Atomic Structure: 17-Phys-B7 Structure of Materials, December 2018, Q1

Question I: Atomic 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.

Part (a) — Bohr vs. quantum-mechanical atomic model. Bohr's 1913 model treats the electron as a point particle moving in a fixed, well-defined circular orbit of radius $r_n$ around the nucleus, with angular momentum quantized as $m_evr_n = nh/2\pi$; the electron's position and momentum are both simultaneously "known," and only the ONE quantum number $n$ (the orbit index) appears. It correctly reproduces the hydrogen emission spectrum ($\Delta E = R_E(1/n_f^2-1/n_i^2)$) but fails for any multi-electron atom. The quantum-mechanical model instead solves the Schrodinger wave equation for the electron; the electron has no definite trajectory, only a probability-density "electron cloud" $|\psi|^2$ around the nucleus, consistent with the Heisenberg uncertainty principle ($\Delta x\,\Delta p \ge h/4\pi$). Four quantum numbers are needed to fully specify a state: $n$ (principal, level/energy), $l$ (azimuthal, sub-level shape s/p/d/f), $m_l$ (magnetic, orbital orientation), $m_s$ (spin, $\pm1/2$). Example: the hydrogen 1s electron is a sharp circular orbit of radius $0.0529$ nm in Bohr's picture, but a spherically symmetric probability cloud (densest near, but not confined to, that same radius) in the quantum-mechanical picture.

Part (b) — Aufbau principle. The Aufbau ("building-up") principle states that, in the ground state of a many-electron atom, electrons fill the available orbitals starting from the lowest energy level and proceeding upward, subject to the Pauli exclusion principle (no two electrons share all four quantum numbers) and Hund's rule (orbitals of equal energy are each singly occupied, with parallel spin, before any is doubly occupied). The filling ORDER follows increasing $n+l$ (and, for equal $n+l$, increasing $n$): $1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,\ldots$ — notice $4s$ fills before $3d$ even though $n=3<4$, because $4s$ ($n+l=4$) sits below $3d$ ($n+l=5$). Example: vanadium ($Z=23$) builds up as $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^3$, i.e. $[\text{Ar}]\,4s^2\,3d^3$.

Part (c) — Transition metals. Transition metals are the elements of the $d$-block (Groups 3–12) whose atoms (or common ions) have a partially filled $d$ sub-level. Because the $3d$/$4d$/$5d$ electrons sit close in energy to the outer $s$ electrons, these elements characteristically show multiple stable oxidation states, form coloured compounds (d–d electronic transitions), and are good catalysts. Example: iron, $Z=26$, $[\text{Ar}]\,3d^6\,4s^2$ — the partially filled $3d^6$ sub-level is what makes Fe a transition metal, and lets it exist as both Fe²⁺ ($3d^6$) and Fe³⁺ ($3d^5$).

Part (d) — electronic structures of the ions. Each ion is built from its parent atom's Aufbau ground state by first removing/adding the OUTERMOST ($4s$, for a first-row transition metal) electrons, then the $3d$ electrons if more must be removed.

Co³⁺ (Z=27). Co atom: $[\text{Ar}]\,4s^2\,3d^7$. Removing 3 electrons removes the two $4s$ electrons first, then one $3d$ electron: $\text{Co}^{3+} = [\text{Ar}]\,3d^6$.
Cu²⁺ (Z=29). Cu atom (anomalous filling): $[\text{Ar}]\,4s^1\,3d^{10}$. Removing 2 electrons removes the single $4s$ electron, then one $3d$ electron: $\text{Cu}^{2+} = [\text{Ar}]\,3d^9$.
Cl⁻ (Z=17, corrected — see the check note above). Cl atom: $[\text{Ne}]\,3s^2\,3p^5$. Adding 1 electron completes the $3p$ sub-level: $\text{Cl}^- = [\text{Ne}]\,3s^2\,3p^6$, i.e. isoelectronic with argon.

Given. The net interatomic potential energy is $E(r) = -1.436/r + 7.32\times10^{-6}/r^8$, with $r$ in nm and $E$ in eV.

Find. The equilibrium interatomic spacing $r_0$.

Approach. At equilibrium the net force is zero, i.e. $dE/dr=0$; solve that condition for $r_0$.

Differentiate and set to zero. $$\frac{dE}{dr} = \frac{1.436}{r^2} - \frac{8(7.32\times10^{-6})}{r^9} = 0 \;\;\Longrightarrow\;\; 1.436\,r^7 = 8(7.32\times10^{-6}).$$
Solve for $r_0$. $$r_0 = \left[\frac{8(7.32\times10^{-6})}{1.436}\right]^{1/7} = \left(4.078\times10^{-5}\right)^{1/7} = \boxed{0.236\ \text{nm}}.$$
Equilibrium bond energy (illustrative). Substituting back, $$E(r_0) = -\frac{1.436}{0.236} + \frac{7.32\times10^{-6}}{0.236^8} = -6.085 + 0.761 = \boxed{-5.32\ \text{eV}},$$ a negative value confirming a bound, stable equilibrium (a potential-energy minimum).

Final results — Question I
Quantity	Value
Co³⁺ electronic structure	[Ar] 3d⁶
Cu²⁺ electronic structure	[Ar] 3d⁹
Cl⁻ electronic structure	[Ne] 3s² 3p⁶
Equilibrium spacing $r_0$	0.236 nm
Equilibrium bond energy $E(r_0)$	−5.32 eV

Topic: Atomic models, electron configuration (Aufbau/Pauli/Hund), and the interatomic-potential picture of bonding.

Key relations: quantum numbers $n,l,m_l,m_s$; Aufbau fill order by increasing $n+l$; equilibrium spacing from $dE/dr=0$; bond stiffness from the curvature $d^2E/dr^2$ at $r_0$.

Why this works: the attractive term ($-1.436/r$, long-ranged Coulombic/ionic attraction) and the repulsive term ($+7.32\times10^{-6}/r^8$, short-ranged electron-cloud overlap) trade off; the repulsive term wins at small $r$ and the attractive term wins at large $r$, so their sum has exactly one minimum, at $r_0$.

Common pitfall: filling $3d$ before $4s$ (reversing the Aufbau order), or forgetting that $4s$ empties BEFORE $3d$ when a transition-metal ATOM is ionized (Cu²⁺ is $3d^9$, not $3d^8 4s^1$); algebraically, setting $E(r)=0$ instead of $dE/dr=0$ to find $r_0$.

Source: Callister & Rethwisch, Materials Science and Engineering, Ch. 2 (atomic structure, electron configurations, bonding forces and energies).

Question 1 of 7: Atomic Structure

Question I: Atomic Structure (20 marks)