\frac{-221.05}{2} =-110.525\ \mathrm{kJ\,mol^{-1}}. \]

Reaction (1) also gives

\[ \Delta_{\mathrm f}H^\circ(\ce{CO2(g)}) =-393.51\ \mathrm{kJ\,mol^{-1}}. \]

Using zero standard formation enthalpy for elemental iron,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=\left[2(0)+3(-393.51)\right] -\left[-824.2+3(-110.525)\right]\\ &=-24.755\ \mathrm{kJ\,mol^{-1}} \approx-24.8\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The standard reaction entropy is

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=\left[2S^\circ(\ce{Fe(s)})+3S^\circ(\ce{CO2(g)})\right]\\ &\quad-\left[S^\circ(\ce{Fe2O3(s)})+3S^\circ(\ce{CO(g)})\right]\\ &=2(27.28)+3(213.74)-87.40-3(197.674)\\ &=15.358\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

No heat capacities or phase-transition data are supplied, so use these enthalpy and entropy values without temperature corrections. At

\[ T=1200+273.15=1473.15\ \mathrm K, \]

the standard Gibbs energy change is

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ\\ &=-24.755-(1473.15)(0.015358)\\ &=-47.3796\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Since

\[ \Delta_{\mathrm r}G^\circ=-RT\ln K, \]

using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\),

\[ \ln K =\frac{47379.6}{(8.314)(1473.15)} \approx3.868, \]

and hence

\[ \boxed{K=e^{3.868}\approx47.9}. \]

For pure solid phases and ideal gases, the dimensionless equilibrium constant is

\[ K= \frac{(p_{\ce{CO2}}/p^\circ)^3} {(p_{\ce{CO}}/p^\circ)^3} = \left(\frac{p_{\ce{CO2}}}{p_{\ce{CO}}}\right)^3. \]

8.3 — Number of available octahedral sites

The oxide ion is \(\ce{O^2-}\), [O-2]. Its SMILES specifies the individual ion; the fcc arrangement is a property of the extended crystal.

A conventional cell of the fcc oxygen array contains

\[ N_{\mathrm O}=8\left(\frac18\right)+6\left(\frac12\right)=4 \]

oxide ions. Thus its oxygen content corresponds to one \(\ce{AB2O4}\) formula unit.

Octahedral sites occur at the body center and at the twelve edge centers. The body-center site belongs entirely to the cell, while each edge-center site is shared by four cells:

\[ N_{\mathrm{oct}} =1+12\left(\frac14\right) =\boxed{4}. \]

These are the available octahedral sites; only two are occupied per formula unit in either spinel arrangement described.

8.4 — Fraction of tetrahedral sites occupied

The fcc oxygen cell has eight tetrahedral sites, all inside the cell, at positions whose fractional coordinates are independently \(1/4\) or \(3/4\). Therefore, there are eight available tetrahedral sites per four oxide ions, or per \(\ce{AB2O4}\) formula unit.

Exactly one tetrahedral site is occupied per formula unit:

  • In normal spinel, its occupant is iron(II), \(\ce{Fe^2+}\), [Fe+2].
  • In inverse spinel, its occupant is iron(III), \(\ce{Fe^3+}\), [Fe+3].

Consequently, the percentage is the same for either arrangement:

\[ \boxed{ \frac{1\text{ occupied tetrahedral site}} {8\text{ available tetrahedral sites}} \times100\%=12.5\%. } \]

8.5 — Crystal-field splitting and electron occupation

In inverse-spinel magnetite, the divalent iron ion occupies an octahedral site. Iron(II), \(\ce{Fe^2+}\), has the configuration \([\mathrm{Ar}]\,3d^6\).

An octahedral crystal field splits the five \(d\) orbitals into:

  • Three lower-energy \(t_{2g}\) orbitals: \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\), each at \(-\frac25\Delta_o\) relative to the barycenter.
  • Two higher-energy \(e_g\) orbitals: \(d_{z^2}\) and \(d_{x^2-y^2}\), each at \(+\frac35\Delta_o\).

Because the pairing energy exceeds \(\Delta_o\), the configuration is high spin. The first five electrons singly occupy the five orbitals with parallel spins; the sixth pairs in a lower \(t_{2g}\) orbital:

\[ \boxed{t_{2g}^{4}e_g^{2}}. \]

The requested drawing should have energy increasing upward, with three equal-height orbital lines below two equal-height orbital lines, separated by \(\Delta_o\). Put an up-arrow and a down-arrow in one lower orbital, and one up-arrow in each of the other four orbitals. Any of the three lower orbitals may contain the pair.

This gives one electron pair and four unpaired electrons. The ionic SMILES [Fe+2] does not encode the octahedral coordination geometry or these orbital occupations.

The solution was checked against the official solution.