Past exam of the materials science course of the University of Cambridge / 2016 / Paper 1 / 1

Use $y/b$ horizontally and $z/c$ vertically, and label each molecular center by its fractional height $x/a$ along the viewing direction. The four centers in one periodic cell are $(0,0)$ with height $0$ and orientation A, $(1/2,0)$ with height $1/2$ and orientation A, $(0,1/2)$ with height $1/2$ and orientation B, and $(1/2,1/2)$ with height $0$ and orientation B. Boundary images complete the projected rectangle. The A and B bonds have opposite slopes; the atoms in each bond are indistinguishable.

The height labels matter: a symmetry of the unlabelled projection need not be a translation or reflection symmetry of the three-dimensional crystal. The drawing uses the given $0.267\mathrm{nm}$ molecular bond length and an illustrative inclination, since no exact inclination angle is specified.

This is a C-centered orthorhombic lattice. The translation $(1/2,1/2,0)$ takes each A molecule to another A molecule and each B molecule to another B molecule. Thus the conventional cell has two lattice points.

A two-molecule crystal motif can be chosen as A at $(0,0,0)$ and B at $(1/2,0,1/2)$. Attaching that same motif to both C-centered lattice points produces all four molecules, hence eight iodine atoms per conventional cell.

The centers alone appear face-centered, but $(1/2,0,1/2)$ exchanges A and B orientations. It is not a translation symmetry of the oriented molecular solid. An F-centered assignment would discard essential motif information.

There are $2_1$ screw axes parallel to $[100]$ through $(y/b,z/c)=(1/4,0),(3/4,0),(1/4,1/2),(3/4,1/2)$, with periodic repeats.

For the first axis the operation in fractional coordinates isIt rotates by $180^{\circ }$ about the axis and translates by $a/2$. For example, it sends the A center at $(0,0,0)$ to the A center at $(1/2,1/2,0)$, and sends B at $(1/2,0,1/2)$ to B at $(0,1/2,1/2)$ modulo integer translations. A half-turn preserves each unoriented diatomic bond. A second application gives one full $a$ translation.

There are also ordinary twofold axes through the molecular-center rows; those should not be confused with the screw axes halfway between the corresponding rows.

The relevant planes contain $[100]$, so their normals lie in the $bc$ plane. The symmetry operations can be checked directly on the height-labelled molecules:

Reflection reverses the bond slope, and the $c/2$ translation carries A to B with the correct height.

This again exchanges A and B. Combining it with the C-centering translation gives the equivalent description with a $b/2$ glide in the same plane.

There are no ordinary mirror planes containing $[100]$ for the illustrated generic bond inclination. Reflection alone would reverse a bond slope without the necessary accompanying translation. A mirror perpendicular to $[100]$ is a different symmetry and is outside the requested zone.

Count atoms before comparing cell volumes. The molecular conventional cell has eight atoms; the metallic face-centered cubic cell has four. ThusRelative to the molecular phase,The transformation therefore produces about 22.2% contraction per atom on the stated extrapolated-volume comparison. Conversely, the molecular volume is about $28.5\%$ larger than the metallic volume; these percentages have different denominators. Compressibility would be needed to turn the extrapolated comparison into exact volumes at the transformation pressure.

At fixed temperature, a phase's Gibbs free energy satisfies $\mathrm{d}G=V\mathrm{d}p$. For the metallic-minus-molecular difference,Increasing pressure favors the smaller-volume metallic phase. With a pressure-independent volume approximation, $\Delta G(p)\simeq \Delta G(p_0)+(p-p_0)\Delta V$, so pressure can overcome an initially positive free-energy difference. The volume comparison explains the direction of stabilization; it does not by itself determine the transition pressure or the electronic origin of metallic bonding.