Question 1 of 8: Electron Structure and Bonding (20 marks)

Question I — Electron Structure and Bonding (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 — Electron configurations and the 4s/3d filling order

Ti (Z = 22): $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^2$ (argon core, $[\text{Ar}]\,4s^2 3d^2$).
Fe (Z = 26): $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$ ($[\text{Ar}]\,4s^2 3d^6$).

The paper reads "3d gets filled earlier than 4s," but that is backwards for the neutral, ground-state atoms asked about here: for K, Ca and every first-row transition metal, the 4s orbital fills before 3d. The Madelung (n + l) rule ranks energy sub-levels by $n+l$, breaking ties by lower $n$: 4s has $n+l = 4+0=4$, while 3d has $n+l = 3+2 = 5$, so 4s sits at lower energy and is occupied first for the elements built up to that point. Physically this is because 4s is a more penetrating orbital — its radial probability density has a lobe close to the nucleus that is poorly shielded by the (still-filling) 3d electrons, lowering its energy below the more compact but strongly interelectron-repelled 3d orbital. (Once 3d starts filling, the increased nuclear charge felt by 3d relative to 4s is why, later, ionization of a transition-metal atom removes the 4s electrons first, e.g. $\text{Fe} \to \text{Fe}^{2+}$ leaves $3d^6$ with no 4s electrons — the ordering is level-dependent, not fixed.)

I.2 — Diamond vs. graphite: bonding and hardness

Both are pure-carbon allotropes, but the geometry of the covalent bonding differs completely. In diamond, every carbon atom is $sp^3$-hybridised and forms four identical, strong covalent bonds to neighbours arranged tetrahedrally, building a rigid three-dimensional network with no weak directions — there is no plane along which the structure can shear without breaking primary covalent bonds. This makes diamond the hardest known natural material (very high resistance to indentation and scratching) and an electrical insulator (all four valence electrons are locked in $\sigma$ bonds).

In graphite, each carbon is $sp^2$-hybridised, forming three strong in-plane covalent bonds to neighbours in a hexagonal sheet (the fourth, unhybridised $p_z$ electron delocalises into a $\pi$ system over the whole sheet, giving graphite its electrical conductivity). The sheets themselves stack via weak van der Waals bonding, with a large interlayer spacing (≈0.335 nm) compared with the strong in-plane C–C bond (≈0.142 nm). Because the interlayer bonding is roughly two orders of magnitude weaker than the in-plane covalent bonding, the sheets slide over each other easily under shear — graphite is soft, cleaves readily along the basal planes, and is used as a solid lubricant, whereas diamond's isotropic covalent network gives it extreme hardness. The mechanical contrast is a direct consequence of dimensionality of bonding (3-D covalent network vs. 2-D covalent sheets held together by weak secondary bonds), not of the atoms themselves, which are identical.

I.3 — Wave-particle duality: electron vs. baseball

Given. Electron speed $v_e = 0.1667c$ with $c = 3\times10^{8}\ \text{m/s}$ and electron mass $m_e = 9.11\times10^{-31}\ \text{kg}$ (standard value); baseball mass $m_b = 0.142\ \text{kg}$ at $v_b = 42.91\ \text{m/s}$; Planck's constant $h = 6.626\times10^{-34}\ \text{J}\cdot\text{s}$.

Find. The de Broglie wavelength of each particle (part a), and the position uncertainty of the baseball given a 1% speed-measurement uncertainty (part b).

Approach. Apply the de Broglie relation $\lambda = h/(mv)$ to both particles, then apply the Heisenberg uncertainty principle $\Delta x\,\Delta p \ge h/4\pi$ to the baseball's momentum uncertainty.

1. Electron speed and wavelength. $v_e = 0.1667 \times 3\times10^{8} = 5.001\times10^{7}\ \text{m/s}$. $$\lambda_e = \frac{h}{m_e v_e} = \frac{6.626\times10^{-34}}{(9.11\times10^{-31})(5.001\times10^{7})} = \boxed{1.454\times10^{-11}\ \text{m} = 14.5\ \text{pm}}$$
2. Baseball wavelength. $$\lambda_b = \frac{h}{m_b v_b} = \frac{6.626\times10^{-34}}{(0.142)(42.91)} = \boxed{1.087\times10^{-34}\ \text{m}}$$ The baseball's wavelength is about $10^{23}$ times shorter than the electron's — some $10^{25}$ times smaller than the baseball's own diameter — so wave behaviour is utterly unobservable for the macroscopic object while it dominates the electron's dynamics at atomic length scales.
3. Momentum uncertainty from a 1% speed error. $\Delta v = 0.01 \times 42.91 = 0.4291\ \text{m/s}$, so $$\Delta p = m_b\,\Delta v = 0.142 \times 0.4291 = 6.093\times10^{-2}\ \text{kg}\cdot\text{m/s}$$
4. Position uncertainty. Taking the Heisenberg relation at its minimum ($\Delta x\,\Delta p = h/4\pi$): $$\Delta x_{\min} = \frac{h}{4\pi\,\Delta p} = \frac{6.626\times10^{-34}}{4\pi(6.093\times10^{-2})} = \boxed{8.65\times10^{-34}\ \text{m}}$$ This is roughly $10^{19}$ times smaller than a proton's radius — for a macroscopic body a 1% speed uncertainty places essentially no meaningful floor on position knowledge, unlike for an electron where the same relation is the dominant physical constraint.