Quantum Chemistry Questions and Answers

Answer: A

The secular determinant is (A), the tridiagonal one with β between neighbouring carbons only.

The allyl radical CH₂=CH-CH₂ has three p orbitals in a chain, C1-C2-C3. In Hückel theory every diagonal element is α - E, an off-diagonal element is β for carbons that are bonded and 0 for carbons that are not. C1 and C2 are bonded, C2 and C3 are bonded, C1 and C3 are not:

\[ \begin{vmatrix} \alpha-E & \beta & 0 \\ \beta & \alpha-E & \beta \\ 0 & \beta & \alpha-E \end{vmatrix} = 0 \]

(B) and (C) each drop one of the two bonds, describing an isolated ethylene plus a lone p orbital. (D) has -β, which flips the sign convention without any physical basis in standard Hückel theory.

Answer: A

The energies are \(\alpha + \sqrt{2}\beta,\ \alpha,\ \alpha - \sqrt{2}\beta\).

Dividing each row of the determinant by \(\beta\) and writing \(x = (\alpha - E)/\beta\):

\[ \begin{aligned} \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} &= x(x^{2}-1) - x \\ &= x^{3} - 2x = 0 \\ x &= 0,\ \pm\sqrt{2} \end{aligned} \]

So \(E = \alpha - x\beta\) gives \(\alpha + \sqrt{2}\beta\) (bonding), \(\alpha\) (non-bonding) and \(\alpha - \sqrt{2}\beta\) (antibonding). The radical's odd electron sits in the non-bonding level.

\(\alpha \pm \beta\) are the ethylene levels, \(\alpha \pm 2\beta\) are the extreme levels of benzene, and \(2\sqrt{2}\) doubles the root of \(x^{2} = 2\).

Answer: D

The uncertainty principle states \(\Delta y\,\Delta p_{y} \ge \hbar/2\).

Heisenberg's relation, in the form derived from the commutator of position and momentum, puts a lower bound on the product of the two standard deviations:

\[ \Delta y\,\Delta p_{y} \ge \frac{\hbar}{2} = \frac{h}{4\pi} \]

The sign must be 'greater than or equal to', because a minimum-uncertainty state (a Gaussian wave packet, or the ground state of a harmonic oscillator) reaches the bound exactly. That rules out both strict inequalities, \(\gt \hbar\) and \(\gt \hbar/2\). The bound is \(\hbar/2\), not \(\hbar\); \(\ge \hbar\) is a looser statement that the oscillator ground state, with a product of exactly \(\hbar/2\), would violate.

Answer: C

The quantum numbers are l = 3 and m = 2.

The azimuthal factor \(e^{im\phi}\) gives m directly: \(e^{2i\phi}\) means m = +2, not -2.

For l, count the powers of the polar functions. A spherical harmonic \(Y_{l}^{m}\) contains \(\sin^{|m|}\theta\) times a polynomial in \(\cos\theta\) of degree \(l - |m|\). Here \(\sin^{2}\theta\) gives |m| = 2 and the single \(\cos\theta\) gives \(l - 2 = 1\), so l = 3. Indeed \(Y_{3}^{2} \propto \sin^{2}\theta\cos\theta\,e^{2i\phi}\), an f orbital.

An l = 2 function with m = 2 is \(\sin^{2}\theta\,e^{2i\phi}\) with no cosine factor, so the 2 and 2 option fails, and the -2 options have the wrong sign of the exponent.

Answer: C

The normalisation constant is \(N = \left(\dfrac{1}{2\sqrt{\pi}}\right)^{1/2}\).

Normalisation requires \(\int \psi^{2}\,dx = 1\). Squaring the function gives \(N^{2}(4x^{4} - 4x^{2} + 1)e^{-x^{2}}\). Using the given integral with n = 2, 1 and 0:

\[ \begin{aligned} \int x^{4}e^{-x^{2}}dx &= \tfrac{3}{4}\sqrt{\pi} \\ \int x^{2}e^{-x^{2}}dx &= \tfrac{1}{2}\sqrt{\pi} \\ \int e^{-x^{2}}dx &= \sqrt{\pi} \end{aligned} \]

\[ \begin{aligned} N^{2}\sqrt{\pi}\,(4 \cdot \tfrac{3}{4} - 4 \cdot \tfrac{1}{2} + 1) &= 1 \\ 2\sqrt{\pi}\,N^{2} &= 1 \\ N &= (2\sqrt{\pi})^{-1/2} \end{aligned} \]

The 8 in the first option is what the Hermite form 4x² - 2, twice this polynomial, would give. 4 is what dropping the cross term -4x² gives (3 + 1), and 3 is the x⁴ term alone.

Answer: A

The wavelength is 2L/n.

The wavefunctions of a particle in a box, \(\psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\), must vanish at both walls, so they are standing waves with a whole number of half-wavelengths fitting into the box:

\[ L = n\frac{\lambda}{2} \Rightarrow \lambda = \frac{2L}{n} \]

For the ground state (n = 1) the box holds half a wavelength, so \(\lambda = 2L\). L/n counts whole wavelengths instead of half-wavelengths, L/2n halves the answer again, and nL would make the wavelength grow with the quantum number, when higher states have more nodes and shorter wavelengths.

Answer: C

The incorrect statement is The time independent Schrödinger equation for Li²⁺ cannot be solved exactly.

Li²⁺ has a single electron, so it is a hydrogen-like ion (Z = 3). Its Schrödinger equation separates exactly, like hydrogen's, giving energies \(E_{n} = -Z^{2}R_{H}/n^{2}\) and the familiar orbitals scaled by Z. Exact solution fails only for systems with two or more electrons, such as He or Li⁺, because of the electron-electron repulsion.

The other statements are true. The harmonic oscillator's zero-point energy is \(h\nu/2\). Rigid-rotor levels are \(E_{J} = J(J+1)\hbar^{2}/2I\), inversely proportional to the moment of inertia. The total angular momentum is the (vector) sum of the orbital and spin angular momenta, \(\mathbf{J} = \mathbf{L} + \mathbf{S}\).

Answer: B

The most probable distance is \(a_{0}/4\).

The wavefunction \(e^{-4r/a_{0}}\) is the 1s function \(e^{-Zr/a_{0}}\) for \(Z = 4\) (Be³⁺ is hydrogen-like). The radial probability density is

\[ P(r) \propto r^{2}\psi^{2} = r^{2}e^{-8r/a_{0}} \]

Setting the derivative to zero:

\[ \begin{aligned} 2r - \frac{8r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{4} \end{aligned} \]

This is the hydrogen result \(a_{0}\) divided by Z. \(a_{0}/8\) uses the exponent of \(\psi^{2}\) as if it were Z, and \(4a_{0}\) and \(8a_{0}\) multiply by Z instead of dividing; a higher nuclear charge pulls the electron closer, not further out.

Answer: A

The correct match is P→III, Q→I, R→IV, S→II.

• Associated Legendre polynomials appear in the spherical harmonics, the angular part of the hydrogen-atom wavefunctions.
• Hermite polynomials multiply a Gaussian in the harmonic oscillator wavefunctions.
• Associated Laguerre polynomials form the radial part of the hydrogen-atom wavefunctions.
• Trigonometric functions (sines) are the particle-in-a-box wavefunctions.

The other options assign Hermite polynomials to the hydrogen atom, swap the radial and angular parts, or give the particle in a box polynomial solutions, all of which contradict the standard solutions.

Answer: C

The order is \(E_{3} \lt E_{1} = E_{4} \lt E_{2}\).

In Hückel theory the energy of a benzene MO is \(\alpha + 2\beta\cos(2\pi k/6)\), with \(\beta \lt 0\), so fewer nodes means lower energy.

• \(\psi_{3}\) has all coefficients equal, no nodes: \(\alpha + 2\beta\), the lowest.
• \(\psi_{1}\) (pattern 0, 1, 1, 0, -1, -1) and \(\psi_{4}\) (2, 1, -1, -2, -1, 1) each change sign twice around the ring: they are the degenerate pair at \(\alpha + \beta\).
• \(\psi_{2}\) alternates in sign at every atom: \(\alpha - 2\beta\), the highest.

So \(E_{1} = E_{4}\) lie between \(E_{3}\) and \(E_{2}\). The other orders either break the degeneracy of \(\psi_{1}\) and \(\psi_{4}\) or put the fully alternating orbital too low.