Question 4 of 7: Crystal Structure II

Question IV: Crystal Structure II (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.

Part 1 — Given. Vanadium, $\rho_{\text{meas}}=5.8\ \text{g/cm}^3$, $a=0.303$ nm, $M=50.94$ g/mol.

Find. Whether V is FCC ($n=4$) or BCC ($n=2$).

1. Predict both structures. $a^3=(3.03\times10^{-8})^3 =2.782\times10^{-23}\ \text{cm}^3$. $$\begin{aligned} \rho_{\text{BCC}}&=\frac{2\times50.94}{6.023\times10^{23}\times2.782\times10^{-23}} =6.08\ \text{g/cm}^3,\\ \rho_{\text{FCC}}&=\frac{4\times50.94}{6.023\times10^{23}\times2.782\times10^{-23}} =12.16\ \text{g/cm}^3. \end{aligned}$$
2. Compare with measured density. $5.8\ \text{g/cm}^3$ is close to $\rho_{\text{BCC}}=6.08$ and far from $\rho_{\text{FCC}}=12.16$, so vanadium is $$\boxed{\text{BCC}}.$$ (This matches vanadium's known room-temperature BCC structure.)

Part 2 — Given. The atomic radius/structure/electronegativity/valence table above for Cu, Zn, Pb.

Find. Which factors govern substitutional solid solubility (Hume-Rothery rules), and a prediction for Zn-in-Cu vs. Pb-in-Cu.

Four Hume-Rothery factors favour extensive substitutional solid solubility: (i) atomic size—solute and solvent atomic radii should differ by less than about $15\%$; (ii) crystal structure—solute and solvent should share the same crystal structure; (iii) electronegativity—a small electronegativity difference favours a solid solution, a large one favours compound formation instead; (iv) valence—similar valence is favourable (a lower-valence solvent tends to dissolve more of a higher-valence solute than the reverse).

1. Atomic-size factor (dominant, and usually decisive on its own). $$\begin{aligned} \Delta r_{\text{Zn}}&=\frac{0.133-0.128}{0.128}\times100=\boxed{3.9\%},\\ \Delta r_{\text{Pb}}&=\frac{0.175-0.128}{0.128}\times100=\boxed{36.7\%}. \end{aligned}$$ Zn easily clears the $15\%$ rule; Pb badly fails it.
2. Remaining factors. Zn: different structure (HCP vs. Cu's FCC, unfavourable) but a very small electronegativity difference ($0.1$) and identical $+2$ valence — three of four factors favourable. Pb: same FCC structure as Cu (favourable) and a modest electronegativity difference ($0.2$), with valence only partially matching ($+2$ common to both, but Pb also takes $+4$) — yet its enormous size mismatch dominates.
3. Prediction. Because the size factor is the most restrictive Hume-Rothery rule, Zn is predicted to have substantially higher solid solubility in Cu than Pb does. This matches real Cu–Zn (brass) alloys, which dissolve up to several tens of weight-percent Zn in $\alpha$-Cu, versus Cu–Pb, in which Pb is essentially insoluble in solid Cu.

Part 3 — Given. Magnesium at $T=700\,{}^{\circ}\text{C}=973\ \text{K}$, vacancy formation energy $Q_v=0.8$ eV, $M=24.304$ g/mol, $\rho=1.74$ $\text{g/cm}^{3}$, $k=8.62\times10^{-5}\ \text{eV/atom-K}$.

Find. Equilibrium vacancy concentration $N_v$ per cubic metre.

1. Atomic (lattice-site) concentration. $\rho=1.74\times10^{6}\ \text{g/m}^3$, $$N=\frac{\rho N_A}{M}=\frac{1.74\times10^{6}\times6.023\times10^{23}}{24.304} =\boxed{4.31\times10^{28}\ \text{atoms/m}^3}.$$
2. Boltzmann factor. $kT=8.62\times10^{-5}\times973=0.0839\ \text{eV}$, $$\frac{N_v}{N}=\exp\!\left(-\frac{Q_v}{kT}\right)=\exp\!\left(-\frac{0.8}{0.0839}\right) =\exp(-9.54)=\boxed{7.21\times10^{-5}}.$$
3. Vacancy concentration. $$N_v=N\times\frac{N_v}{N} =4.31\times10^{28}\times7.21\times10^{-5}=\boxed{3.11\times10^{24}\ \text{vacancies/m}^3}.$$