Question 2 of 8: Crystal Structure I: HCP Cobalt (20 marks)

Question II — Crystal Structure I: HCP Cobalt (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.

II.1 — Theoretical density of Co

Given. HCP, $c/a=1.623$, atomic radius $r=0.1253$ nm, $A_w=58.933$ g/mol.

Find. Theoretical density $\rho$.

Approach. For close-packed spheres, $a=2r$; the full hexagonal-prism unit cell contains $n=6$ atoms and has volume $V=\tfrac{3\sqrt3}{2}a^2c$. Then $\rho=nA_w/(VN_A)$.

1. Lattice parameters. $$a=2r=2(0.1253)=0.2506\ \text{nm}, \qquad c=1.623a=0.4067\ \text{nm}$$
2. Unit-cell volume. $$V=\frac{3\sqrt3}{2}a^2c=\frac{3\sqrt3}{2}(0.2506\times10^{-7}\text{cm})^2(0.4067\times10^{-7}\text{cm})=6.636\times10^{-23}\ \text{cm}^3$$
3. Density. $$\rho=\frac{nA_w}{VN_A}=\frac{6(58.933)}{(6.636\times10^{-23})(6.023\times10^{23})}=\boxed{8.85\ \text{g/cm}^3}$$

This matches the accepted density of cobalt (8.90 g/cm³) to within 0.6% — the small residual gap is typical of the theoretical/measured comparison (real Co carries a small equilibrium vacancy population and the room-temperature $c/a$ deviates slightly from the ideal close-packed 1.633).

II.2 — Directions [10̄10] and [12̄11]

Approach. Using the hexagonal $a_1,a_2,a_3,c$ axis system ($a_1,a_2$ at 120° in the basal plane, $a_3=-(a_1+a_2)$, $c$ vertical), a direction $[uvtw]$ is the vector $u\vec a_1+v\vec a_2+t\vec a_3+w\vec c$.

[10̄10] $=\vec a_1-\vec a_3=2\vec a_1+\vec a_2$ lies entirely in the basal plane, normal to a $\{10\bar10\}$ prism face. [12̄11] $=\vec a_1-2\vec a_2+\vec a_3+\vec c=-3\vec a_2+\vec c$ rises the full unit-cell height while translating $3a$ in-plane — a long, steeply-inclined pyramidal-type direction.

II.3 — Planar densities of the shaded planes

Approach. The figure's shaded faces (of the smaller rhombic sub-cell used to define the $a_1,a_2,c$ axes) are the top rhombic face — a $(0001)$-type basal plane — and the front $a\times c$ face — a $\{10\bar10\}$-type prism plane. Planar density $=$ (atoms centred on the plane)/(plane area), using only the plane's own primitive repeat area.

1. (0001) basal plane. The rhombic primitive cell (edges $a_1,a_2$ at 120°) has area $a^2\sin60^{\circ}$ and contains exactly 1 lattice point (a primitive 2-D cell always encloses one lattice point, regardless of its corner angles): $$PD_{(0001)}=\frac{1}{a^2\sin60^{\circ}}=\frac{2}{\sqrt3\,a^2}=\frac{2}{\sqrt3\,(0.2506\ \text{nm})^2}=\boxed{18.39\ \text{atoms/nm}^2}$$ (Cross-check: the full hexagonal top face has 6 corner atoms shared 1/3 each (3 hexagons meet at each vertex of the honeycomb tiling) $+$ 1 centre atom $=3$ atoms over area $\tfrac{3\sqrt3}{2}a^2$, giving the identical $2/(\sqrt3a^2)$.)
2. {10̄10} prism plane. The front $a\times c$ rectangle has 4 corner atoms, each shared among 4 such rectangles tiling that face, so 1 net atom per rectangle (the HCP mid-layer atom is offset $\tfrac13,\tfrac23$ in-plane and does not lie on this particular face): $$PD_{\{10\bar10\}}=\frac{1}{ac}=\frac{1}{(0.2506)(0.4067)\ \text{nm}^2}=\boxed{9.81\ \text{atoms/nm}^2}$$

The basal (0001) plane is nearly twice as densely packed as the {10̄10} prism plane (18.39 vs. 9.81 atoms/nm²) — consistent with (0001) being the true close-packed plane of the HCP structure (packing fraction 0.74), while prism planes are comparatively open.

II.4 — Linear densities of [10̄10] and [12̄11]

Approach. Each vector above has integer, mutually-coprime $a_1,a_2$ coefficients, so it is a primitive lattice translation: exactly one atom's worth of length is spanned (the two endpoint atoms are each shared 50/50 with the adjacent repeat), and no third lattice atom falls exactly on the segment between them (checked directly: at the vector's mid-point neither direction's in-plane fractional position coincides with the HCP mid-layer's $(\tfrac13,\tfrac23)$ offset). So $LD=1/(\text{vector length})$.

1. Vector lengths. $$|[10\bar10]|=|2\vec a_1+\vec a_2|=\sqrt3\,a=\sqrt3(0.2506)=0.4341\ \text{nm}$$ $$|[1\bar211]|=|{-3}\vec a_2+\vec c|=\sqrt{9a^2+c^2}=\sqrt{9(0.2506)^2+(0.4067)^2}=0.8548\ \text{nm}$$
2. Linear densities. $$LD_{[10\bar10]}=\frac{1}{0.4341\ \text{nm}}=\boxed{2.30\ \text{atoms/nm}}\qquad LD_{[1\bar211]}=\frac{1}{0.8548\ \text{nm}}=\boxed{1.17\ \text{atoms/nm}}$$