Question 2 of 8: Crystal Structure and Point Defects (20 marks)

Question II — Crystal Structure and Point Defects (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.

Numbering note. The source paper numbers this question's three parts 1, 2, 1 (the last sub-question repeats "1"); they are labelled II.1–II.3 below for clarity, in the order printed.

II.1 — Crystallographic planes and directions

Method. For a plane $(hkl)$, take the reciprocal intercepts $1/h,\,1/k,\,1/l$ on the $a,b,c$ axes (a zero index means the plane is parallel to that axis, an infinite intercept), mark those points, and connect them — a negative index (barred) is an intercept on the negative axis. For an HCP direction $[uvtw]$ (Miller–Bravais, four axes $a_1,a_2,a_3,c$ with $t=-(u+v)$), the vector is the sum $u\mathbf{a}_1+v\mathbf{a}_2+t\mathbf{a}_3+w\mathbf{c}$.

II.2 — Vanadium: FCC or BCC?

Given. Vanadium ($Z=23$), measured density $\rho = 5.8\ \text{g/cm}^3$, cubic unit-cell edge $a=0.303$ nm, molar mass $A_{wt}=50.94$ g/mol.

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

Approach. Compute the theoretical density $\rho = nA_{wt}/(V_c N_A)$ for both candidate structures and see which matches the measured 5.8 g/cm$^3$.

1. Unit-cell volume. $V_c = a^3 = (0.303\times10^{-7}\ \text{cm})^3 = 2.782\times10^{-23}\ \text{cm}^3$.
2. BCC candidate ($n=2$). $$\rho_{BCC} = \frac{2(50.94)}{(2.782\times10^{-23})(6.023\times10^{23})} = \boxed{6.08\ \text{g/cm}^3}$$
3. FCC candidate ($n=4$). $$\rho_{FCC} = \frac{4(50.94)}{(2.782\times10^{-23})(6.023\times10^{23})} = \boxed{12.16\ \text{g/cm}^3}$$
4. Compare. $|\rho_{BCC}-5.8| = 0.28$ vs. $|\rho_{FCC}-5.8|=6.36$ g/cm$^3$ — the BCC prediction is far closer to the measured value (within 5%, vs. FCC over-predicting by more than 2$\times$), so vanadium is BCC. This matches vanadium's known room-temperature structure.

II.3 — Solid-solubility factors and Zn/Pb in Cu

The Hume-Rothery rules govern how much of one element can dissolve substitutionally in another before a second phase must appear: (1) atomic size factor — radii should differ by less than about 15%, or lattice strain limits solubility; (2) crystal structure — extensive (complete) solubility requires the same crystal structure; (3) electronegativity — a large difference favours compound formation over solid solution; (4) valence — a higher-valence solute dissolves more easily in a lower-valence solvent than the reverse (relative valence effect).

Applying these to the table: zinc in copper — size difference $(0.133-0.128)/0.128 = 3.9\%$ (favourable, $<15\%$), electronegativity difference $|1.8-1.7|=0.1$ (small, favourable), same valence (+2, favourable), but Zn is HCP against Cu's FCC (unfavourable — three of four factors favourable). Lead in copper — size difference $(0.175-0.128)/0.128 = 36.7\%$ (well over 15%, strongly unfavourable), electronegativity difference $0.2$, multivalent (+2/+4) against Cu's +2, and although Pb is nominally FCC the size mismatch dominates (only one of four factors favourable). Prediction: zinc is substantially more soluble in copper than lead is — consistent with the real Cu–Zn system (the $\alpha$ brass field extends to roughly 35 wt% Zn) versus Cu–Pb, where Pb is practically insoluble in solid Cu and instead precipitates as discrete globules (exploited industrially in free-machining leaded brasses/bronzes).