Practice paper 1: Lectures 1 to 11, ten questions with model answers

Working, in plain words. For (i), every moving thing has a wave size, $\lambda = h/p$. If that wave is tiny next to the thing you are studying, you can treat it as a ball with a path. If the wave is as big as the system, you cannot. Put the wave size beside the system size for each case.

For (ii), picture squeezing a spring that is also being pulled in. The pull-in is the Coulomb attraction. The push-back is that a squeezed wave carries more kinetic energy. Write both as a function of $\lambda$ and see where the total is lowest.

Model answer.

(i) Classical mechanics is valid when $\lambda$ is much smaller than the characteristic length of the phenomenon (1). For the electron in H₂, $\lambda$ (3.3e-10 m) is larger than the bond length (7.4e-11 m), so a quantum treatment is needed (1). For C₆₀, $\lambda$ (4.0e-12 m) is far smaller than the diameter (0.7e-9 m), so classical mechanics is acceptable (1).

(ii) The kinetic energy is $h^2/(2m\lambda^2)$, so squeezing the electron wave to a small $\lambda$ raises the kinetic energy as $1/\lambda^2$ (1). The potential energy is $-Ze^2/(4\pi\epsilon_0 r)$ and, taking $r \approx \lambda$, falls only as $-1/\lambda$ (1). The total $E \propto 1/\lambda^2 - 1/\lambda$ is large and positive at small $\lambda$, so it has a minimum at an intermediate size. That minimum is the size of the atom, so the electron does not collapse (1).

Where this is taught. The test and the table of de Broglie lengths: the role of models. Kinetic and potential energy and the energy minimum: interparticle forces.

Working, in plain words. For (i), turn the molecule and flip it. Every turn or flip that leaves it looking the same is an operation. Doing nothing counts. A three-fold axis gives two distinct turns, a third and two thirds. Count the vertical mirror planes through that axis. Then use the decision path: no axis of order above 3, one $C_3$, no perpendicular $C_2$, vertical planes present.

For (ii), the two molecules share a long axis, so ask what else one has that the other lacks. For (iii), a chiral molecule is one you cannot superimpose on its mirror image. Look for a symmetry operation that is neither a mirror nor an inversion but still includes a reflection.

Model answer.

(i) The operations are the identity $E$ (1), the two rotations $C_3$ (120°) and $C_3^2$ (240°) (1), and three vertical mirror planes $\sigma_v$ (1). The point group is $C_{3v}$.

(ii) CO₂ is $D_{\infty h}$ and N₂O is $C_{\infty v}$ (1). CO₂ has an inversion centre and N₂O does not (1).

(iii) The molecule has an improper rotation axis, $S_4$ (1). Any molecule with an improper element, a reflection or an inversion in some combination, cannot be chiral, so it is achiral even without $\sigma$ or $i$ (1).

Where this is taught. Rotation, $E$, $C_n$ and the $n - 1$ count: symmetry and rotation. Improper rotation and chirality: mirrors, inversion and improper rotation. The decision path, $C_{3v}$, $D_{\infty h}$ and $C_{\infty v}$: point groups.

Working, in plain words. A glide is a mirror followed by a slide. A screw is a turn followed by a slide, like a spiral staircase. Ask which one contains a reflection, because only reflections swap left and right. For the symbol, the first letter names the cell type. The next letters are read along $a$, then $b$, then $c$. Planes sit perpendicular to the viewing direction and axes sit parallel to it.

Model answer.

(i) A glide plane is a reflection in a plane combined with a translation, usually half a cell edge or diagonal (1). It changes handedness because it contains a reflection (1).

(ii) The symbol $m_n$ means translate $n/m$ of the cell and rotate by $360^\circ/m$, so $4_2$ translates 2/4 of the cell and rotates 90° (1). It does not change handedness, because it contains a rotation, not a reflection (1).

(iii) $P$ means a primitive cell (1). Read along $a$, $b$, $c$: a $c$ glide plane perpendicular to $a$, an $a$ glide plane perpendicular to $b$, and a $2_1$ screw axis parallel to $c$ (1).

Where this is taught. Translational symmetry and the 14 Bravais lattices: unit cells and lattices. Glides, screws and handedness: glides, screws and space groups. Reading $Pca2_1$ and which elements are parallel or perpendicular: reading space group symbols.

Working, in plain words. Think of big anions in a crate with small cations in the gaps. Say which gaps are filled. For the count, an atom on a corner is shared by 8 boxes, on a face by 2, on an edge by 4, and in the middle by none. Add up the shares.

Model answer.

(i) Cl⁻ anions are cubic close-packed and Na⁺ cations fill all the octahedral holes (1). The coordination is 6:6 (1). F means face-centred (1).

(ii) Na: $8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 1 + 3 = 4$ (1). Cl: $12 \times \tfrac{1}{4} + 1 \times 1 = 3 + 1 = 4$, so the cell holds Na₄Cl₄, four formula units (1).

(iii) In zinc blende half of the tetrahedral holes are filled and the coordination is 4:4. In antifluorite all of them are filled and the coordination is 4:8 (2 together, 1 for each structure).

Where this is taught. Packing, holes and the filling table: close packing and holes. Coordination numbers, space groups and the atom count: ionic structures. F as a lattice type: reading space group symbols.

Working, in plain words. For (i), the waves bounced off neighbouring planes only add up when the extra distance one travels is a whole number of wavelengths. That extra distance is $2d\sin\theta$. Rearrange for $\sin\theta$ and read off the angle. For (ii), compare the wave size with the size of an atom. For (iii), each Miller index is the reciprocal of the intercept, and a zero index means the plane never meets that axis. For (iv), the pattern is the reciprocal of the crystal, so big in the crystal means small in the pattern.

Model answer.

(i) The Bragg equation is $n\lambda = 2d\sin\theta$ (1). For $n = 1$: $\sin\theta = 1.5/(2 \times 3.0) = 0.25$, so $\theta = 14.5^\circ$ (1). For $n = 2$: $\sin\theta = 2 \times 1.5/6.0 = 0.5$, so $\theta = 30^\circ$ (1).

(ii) Atoms are about 1 Å across and typical interatomic distances are 1 to 3 Å, which matches the wavelength of X-rays (1). Visible light has a wavelength of about 1 µm, far too large to resolve atoms (1).

(iii) The intercept is $1/h$ on $a$, so $1/2$. With $k = 0$ the plane never meets $b$ (it is parallel to it). The intercept on $c$ is $1/1 = 1$ (2 together, 1 for the $a$ and $c$ values and 1 for $b$).

(iv) The spots sit closer together (1). Spot spacing is inversely proportional to the d-spacing, so planes further apart give spots closer together.

Where this is taught. The Bragg equation, the worked angle, and why X-rays are used: diffraction, Miller indices and Bragg’s law. The inverse relation between plane spacing and spot spacing is in the section on the reciprocal lattice on the same page.

Working, in plain words. Same structure means the Madelung constant is the same for all five, so only the charges and the distance $d$ differ. Charge counts most, so sort by $\lvert z_A z_B \rvert$ first. Break ties by size: bigger ions sit further apart and the pull is weaker. For (ii), find 0.80 in the radius ratio table.

Model answer.

(i) Order of increasing $\Delta H^0_L$: KF < NaF < LiF < CaO < MgO (1). The Born-Mayer prefactor contains $\lvert z_A z_B \rvert$, which is 1 for the three fluorides and 4 for the two oxides, so the oxides come out highest (1). Within each group the larger ions give a larger $d$ and a lower enthalpy: K⁺ > Na⁺ > Li⁺ so KF < NaF < LiF, and Ca²⁺ > Mg²⁺ so CaO < MgO (1).

(ii) A ratio of 0.80 lies between 0.732 and 1.000, so the coordination number is 8 and the structure is CsCl (1).

(iii) (a) The radius increases as the coordination number increases (1). (b) For isoelectronic cations the radius decreases as the oxidation number increases (1).

Where this is taught. Born-Mayer, charge and size trends and the values in the table (LiF 1024, NaF 911, KF 815, CaO 3566, MgO 3938 kJ mol⁻¹): lattice enthalpy. The radius ratio table and the radius trends: ionic bonding and radii.

Working, in plain words. For (i), a band conducts when it is partly full. Ask where the potassium electrons go. For (ii), follow the chain from cation size to $T_c$ one link at a time: cell size, overlap, band width, number of seats at the Fermi level, then the exponent.

Model answer.

(i) Charge moves from K to the carbon sheet, which leaves a partly filled band (1). A partly filled band has many electrons near the Fermi level that can move, so the in-plane conductivity rises (from about 20,000 S cm⁻¹ for graphite to about 1,000,000 S cm⁻¹ for KC₈) (1).

(ii) A larger cation pushes the C₆₀ molecules further apart, so the cell edge $a$ is larger (1). Less orbital overlap gives a narrower band (1). The same electrons are packed into a narrower range of energies, so $N(E_f)$ is larger (1). A larger $N(E_f)$ makes the exponent less negative, so $T_c$ is higher (1).

A very good answer adds: the effect has a limit. If the molecules are pulled too far apart the band picture fails and the material becomes a Mott-Hubbard insulator.

Where this is taught. Charge transfer, graphite and KC₈: conductors in 1D, 2D and 3D. Partly filled bands: bands and the Fermi level. BCS, $N(E_f)$, band width and Mott-Hubbard localisation: superconductivity and localisation.

Working, in plain words. For (i), count the electrons available to go into the band. For (ii), conductivity is how many carriers there are times how easily each one moves. Heating changes both. Decide which change wins. For (iii), compare the outer electrons of Ga and Si.

Model answer.

(i) Re⁶⁺ is 5d¹, so there is one electron per metal to put in the band formed by overlap of the metal $t_{2g}$ and oxygen p orbitals. The band is partly filled, so ReO₃ is metallic (1). W⁶⁺ is 5d⁰, so the band is empty and WO₃ is an insulator (1).

(ii) A semiconductor (1). The carrier concentration $n$ rises with temperature, as $n \propto \exp\{-E_g/2kT\}$, because more electrons cross the gap. The relaxation time $\tau$ falls as phonon scattering grows, but the exponential rise in $n$ wins. In a metal $n$ barely changes and $\tau$ falls, so the conductivity would fall (1).

(iii) Ga has one fewer outer electron than Si, so the result is a p-type semiconductor (1). Ga gives an unoccupied level just above the filled valence band, and the carriers are positive holes (1).

Where this is taught. ReO₃, WO₃ and the electron count: bonding sets the bands. $\sigma = ne^2\tau/m^*$ and the metal against semiconductor trend: conductivity and temperature. n-type, p-type and the dopant level: doping and p-n junctions.

Working, in plain words. For (i), the capacitor with the filling is the empty one times $\varepsilon_r$. Susceptibility is $\varepsilon_r$ minus the empty-capacitor value of 1. For (ii), ask where the Ti sits in its octahedron on each side of 120 °C and whether the structure then has an inversion centre. For (iii), read the loop: one landmark is on the polarisation axis and one on the field axis.

Model answer.

(i) $C = \varepsilon_r\varepsilon_0 A/d$, so $C/C_0 = \varepsilon_r = 3.9$ and the capacitance rises by a factor of 3.9 (1). $\chi_e = \varepsilon_r - 1 = 3.9 - 1 = 2.9$ (1).

(ii) Below $T_C$ the Ti ions sit off-centre in the TiO₆ octahedra. The structure is polar ($P4mm$) with no inversion centre, so it has a spontaneous polarisation and is ferroelectric (1). Above 120 °C the Ti is centred, the structure is $Pm\bar{3}m$ with inversion symmetry, and it is paraelectric. Heating through a transition gives the higher symmetry structure (1).

(iii) The remnant polarisation $P_r$ is the polarisation left when the field is brought back to zero (1). The coercive field $E_c$ is the reverse field needed to bring the polarisation to zero (1).

Where this is taught. Capacitance, $\varepsilon_r$ and BaTiO₃: dielectrics and capacitors. $\chi_e$, loop landmarks: polarisation loops and distortions.

Working, in plain words. For (i), match the sign and temperature behaviour of $\chi$ to the four types. For Z, put the straight line into the Curie-Weiss form and see where $1/\chi$ reaches zero. For (ii), picture two metal ions with an oxygen between them. The metal atoms never touch, so the oxygen passes the message. Ask what the Pauli principle does to two electrons that share an orbital.

Model answer.

(i) X is diamagnetic (1). Y is paramagnetic, following the Curie law $\chi = C/T$ (1). Z follows $\chi = C/(T - \theta)$ with $\theta < 0$, so it is an antiferromagnet with $\theta = -T_N$ (1).

(ii) Superexchange is indirect: it acts between magnetic atoms that are not neighbours, through a non-magnetic atom such as O (1). A partly filled $e_g$ orbital on each metal overlaps a full oxygen p orbital (1). By the Pauli principle, the two spins in the shared orbital must be opposite, so neighbouring moments point opposite ways and the order is antiferromagnetic (1).

Where this is taught. Susceptibility, dia and paramagnetism, the Curie law: magnetic moments, diamagnetism and paramagnetism. Curie-Weiss, $\theta$ and the Néel temperature, superexchange: magnetic order and exchange.