Question 1 of 8: Electron Structure and Bonding

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.

Part 1 — Given. Titanium, $Z=22$; iron, $Z=26$.

Find. The ground-state electron configuration of each atom, and why 4s fills before 3d in the neutral-atom build-up despite 3d becoming the lower-energy orbital once occupied.

Approach. Apply the Madelung ($n+\ell$) filling rule atom by atom, then explain the energy-ordering reversal that occurs once 3d starts to fill.

Ti ($Z=22$): $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^2$, conventionally written $[\text{Ar}]\,3d^2\,4s^2$. Fe ($Z=26$): $[\text{Ar}]\,3d^6\,4s^2$. Building up either atom orbital by orbital, the Madelung rule orders sub-orbitals by increasing $n+\ell$ (ties broken by lower $n$): $4s$ has $n+\ell=4+0=4$ while $3d$ has $n+\ell=3+2=5$, so 4s is lower in energy and fills first for the neutral, un-ionized atom — both electrons enter $4s$ before any enter $3d$. Once electrons occupy 3d, however, the 3d orbitals penetrate the core poorly and are screened inefficiently from the increasing nuclear charge, so their energy drops sharply relative to 4s; by the time the 3d level is partly filled, its orbitals sit below 4s. This is why ionization removes the 4s electrons first (the configuration that is lowest in energy for the ion is $[\text{Ar}]\,3d^n$, not $[\text{Ar}]\,3d^{n-2}4s^2$) even though 4s was populated first when the neutral atom was built up.

Part 2. Diamond and graphite are both pure carbon, differing only in bonding geometry. In diamond, every carbon atom is $sp^3$-hybridized and forms four identical, strong covalent bonds to its neighbours, building a rigid three-dimensional network with no weak directions — this makes diamond the hardest known natural material, an excellent electrical insulator (all four valence electrons are locked in $\sigma$ bonds, none delocalized) and a very high melting point. Graphite instead has $sp^2$-hybridized carbon arranged in strongly-bonded two-dimensional hexagonal sheets; within a sheet the in-plane covalent bonds are comparable in strength to diamond's, but adjacent sheets are held together only by weak van der Waals forces. Under shear, whole sheets glide past one another with little resistance, which is why graphite is soft, slippery and used as a solid lubricant. The unhybridized $p_z$ electron on each carbon delocalizes into a $\pi$ system spanning each sheet, giving graphite (unlike diamond) in-plane electrical conductivity. The mechanical contrast — extreme hardness vs. easy interlayer shear — is therefore a direct consequence of isotropic 3-D covalent bonding (diamond) vs. strongly-bonded 2-D sheets held together only by weak secondary bonding (graphite).

Part 3(a) — Given. Electron speed $v_e=0.1667\,c$, $m_e=9.11\times 10^{-31}$ kg; baseball mass $m=0.142$ kg at $v=42.91$ m/s.

Find. The de Broglie wavelength of each and their ratio.

1. Electron momentum and wavelength. $v_e=0.1667\times(2.998\times10^{8}) =4.998\times10^{7}\ \text{m/s}$ (non-relativistic treatment; $\gamma\approx1.014$ at this speed, a <2% correction, negligible for the comparison asked). $p_e=m_ev_e=9.11\times10^{-31}\times 4.998\times10^{7}=4.553\times10^{-23}\ \text{kg}\cdot\text{m/s}$, so $$\lambda_e=\frac{h}{p_e}=\frac{6.626\times10^{-34}}{4.553\times10^{-23}} =\boxed{1.455\times10^{-11}\ \text{m}=14.55\ \text{pm}}.$$
2. Baseball momentum and wavelength. $p=mv=0.142\times42.91=6.093\ \text{kg}\cdot\text{m/s}$, so $$\lambda_{\text{ball}}=\frac{h}{p}=\frac{6.626\times10^{-34}}{6.093} =\boxed{1.087\times10^{-34}\ \text{m}}.$$
3. Comparison. $$\frac{\lambda_e}{\lambda_{\text{ball}}} =\frac{1.455\times10^{-11}}{1.087\times10^{-34}}\approx\boxed{1.34\times10^{23}}.$$ The electron's wavelength is comparable to atomic dimensions (tens of picometres) and its wave nature is essential to how it behaves; the baseball's wavelength is $10^{23}$ times smaller than a proton, utterly unobservable, which is why macroscopic objects show no measurable wave behaviour despite de Broglie's relation applying to both.

Part 3(b) — Given. Speed uncertainty $\Delta v=1\%$ of 42.91 m/s.

Find. The minimum position uncertainty $\Delta x$ implied by the Heisenberg uncertainty principle.

1. Momentum uncertainty. $\Delta v=0.01\times42.91=0.4291\ \text{m/s}$, so $\Delta p=m\,\Delta v=0.142\times0.4291=0.06093\ \text{kg}\cdot\text{m/s}$.
2. Position uncertainty. Using $\Delta x\,\Delta p\ge\hbar/2$, $$\Delta x_{\min}=\frac{\hbar}{2\,\Delta p}=\frac{1.0546\times10^{-34}}{2\times0.06093} =\boxed{8.65\times10^{-34}\ \text{m}}.$$ This is some 20 orders of magnitude smaller than a proton's radius — the uncertainty principle places no practically meaningful limit on locating a macroscopic baseball, the opposite of its role for an electron.