Past exam of the chemistry A course of the University of Cambridge / 2024

Inside the infinite square well, the stationary wavefunctions can be chosen real:They vanish outside the well and at its impenetrable endpoints. A sine meets the left boundary condition; the right one requires $\sin (n\pi )=0$. The putative $n=0$ function is identically zero, so it is not a quantum state. The constants $C_n$ are fixed by wavefunction normalization.

Wavefunction normalization requires unit total probability:Choosing an irrelevant overall phase to be zero givesThe amplitude has units of inverse square root length, so $|{\Psi }_n{|}^2\mathrm{d}x$ is dimensionless.

Within the well, the potential is zero and the Hamiltonian is purely kinetic:The infinite walls are encoded in the operator's boundary conditions $\Psi (0)=\Psi (a)=0$, rather than differentiated as part of the interior expression. Sincewe obtainThese positive energy eigenvalues grow quadratically with the quantum number; the ground state has a nonzero confinement energy.

For distinct positive integers $n,m$, use the product-to-sum identity in the inner product:Every endpoint sine vanishes. For $m=n$, the normalization integral is one; the expression with a denominator $m-n$ should not be substituted directly. Thus the normalized functions form an orthonormal set, $⟨{\Psi }_m|{\Psi }_n⟩={\delta }_{mn}$.

Project the expansion onto one normalized basis function. Orthogonality removes every other coefficient:Consequently,This is a Fourier sine series in normalized form. For a square-integrable $f$, completeness gives convergence in the norm of the function space; pointwise convergence requires the usual additional regularity. If $f$ is a normalized wavefunction, the Born rule gives probabilities $|a_n{|}^2$ and ${\sum }_n|a_n{|}^2=1$.

Set $x=r\varphi$ and $a=2\pi r$. The wrapped spatial form isHence the differential expression for the quantum rotor acts asThe box expression becomes $E_n=n^2{\hslash }^2/(8m{r}^2)$, so consistency givesthe moment of inertia of a point mass a distance $r$ from the axis. Equivalently, ${\partial }_x=r^{-1}{\partial }_{\varphi }$ transforms the kinetic operator directly. Being an eigenfunction of this differential expression is not yet sufficient: the ring boundary conditions must also hold.

For a wavefunction normalized with the angular measure $\mathrm{d}\varphi$, use ${\Phi }_n(\varphi )=\sqrt{r}{\Psi }_n(r\varphi )={\pi }^{-1/2}\sin (n\varphi /2)$. This change of normalization convention does not change the energy eigenvalue.

A smooth quantum rotor with no flux or twist requires bothFor the sine family, ${\Phi }_n(0)={\Phi }_n(2\pi )=0$ holds for every positive integer $n$, butThe derivatives match only for $\boxed{n=2k,k=1,2,3,\dots }$. Equivalently, $\sin [n(\varphi +2\pi )/2]=(-1{)}^n\sin (n\varphi /2)$ must be periodic for every $\varphi$, not merely vanish at the chosen join. Odd $n$ would produce a cusp in a periodic continuation and do not belong to the smooth free-ring Hamiltonian's domain.

Solving $-i\hslash u^{\prime }(\varphi )=\lambda u(\varphi )$ gives an exponential. The periodic boundary condition quantizes its phase winding:These are normalized with respect to $\mathrm{d}\varphi$. The corresponding complex Fourier series isUnlike the wrapped sine subset, this orthonormal basis includes the constant $k=0$ state and both sine and cosine combinations. If one retains arc-length normalization instead, replace $1/\sqrt{2\pi }$ by $1/\sqrt{a}$ and integrate with $r\mathrm{d}\varphi$.

For the allowed $n=2k$ states, the normalized standing wave is a quantum superposition of two opposite angular momentum eigenstates:The Born rule therefore predictsThe expectation value is zero, but a single measurement never gives zero for these $k\ge 1$ states. Their uncertainty is $\Delta l_z=k\hslash$. The two components have the same rotor energy ${\hslash }^2{k}^2/(2I)$, so a definite energy does not imply definite angular momentum.

Differentiate the Gaussian before substituting into the Hamiltonian:ThusAn energy eigenvalue must be independent of $x$, so the coefficient of $x^2$ vanishes. Normalizability requires $\alpha >0$, givingThe constant term is the quantum harmonic oscillator's zero-point energy; a Gaussian width ${\alpha }^{-1/2}$ balances its kinetic and potential contributions.

For a normalized state and a Hermitian quantum observable $\hat{A}$, its quantum uncertainty is the standard deviation of repeated measurement outcomes:Here $⟨A⟩=⟨\Psi |\hat{A}|\Psi ⟩$. It is a property of the state and observable, not an instrumental error; the required moments must be finite.

In the position representation, $\hat{x}$ multiplies by $x$ and ${\hat{p}}_x=-i\hslash \mathrm{d}/\mathrm{d}x$. Acting on an arbitrary differentiable test function $f$ shows the commutator explicitly:Therefore $[\hat{x},{\hat{p}}_x]=i\hslash \hat{1}$, whose expectation value is $i\hslash$ in every normalized state in the relevant domains. The Robertson uncertainty principle givesThis is a lower bound, not a claim that every state saturates it.

The density $|{\Psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$ is even. Hence $⟨x⟩=0$; also ${\Psi }_0^{\prime }=-\alpha x{\Psi }_0$ gives $⟨p_x⟩=i\hslash \alpha ⟨x⟩=0$.

To evaluate the second position moment, differentiate the Gaussian integral with respect to its positive parameter:For the momentum second moment, apply the momentum operator twice:Both means vanish, soThe ground state attains the bound proved above and is therefore a minimum uncertainty state. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.

Write $\hat{N}={\hat{a}}^{†}\hat{a}$. The creation and annihilation operators obeyConsequently,The resulting function is not zero: its squared norm isThus ${\hat{a}}^{†}{\Psi }_0$ is the normalized first excited state, up to phase. More generally the creation operator raises the energy by one quantum $\hslash \omega$: ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The factor is required because applying it to a higher state does not preserve normalization.

A polar diatomic molecule can absorb microwaves through its permanent electric dipole moment. In the linear rigid rotor approximation,The electric-dipole selection rule is $\Delta J=+1$ in absorption. It givesso adjacent lines are separated by $2B_0$. ThereforeAssign a line near $m\times 21.999\mathrm{GHz}$ to $J=m-1\to m$ for positive integer $m$. The first member is $0\to 1$, the next $1\to 2$, and so on; if the recorded window misses the lowest lines, their absolute frequencies fix the integer offset. At high temperature many rotational states are populated, so the strongest line need not be $0\to 1$.

This is the leading rotational pattern, not a claim that real ZnF lacks spin or hyperfine structure. Fine splittings and centrifugal distortion can perturb the ideal lines; they do not explain the three systematically displaced progressions as economically as vibrational excitation.

The two extra progressions are rotational transitions within thermally populated $v=1$ and $v=2$ vibrational levels. Isotopic purity rules out attributing them to different isotope masses. Their rotational constants areThe Boltzmann distribution makes each higher vibrational manifold less populated, so its rotational spectrum is weaker. A Morse oscillator explores larger separations as it is excited: the bond's vibrationally averaged moment of inertia increases and the rotational spacing decreases. More precisely, $B_v$ involves an average of the inverse square separation; it is not simply the inverse square of the mean separation.

The observed decrement is constant to the given precision:This fits the leading vibration-rotation interaction law $B_v=B_e-{\alpha }_e(v+\frac{1}{2})$.

An equilibrium bond length requires $B_e$, not merely the ground-vibrational-state constant. Extrapolating the linear vibration-rotation interaction law givesUsing integer isotope masses, the reduced mass isFor two nuclei $I_e=\mu r_e^2$, soUsing $B_0$ instead gives $r_0\simeq 177.1$ pm, an effective vibrational separation. The small difference is precisely why the weaker vibrational progressions matter when the target is equilibrium geometry. Higher terms in $B_v$ and the rigid-rotor approximation limit this estimate's precision.

A vibrational fundamental is the transition $v=0\to 1$. A first vibrational overtone is $v=0\to 2$. A hot band starts from a thermally populated vibrationally excited level, such as $v=1\to 2$.

For a strictly harmonic potential and a dipole linear in displacement, the allowed transitions have $\Delta v=\pm 1$. A fundamental and a $1\to 2$ hot band then share the same frequency, while the overtone has zero intensity. Anharmonicity separates the adjacent gaps and can give overtone intensity; nonlinear displacement dependence of the dipole can also contribute. A hot band is defined by its lower state, not merely by being weak.

The strong band is $0\to 1$, while the weak band is the first vibrational overtone $0\to 2$. Let $w={\tilde{\omega }}_e$ and $q={\tilde{\omega }}_e{x}_e$. Subtracting the Morse oscillator levels givesTwice the fundamental is $881.2\mathrm{c}{\mathrm{m}}^{-1}$, slightly above the overtone because the higher vibrational gap is smaller. Eliminate $w$:soThe implied dimensionless anharmonicity constant is $x_e\simeq 0.004164$. The overtone's weakness follows from its being forbidden in the leading harmonic/linear-dipole model, rather than a lack of population in its lower state.

The relevant allowed absorption is $1\to 2$. Under the stipulated population-only intensity model, the vibrational levels are nondegenerate andSet this ratio to $0.10$ and solve:The energy in the exponential is the measured $0\to 1$ gap, not the harmonic constant $444.3\mathrm{c}{\mathrm{m}}^{-1}$. At this temperature, $N_2/N_0\simeq 0.0102$, so considering the first thermally excited manifold is sensible. Actual band intensity ratios also involve transition moments and rotational structure; the calculation follows the question's simplifying assumption.

Anharmonicity places the $1\to 2$ hot band atonly $3.7\mathrm{c}{\mathrm{m}}^{-1}$ below the fundamental. It can be detected with adequate spectral resolution and signal-to-noise; it is not guaranteed to appear as a separate peak in an unresolved band contour. The predicted ten-percent population-based intensity is appreciable, but overlapping rotational branches and linewidth can hide it beneath the stronger fundamental. No instrumental resolution or linewidth is supplied, so an unconditional assertion of visibility or invisibility would go beyond the information given.

The molecular symmetry operations must be closed under composition. Write the proposed nonidentity operations as $R=C_2$ and $S=\sigma$. Their product $RS$ cannot equal any proposed member:Each conclusion contradicts the distinct operations stipulated. Therefore at least a fourth symmetry operation must exist. For a mirror plane containing the twofold axis, the product generates another mirror plane; for a plane perpendicular to it, the product is inversion. A tilted plane may generate still more operations, but cannot evade group closure.

One choice is the C2v molecular point group, $\{E,C_2(z),{\sigma }_{xz},{\sigma }_{yz}\}$. The additional operation is a second reflection, sinceA bent water molecule has this symmetry. Choose the $z$ axis along its bond-angle bisector and put the molecule in the $xz$ plane: reflection in $yz$ exchanges its two hydrogen atoms, while reflection in $xz$ fixes their positions.

With $R=C_2(z)$, $S={\sigma }_{xz}$ and $T={\sigma }_{yz}$, composition givesEvery entry belongs to the set, proving closure. $E$ is the identity element; every element is its own inverse. Associativity follows because these are actual spatial transformations: composing functions is associative. Thus all group axioms hold, and the group is isomorphic to the Klein four-group.

For a concrete matrix check, use $E=\mathrm{diag}(1,1,1)$, $R=\mathrm{diag}(-1,-1,1)$, $S=\mathrm{diag}(1,-1,1)$ and $T=\mathrm{diag}(-1,1,1)$. Their matrix products reproduce the table.

A symmetry-closed orbital basis, often called a complete basis in this context, is a linearly independent set whose span is carried into itself by every group operation. No transformed basis function requires a function outside that span. This is completeness for the chosen group representation, not a claim that a small atomic-orbital set spans the full electronic Hilbert space.

A representation matrix $D(g)$ specifies the linear transformation produced by a molecular symmetry operation $g$. With the column convention,Its character of a representation is the matrix trace,The character is unchanged by a simultaneous change of basis and is constant on a conjugacy class, which is why character tables have one column per class.

Use the ordered basis $(p_x,p_y,p_z)$ at the origin. Choose $z$ as the principal improper-rotation axis, $x,y$ along the two $C_2^{\prime }$ axes, and the diagonal mirror planes as $x=\pm y$. Use active operations with $\hat{U}(g)f(\mathbf{r})=f(g^{-1}\mathbf{r})$, so the matrix columns give the transformed orbitals. For these real p orbitals the matrices are the ordinary vector transformation matrices.

One representative from each D2d molecular point group class isFor example $S_4^{+}$ sends $p_x\mapsto p_y$, $p_y\mapsto -p_x$ and $p_z\mapsto -p_z$. Choosing the opposite sense of $S_4$ transposes its xy block but leaves every character and the final decomposition unchanged.

Taking the matrix trace of those matrices givesThe one-dimensional $p_z$ subspace has characters $(1,-1,1,-1,1)$, identifying $B_2$. The pair $(p_x,p_y)$ has characters $(2,0,-2,0,0)$, identifying $E$. ThereforeThe dimensions add to three, as required. The $p_x,p_y$ pair must be retained together: a ninety-degree improper rotation mixes them, so either one alone is not a complete symmetry basis.^[ref]

Ideal gaseous phosphorus pentachloride is a trigonal bipyramid with D3h molecular point group symmetry. There are five chlorine sites and three 3p orbitals per site, hence a fifteen-dimensional basis. A moved atom contributes zero to the character; each fixed chlorine contributes the trace of the three-dimensional polar-vector matrix acting on its local p orbitals.For example, a $C_2^{\prime }$ axis through one equatorial chlorine fixes that site but rotates its p-vector basis with trace $1-1-1=-1$. Counting that atom as $+1$ would be incorrect.

Apply molecular character reduction with class sizes $(1,2,3,1,2,3)$ and group order twelve:For $A_1^{\prime }$, this gives $(15-3+3+9)/12=2$; for $E^{\prime }$, $(30+6)/12=3$. The full result isThe dimension check is $2+1+6+2+4=15$. As an additional transparent separation, the two axial chlorines supply $A_1^{\prime }+E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$, while the three equatorial ones supply $A_1^{\prime }+A_2^{\prime }+2E^{\prime }+A_2^{\prime \prime }+E^{\prime \prime }$. These are all chlorine 3p orbitals, including both radial and tangential functions; a five-orbital ligand-sigma basis would answer a different question.

The highest expected orbital degeneracy is five. Ideal buckminsterfullerene has Ih molecular point group symmetry. Its irreducible orbital representations include five-dimensional $H_g$ and $H_u$ representations; no single-valued orbital irrep has higher dimension.

This counts distinct spatial molecular orbitals. It does not multiply by electron spin, and it excludes accidental coincidences between unrelated irreps. In a real environment, symmetry lowering can split these degeneracies.^[ref]

The ideal eclipsed ion has D4h molecular point group symmetry. Put the Mo–Mo axis along $z$, with the metal atoms at $(0,0,\pm c)$, and orient the four chloride positions around each metal along $\pm x,\pm y$.

There is a $C_4(z)$ axis, perpendicular twofold axes, a horizontal plane through the midpoint, vertical mirror planes and an inversion centre at the bond midpoint. The eclipsed arrangement makes the two chloride squares coincide in projection along $z$. These labels refer to the ideal molecular geometry, rather than small crystal-packing distortions.

Let $d_{\mu ,A}$ and $d_{\mu ,B}$ be corresponding real 4d orbitals on the upper and lower metals, with identical global-axis phases. Their symmetry-adapted linear combinations arewith the normalization shown in the negligible-overlap approximation; including overlap changes the denominator but not the symmetry. Inversion exchanges the centres and each local d orbital has even intrinsic parity, so the plus combinations are gerade and the minus combinations ungerade.The reversal of bonding phase for $d_{xz},d_{yz}$ is important: the facing lobes have opposite signs when the two atomic orbitals share the same global phase convention, so their difference is bonding. Each $e$ irrep is two-dimensional. Thuswith total dimension ten. Rotating $x,y$ by $45^{\circ }$ interchanges the names of the two delta families; all labels here retain the chloride-aligned axes fixed above.