Quantum Chemistry -- Practice Problems

Quantum Chemistry — Practice Problems

10 MCQ questions covering the Schrodinger equation, particle in a box, atomic orbitals, angular momentum, vibrational/rotational spectroscopy, and molecular orbital theory.

Worked Examples

Example 1: Particle in a Box — Conjugated Polyene

Problem: For 1,3-butadiene (4 $\pi$ electrons), model the conjugated system as a particle in a box. The C-C bond length is 1.34 Å, and the box extends 0.5 Å beyond each end carbon. Calculate the wavelength of the HOMO→LUMO transition.

Solution:

Step 1: Determine box length

$L = 3 \times 1.34 + 2 \times 0.5 = 5.02$ Å = $5.02 \times 10^{-10}$ m

Step 2: Identify HOMO and LUMO

4 $\pi$ electrons fill $n = 1$ and $n = 2$ (2 electrons per level). HOMO: $n = 2$ . LUMO: $n = 3$ .

Step 3: Calculate energy gap

$\Delta E = \frac{h^2}{8mL^2}(3^2 - 2^2) = \frac{h^2}{8mL^2}(5)$

$\Delta E = \frac{(6.626 \times 10^{-34})^2 \times 5}{8 \times (9.109 \times 10^{-31}) \times (5.02 \times 10^{-10})^2}$

$\Delta E = \frac{2.195 \times 10^{-66}}{1.838 \times 10^{-48}} = 1.194 \times 10^{-18} \text{ J}$

Step 4: Convert to wavelength

$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.194 \times 10^{-18}} = 1.664 \times 10^{-7} \text{ m} = 166 \text{ nm}$

Comparison: Experimental $\lambda_{\max} = 217$ nm. The particle-in-a-box model underestimates the wavelength because it assumes constant potential (no electron-nuclear attraction).

Example 2: Hydrogen Atom — Transition Wavelengths

Problem: Calculate the wavelength of the photon emitted when an electron in hydrogen transitions from $n = 4$ to $n = 2$ (Balmer series).

Solution:

$\Delta E = E_4 - E_2 = -13.6\left(\frac{1}{16} - \frac{1}{4}\right) = -13.6 \times (-0.1875) = +2.55 \text{ eV}$

Converting to wavelength:

$\lambda = \frac{hc}{\Delta E} = \frac{1240 \text{ eV·nm}}{2.55 \text{ eV}} = 486 \text{ nm}$

This is the H $\beta$ line in the visible Balmer series (blue-green light).

Balmer series formula: $\frac{1}{\lambda} = R_H\left(\frac{1}{4} - \frac{1}{n^2}\right)$ for $n = 3, 4, 5, \ldots$

Transition	$n_i \to n_f$	$\lambda$ (nm)	Colour
H $\alpha$	$3 \to 2$	656	Red
H $\beta$	$4 \to 2$	486	Blue-green
H $\gamma$	$5 \to 2$	434	Violet
H $\delta$	$6 \to 2$	410	Violet

Example 3: Harmonic Oscillator — Vibrational Spectroscopy

Problem: HCl has a fundamental vibrational frequency of $\tilde{\nu} = 2990$ cm⁻¹. Calculate the force constant and the zero-point energy.

Solution:

Step 1: Calculate the force constant

$k = 4\pi^2 c^2 \tilde{\nu}^2 \mu$

Reduced mass of HCl:

$\mu = \frac{m_H \cdot m_{Cl}}{m_H + m_{Cl}} = \frac{1.008 \times 35.45}{1.008 + 35.45} \times \frac{1}{N_A} = 1.628 \times 10^{-27} \text{ kg}$

$k = 4\pi^2 (3.00 \times 10^{10} \text{ cm/s})^2 (2990 \text{ cm}^{-1})^2 (1.628 \times 10^{-27} \text{ kg})$

$k = 516 \text{ N/m}$

Step 2: Calculate zero-point energy

$E_0 = \frac{1}{2}h c \tilde{\nu} = \frac{1}{2}(6.626 \times 10^{-34})(3.00 \times 10^{10})(2990)$

$E_0 = 2.97 \times 10^{-20} \text{ J} = 0.186 \text{ eV}$

Key insight: The zero-point energy means HCl continues to vibrate even at absolute zero. This is a direct consequence of the uncertainty principle — confining the oscillator to its equilibrium position would violate $\Delta x \cdot \Delta p \geq \hbar/2$ .

The Schrodinger Equation and Model Systems

Atomic Structure and Molecular Orbital Theory

Intuition

Quantum chemistry replaces classical orbits with probability clouds: Electrons do not orbit the nucleus like planets around the sun. Instead, they exist as wavefunctions — mathematical descriptions of where an electron is likely to be found. The Schrödinger equation is the master equation that determines these probability distributions.

Why it matters: Quantum chemistry explains why atoms bond, why metals conduct electricity, and why semiconductors have band gaps. Without it, modern electronics, solar cells, and LED technology would be impossible to design.

The key insight: The Pauli exclusion principle — no two electrons can occupy the same quantum state — is why the periodic table has its structure and why matter takes up space.

Common Mistakes

Confusing the rigid rotor with the harmonic oscillator: The rigid rotor model ( $E_J = J(J+1)\hbar^2/2I$ ) describes rotation with equally spaced transitions ( $\Delta E = 2B$ ). The harmonic oscillator ( $E_v = (v+1/2)\hbar\omega$ ) describes vibration with equally spaced transitions ( $\Delta E = \hbar\omega$ ). Mixing up these models gives incorrect spectral predictions.

Assuming all transitions are allowed: The selection rule $\Delta J = \pm 1$ for rotational spectroscopy comes from conservation of angular momentum during photon absorption. Transitions with $\Delta J \neq \pm 1$ are forbidden in the electric dipole approximation. Ignoring selection rules leads to predicting spectral lines that do not exist.

Forgetting that the reduced mass $\mu$ differs from the atomic mass: For a diatomic molecule AB, the reduced mass is $\mu = m_A m_B / (m_A + m_B)$ , not directly $m_A$ or $m_B$ . Using the wrong mass in rotational or vibrational calculations gives incorrect energy levels and spectral frequencies.

Cross-References

Quantum Chemistry: Detailed notes on quantum mechanics, wave functions, and molecular orbital theory.
Statistical Mechanics: Covers partition functions and thermodynamic properties that connect to quantum energy levels.
Practice Physical Chemistry: Interactive practice problems covering quantum chemistry and statistical mechanics.

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