Crystal Lattices and Symmetry Questions and Answers

Answer: C

The primitive cells are ①, ②, ③ and ④.

A primitive cell contains exactly one lattice point and has the smallest possible area, one lattice spacing squared for this lattice; it can have any shape and need not have lattice points at its corners.

• ①, a unit square with points at its four corners, holds 4 × 1/4 = 1 point.
• ②, a parallelogram on four lattice points with the same base and height as the unit square, also has unit area and one point.
• ③ and ④ are unit squares shifted off the lattice, each enclosing exactly one lattice point inside: still one point per cell.
• ⑤ is two units long and holds 4 × 1/4 + 2 × 1/2 = 2 points, so it is a conventional (non-primitive) cell.

Options stopping at ② or ③ leave out valid primitive cells, and the option including ⑤ adds a cell of double area.

Answer: B

The formula is \(\mathrm{XYZ_{3}}\).

An FCC unit cell holds 4 lattice points, here 4 Z ions. It contains 8 tetrahedral voids and 4 octahedral voids.

\[ \begin{aligned} X &= \frac{1}{6} \times 8 = \frac{4}{3}, \quad Y = \frac{1}{3} \times 4 = \frac{4}{3} \\ X : Y : Z &= \frac{4}{3} : \frac{4}{3} : 4 = 1 : 1 : 3 \end{aligned} \]

The other formulas do not follow from these counts; they come from miscounting the ions or voids per cell, for example forgetting that an FCC cell has twice as many tetrahedral as octahedral voids.

Answer: C

The key's answer is \(\frac{4}{3\sqrt{3}a^{2}}\), which treats the hexagonal lattice as a honeycomb of regular hexagons of side \(a\).

In that reading, atoms sit at the corners of hexagons of side \(a\). Each corner is shared by three hexagons, so one hexagon holds \(6 \times \frac{1}{3} = 2\) atoms, and its area is \(\frac{3\sqrt{3}}{2}a^{2}\):

\[ \begin{aligned} n &= \frac{2}{\frac{3\sqrt{3}}{2}a^{2}} \\ &= \frac{4}{3\sqrt{3}a^{2}} \approx \frac{0.77}{a^{2}} \end{aligned} \]

Strictly, the honeycomb is not a Bravais lattice; the triangular Bravais lattice with spacing \(a\) has density \(\frac{2}{\sqrt{3}a^{2}}\), which is not among the options, so the paper intends the honeycomb arrangement. \(\frac{1}{3\sqrt{3}a^{2}}\) counts only one-third of an atom per hexagon, and the other options use wrong areas.

Answer: A

It is BCC, with a primitive cell volume of \(\frac{a^{3}}{2}\).

The primitive cell volume is the scalar triple product of the primitive vectors:

\[ \begin{aligned} V &= \vec{a_{1}} \cdot (\vec{a_{2}} \times \vec{a_{3}}) \\ &= a\hat{i} \cdot \left(a\hat{j} \times \tfrac{a}{2}(\hat{i} + \hat{j} + \hat{k})\right) = \frac{a^{3}}{2} \end{aligned} \]

Lattice points generated by these vectors include all cube corners \((a\hat{i}, a\hat{j}\) and, since \(2\vec{a_{3}} - \vec{a_{1}} - \vec{a_{2}} = a\hat{k}\), also \(a\hat{k}\)) and the cube centre \(\vec{a_{3}}\). Corners plus body centre of a cube of side \(a\) is the BCC lattice, whose conventional cell of volume \(a^{3}\) contains 2 points, so the primitive cell has \(\frac{a^{3}}{2}\).

An FCC lattice has 4 points per conventional cell, giving \(\frac{a^{3}}{4}\), and needs face-centre vectors such as \(\frac{a}{2}(\hat{i} + \hat{j})\), which this set does not contain.

Answer: B

The packing fraction of BCC is close to 68 %.

A BCC cell has 2 atoms (8 corners shared eight ways, plus 1 at the centre). The atoms touch along the body diagonal, so \(4r = \sqrt{3}\,a\):

\[ \begin{aligned} \text{PF} &= \frac{2 \times \frac{4}{3}\pi r^{3}}{a^{3}}, \quad r = \frac{\sqrt{3}}{4}a \\ &= \frac{\sqrt{3}\,\pi}{8} \approx 0.680 \end{aligned} \]

74 % is the close-packed value for FCC and HCP, and 53 % is near the simple cubic value of \(\pi/6 \approx 52\) %. 81 % exceeds the close-packing limit for equal spheres, so no lattice of identical atoms can reach it.

Answer: D

The most loosely packed is the diamond structure.

Packing fractions (the fraction of volume filled by touching spheres) are: diamond \( \frac{\sqrt{3}\pi}{16} \approx 0.34 \), simple cubic \( \frac{\pi}{6} \approx 0.52 \), body-centred cubic \( \frac{\sqrt{3}\pi}{8} \approx 0.68 \), and face-centred cubic \( \frac{\sqrt{2}\pi}{6} \approx 0.74 \).

Diamond's atoms have only four nearest neighbours, set by its directed covalent bonds, which leaves much empty space. Simple cubic, with six neighbours, is the loosest of the three Bravais cubic lattices, and FCC is the most closely packed of all.