Quantum Chemistry Questions and Answers

Answer: C

The result is \(\hbar k\,e^{ikx}\), option C.

Apply the momentum operator to the function:

\[ \begin{aligned} \widehat{A}\,e^{ikx} &= -i\hbar\,\frac{d}{dx}e^{ikx} \\ &= -i\hbar\,(ik)\,e^{ikx} \\ &= \hbar k\,e^{ikx} \end{aligned} \]

since \(-i \times i = 1\). So e^ikx is an eigenfunction of the momentum operator with eigenvalue \(\hbar k\). Option A would mean the operator had no effect, B leaves out the factor \(\hbar\), and D changes the exponent to kx², which differentiation of e^ikx never produces.

Answer: D

The incorrect statement is D, 'Velocity cannot be measured accurately'.

The position is fixed to \(\Delta x = 0.1\ \text{Å} = 10^{-11}\ \text{m}\). The uncertainty principle gives

\[ \begin{aligned} \Delta p &\ge \frac{h/4\pi}{\Delta x} = \frac{5.27\times 10^{-35}}{10^{-11}} \\ &= 5.27\times 10^{-24}\ \text{kg m s}^{-1} \\ \Delta v &= \frac{\Delta p}{m} = \frac{5.27\times 10^{-24}}{0.1} \\ &\approx 5\times 10^{-23}\ \text{m s}^{-1} \end{aligned} \]

This is negligible, so statement A is true. Both position and velocity can therefore be known to very high precision (B), and a position known to 0.1 Å is precisely known for a stone (C). Only D contradicts the calculation.

Answer: B

The variational ground-state energy of hydrogen is −0.5 a.u.

With the trial function e^−αr, the energy expectation value in atomic units is a function of α that is minimised as follows:

\[ \begin{aligned} E(\alpha) &= \frac{\alpha^{2}}{2} - \alpha \\ \frac{dE}{d\alpha} &= \alpha - 1 = 0 \Rightarrow \alpha = 1 \\ E &= \frac{1}{2} - 1 = -0.5\ \text{hartree} \end{aligned} \]

This equals the exact energy, −13.6 eV, because the trial function has the form of the true 1s function. −2 a.u. is the energy of He⁺, −2.75 a.u. is the simple variational estimate for helium with an unscreened charge, and 0 is the ionisation limit.

Answer: A

The correct pair is the first one: x̂P̂_xΨ gives only the derivative term, while P̂_xx̂Ψ also gives Ψ.

With \(\hat{P}_{x} = \frac{\hbar}{i}\frac{d}{dx}\):

\[ \begin{aligned} \hat{x}\hat{P}_{x}\Psi &= x\,\frac{\hbar}{i}\frac{d\Psi}{dx} \\ \hat{P}_{x}\hat{x}\Psi &= \frac{\hbar}{i}\frac{d(x\Psi)}{dx} \\ &= \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right) \end{aligned} \]

The product rule produces the extra \(\Psi\), so the two orders differ and \([\hat{x}, \hat{P}_{x}] = i\hbar\). The second option assigns the two results to the wrong orders, the third claims they are equal, and the fourth drops the factor x.

Answer: D

n = 0 is ruled out because it implies \(\Psi = 0\) everywhere in the box.

The wave functions of a particle in a box of length L are \(\Psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\). Putting n = 0 gives \(\sin 0 = 0\) at every x, so \(\Psi = 0\) throughout. Then \(|\Psi|^{2}\) is zero everywhere and the probability of finding the particle anywhere is zero, which means there is no particle at all.

The energy \(E_{n} = n^{2}h^{2}/8mL^{2}\) would simply be zero, not divergent, and the wave function would vanish rather than diverge. Nothing about n = 0 makes the particle classical; the lowest allowed state is n = 1, with non-zero zero-point energy.

Answer: B

The electron-nucleus term is \(-\sum_{i} Ze^{2}/r_{i}\).

Each electron i is attracted by the nucleus of charge +Ze at distance r_i. The potential energy of that attraction is -Ze²/r_i (times 1/4πε₀ in SI units), and the Hamiltonian sums it over all electrons. The minus sign shows attraction.

Terms in e²/r_ij are the electron-electron repulsions, summed over pairs, and are positive. Ze/r_i lacks one factor of the charge and has the wrong sign, so it is not an energy of interaction.

Answer: C

The order of increasing energy is (2), (3), (4), (1).

In Huckel theory the resonance integral β is negative, so adding a multiple of β to α lowers the energy. For butadiene the four levels are α + 1.62β, α + 0.62β, α - 0.62β and α - 1.62β, in that order from lowest to highest.

• Lowest: α + 1.6β (2), the fully bonding orbital.
• Then α + 0.6β (3), the HOMO.
• Then α - 0.6β (4), the LUMO.
• Highest: α - 1.6β (1).

Option A lists them as numbered, and option D treats β as if it were positive, which reverses the order.