Question 1 of 8: Electron Structure (20 marks)

Question I — Electron Structure (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

I.1 — Bohr's model: description and limitations

Bohr's model (1913) treats the electron as a point particle orbiting the nucleus in fixed, quantised circular orbits, each with a discrete allowed energy $E_n = -Z^2R_E/n^2$ (with $R_E = 13.61$ eV, the appendix's Rydberg-type constant, and $n = 1,2,3,\ldots$). Angular momentum is quantised, $L = n\hbar$, and an electron radiates only when it jumps between orbits, emitting or absorbing a photon of energy $\Delta E = E_f - E_i$. This correctly predicted the hydrogen line spectrum and introduced the idea that atomic energy is quantised at all.

Its limitations are structural, not just numerical: (i) it works only for one-electron (hydrogen-like) systems and fails for multi-electron atoms, where electron–electron repulsion is not captured; (ii) it gives the electron a definite classical orbit and trajectory, in violation of the Heisenberg uncertainty principle; (iii) it cannot explain the fine structure of spectral lines, the Zeeman effect (splitting in a magnetic field), or chemical bonding and molecular geometry; and (iv) it offers no reason why angular momentum should be quantised — the quantisation is asserted, not derived.

The wave-mechanical (Schrödinger) model resolves these by treating the electron as a three-dimensional standing wave described by a wavefunction $\psi$, whose solution for a given potential yields four quantum numbers ($n$, $l$, $m_l$, $m_s$, per the appendix) instead of Bohr's one. $|\psi|^2$ gives a probability density (an orbital, not a fixed orbit), consistent with Heisenberg uncertainty; $l$ and $m_l$ emerge naturally from the boundary conditions on the wave equation rather than being assumed, correctly predicting orbital shapes, multi-electron structure (via the Pauli exclusion principle applied to the four quantum numbers), and the periodic table itself.

I.2 — Element count, natural vs. synthetic elements

a. Including ununpentium (now officially named moscovium, Mc, Z = 115), the periodic table extends through 115 confirmed elements at the time this exam was set (December 2013); several heavier elements up to Z = 118 had been synthesised and were pending formal IUPAC confirmation/naming (completed only in 2016), so 115 is the defensible "confirmed" count for this sitting.

b. Naturally occurring example: iron (Fe, Z = 26) or uranium (Z = 92), the heaviest element found in appreciable quantity in nature. Man-made (synthetic) example: plutonium (Pu, Z = 94) or ununpentium/moscovium itself (Z = 115), produced only by particle accelerators fusing heavy nuclei.

c. Not strictly true. The general trend — heavier, transuranium elements (Z > 92) are all synthetic because no natural process on Earth produces or preserves them — is correct as a trend, but it is not true that all lighter elements are natural. Two light elements, technetium (Z = 43) and promethium (Z = 61), have no stable isotopes and do not occur naturally in usable quantities despite being far lighter than uranium; both were first made artificially. So the rule is "elements past Z = 92 are synthetic" (true, since nature has no stable path to build them and they are not primordial), while "light = natural" has these two well-known exceptions.

I.3 — Secondary-bond (6-12 potential) for argon: bond energy and length

Given. $E(r) = -A/r^6 + B/r^{12}$, with $A = 10.37\times10^{-78}\ \text{J}\cdot\text{m}^6$ and $B = 16.16\times10^{-135}\ \text{J}\cdot\text{m}^{12}$ (argon).

Find. The equilibrium bond length $r_0$ (minimum of $E$) and the bond energy $E_0 = E(r_0)$.

Approach. The equilibrium spacing is where the net force is zero, i.e. $dE/dr = 0$; substitute that $r_0$ back into $E(r)$ for the bond energy.

1. Set the derivative to zero. $$\frac{dE}{dr} = \frac{6A}{r^7} - \frac{12B}{r^{13}} = 0 \ \Rightarrow\ 6A\,r^6 = 12B \ \Rightarrow\ r_0^6 = \frac{2B}{A}$$
2. Evaluate $r_0$. $$r_0 = \left(\frac{2(16.16\times10^{-135})}{10.37\times10^{-78}}\right)^{1/6} = \left(3.117\times10^{-57}\right)^{1/6} = \boxed{3.82\times10^{-10}\ \text{m} = 0.382\ \text{nm}}$$
3. Bond energy at $r_0$. Since $r_0^6 = 2B/A$, $r_0^{12} = 4B^2/A^2$, so $$E_0 = -\frac{A}{r_0^6} + \frac{B}{r_0^{12}} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$$ $$E_0 = -\frac{(10.37\times10^{-78})^2}{4(16.16\times10^{-135})} = \boxed{-1.66\times10^{-21}\ \text{J} = -1.04\times10^{-2}\ \text{eV per bond}}$$

Both figures are physically sensible for a noble-gas van der Waals bond: the equilibrium spacing (≈0.38 nm) and the very shallow well depth (≈10 meV) match the known weak, purely-dispersion secondary bonding of solid argon, in contrast to primary (ionic/covalent/metallic) bonds, which are 1–2 eV or more.