Question 1 of 7: FeO Rock-Salt Structure, Packing Factor, Density, and Wustite Semiconductivity

Question 1: FeO Rock-Salt Structure, Packing Factor, Density, and Wustite Semiconductivity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ionic radii $r_{Fe^{2+}}=0.074$ nm, $r_{O^{2-}}=0.140$ nm; atomic masses $M_{Fe}=55.85$, $M_O=16.0$ g/mol (page-1 table); Avogadro's number $N_A=6.02\times10^{23}$ mol$^{-1}$ (page-1 gives this as $0.602\times10^{24}$, the same value in a shifted mantissa/exponent split, a house style of this subject’s constants table, not an error); non-stoichiometry parameter $x=0.5$ for part (c).

Find. (a) A structural justification for the rock-salt structure, the atomic packing factor (APF), and the theoretical density of FeO. (b) A qualitative explanation of why Fe$_{1-x}$O conducts as a semiconductor. (c) The number of charge carriers per cm$^3$ at $x=0.5$.

Approach

The cation/anion radius ratio predicts the coordination number (CN) and hence the crystal structure by the standard radius-ratio rules; this ratio, together with the touching condition along the unit-cell edge, fixes the lattice parameter, from which the APF and density follow directly. Non-stoichiometry (missing Fe$^{2+}$ ions) forces some remaining iron ions to change valence to Fe$^{3+}$ to preserve electrical neutrality, and it is the resulting mixed-valence lattice — not free electrons in a band — that gives wustite its semiconducting behaviour.

1. (a) Radius ratio and coordination. $$\frac{r_{Fe^{2+}}}{r_{O^{2-}}}=\frac{0.074}{0.140}=0.529$$ This falls in the $0.414$–$0.732$ band, which predicts octahedral (CN 6) coordination — exactly the coordination of the rock-salt (NaCl-type) structure, in which each cation is surrounded by 6 anions and vice versa. This is why FeO is expected to adopt the rock-salt structure rather than, say, a tetrahedral (CN 4, zinc-blende-type) arrangement.
2. (a) Lattice parameter. In the rock-salt structure, cations and anions touch along the cube edge: $$a=2(r_{Fe^{2+}}+r_{O^{2-}})=2(0.074+0.140)=\boxed{0.428\ \text{nm}}$$
3. (a) Atomic packing factor. The conventional rock-salt cell contains $Z=4$ formula units (4 cations + 4 anions). The APF is the ion volume divided by the cell volume, which simplifies (using $a=2(r_c+r_a)$) to a ratio depending only on the two radii: $$\text{APF}=\frac{4\left(\tfrac43\pi r_{Fe}^3\right)+4\left(\tfrac43\pi r_O^3\right)}{a^3} =\frac{2\pi}{3}\cdot\frac{r_{Fe}^3+r_O^3}{(r_{Fe}+r_O)^3}$$ $$\text{APF}=\frac{2\pi}{3}\cdot\frac{(0.074)^3+(0.140)^3}{(0.214)^3} =\frac{2\pi}{3}\cdot\frac{0.003149}{0.009800}$$ $$\boxed{\text{APF}\approx0.673}$$
4. (a) Theoretical density. With $Z=4$ FeO formula units per cell, $M_{FeO}=55.85+16.0=71.85$ g/mol, and $a=0.428$ nm $=4.28\times10^{-8}$ cm: $$\rho=\frac{ZM_{FeO}}{N_Aa^3}=\frac{4(71.85)}{(6.02\times10^{23})(4.28\times10^{-8}\,\text{cm})^3} =\frac{287.4}{(6.02\times10^{23})(7.838\times10^{-23}\,\text{cm}^3)}$$ $$\boxed{\rho\approx6.09\ \text{g/cm}^3}$$ This is close to the handbook wustite density ($\approx5.7$–$5.9$ g/cm$^3$ for the naturally vacancy-deficient material) — reasonable agreement for the idealized, fully-stoichiometric hard-sphere model used here.
5. (b) Why Fe$_{1-x}$O is a semiconductor. In stoichiometric FeO every iron ion is Fe$^{2+}$ and there is no mechanism for electrons to move between lattice sites — the material would be an insulator. In wustite, however, a fraction of the Fe$^{2+}$ sites are simply vacant (missing cations). Because the crystal must remain electrically neutral overall, for every Fe$^{2+}$ vacancy removed (a local deficit of $+2$), two of the remaining iron ions must convert from Fe$^{2+}$ to Fe$^{3+}$ to make up the missing positive charge. The lattice therefore contains a mixture of Fe$^{2+}$ and Fe$^{3+}$ ions on equivalent sites. An electron can hop from a neighbouring Fe$^{2+}$ ion onto an Fe$^{3+}$ ion (equivalently, a positive "hole" hops the other way, from the Fe$^{3+}$ site to the Fe$^{2+}$ site), and this ion-to-ion electron hopping conducts charge through the crystal. Because conduction proceeds by hopping of a small number of these localized carriers (holes centred on Fe$^{3+}$ sites) rather than by free electrons in a conduction band, wustite is a defect (extrinsic), $p$-type semiconductor, and its conductivity scales directly with the vacancy (i.e. Fe$^{3+}$) concentration — which is exactly what part (c) quantifies.
6. (c) Charge carriers per cm$^3$ at $x=0.5$. The rock-salt unit cell has 4 cation (Fe) sites per cell. In Fe$_{1-x}$O, a fraction $x$ of those sites is vacant, so the number of vacancies per unit cell is $$n_{vac,cell}=4x=4(0.5)=2\ \text{vacancies/cell}$$ Using the same cell volume found in part (a), $a^3=7.838\times10^{-23}$ cm$^3$, and the "one charge carrier per vacancy" assumption given in the question: $$n_{carrier}=\frac{n_{vac,cell}}{a^3}=\frac{2}{7.838\times10^{-23}\,\text{cm}^3}$$ $$\boxed{n_{carrier}\approx2.55\times10^{22}\ \text{carriers/cm}^3}$$