21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2018 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).
Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI.2(c) (grain-size strengthening) touches mechanical/deformation properties directly; the paper as a whole is a broad introductory materials-science survey — atomic structure, bonding, crystal structure/directions/planes, point defects, XRD, dislocations and phase diagrams — and is answered as such below.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
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.
Bohr's model (1913) treats the electron as a point particle in a fixed, well-defined circular orbit of radius $r_n\propto n^2$ around the nucleus, with quantized energy $E_n=-Z^2R_E/n^2$ (given in the appendix, $R_E=13.61$ eV). It correctly predicts hydrogen's line spectrum but fails for multi-electron atoms and is inconsistent with the Heisenberg uncertainty principle, since it assigns the electron a simultaneously exact position and momentum. The quantum-mechanical model instead solves the Schrödinger wave equation for the electron, yielding not a fixed orbit but a three-dimensional probability-density cloud (an orbital) describing where the electron is likely to be found, with the electron's state fully specified by four quantum numbers ($n,l,m_l,m_s$, all listed in the appendix). The quantum-mechanical model is the more accurate of the two: it correctly predicts multi-electron atom behaviour, chemical bonding trends and periodicity, none of which the Bohr model can do beyond hydrogen.
The Aufbau ("building-up") principle states that as electrons are added to a many-electron atom in its ground state, they fill the available orbitals in order of increasing orbital energy, lowest first, subject to the Pauli exclusion principle (each orbital holds at most two electrons, of opposite spin). The practical filling order is $1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,\dots$ — note $4s$ fills before $3d$ because it is lower in energy for the neutral atom, which is exactly why the transition-metal configurations in Question I.2 remove $4s$ electrons before $3d$ electrons on ionization (Question I.2 notes below).
Transition metals are the elements of the $d$-block (Groups 3–12) whose atoms, or common ions, have a partially filled $d$ sublevel — e.g. iron ($[Ar]3d^64s^2$), copper ($[Ar]3d^{10}4s^1$) and the Co/Cu examples of Question I.2 below. The partially filled, relatively compact $d$ sublevel gives this group its characteristic properties: multiple stable oxidation states (electrons can be removed from both $4s$ and $3d$), coloured compounds (d–d electronic transitions absorb visible light), and, in the metallic state, strong metallic bonding from both $4s$ and $3d$ electrons contributing to the delocalized electron sea — the structural basis of transition metals' generally high melting points and strength (e.g. W, Mo, Fe) relative to the main-group metals.
Each configuration follows the Aufbau order for the parent neutral atom, then removes (cation) or adds (anion) electrons from the highest-energy occupied sublevel first — for the $3d$-block ions this means the $4s$ electrons are removed before any $3d$ electron (Question I.1(b)).
| Ion | Parent atom (ground state) | Ion configuration |
|---|---|---|
| Co³⁺ ($Z=27$) | Co: $[Ar]3d^7 4s^2$ | remove $4s^2$, then one $3d$: $\boxed{[Ar]3d^6}$ |
| Cu²⁺ ($Z=29$) | Cu: $[Ar]3d^{10}4s^1$ | remove $4s^1$, then one $3d$: $\boxed{[Ar]3d^9}$ |
| Cl⁻ | Cl ($Z=17$): $[Ne]3s^2 3p^5$ | add one $3p$ electron: $\boxed{[Ne]3s^23p^6}$ (isoelectronic with Ar) |
Given. $E(r) = -\dfrac{1.436}{r} + \dfrac{7.32\times10^{-6}}{r^8}$ (the coefficients are of the standard Callister-style ionic-bonding form, consistent with $E$ in eV and $r$ in nm).
Find. The equilibrium spacing $r_0$.
Approach. At equilibrium the net force is zero, i.e. $dE/dr=0$ (the appendix gives $F=-\partial E/\partial r$) — this is the minimum of the potential-energy curve.
| Item | Result |
|---|---|
| I.2(a) Co³⁺ | $[Ar]3d^6$ |
| I.2(b) Cu²⁺ | $[Ar]3d^9$ |
| I.2(c) Cl⁻ (correct $Z=17$) | $[Ne]3s^23p^6$ |
| I.3 Equilibrium spacing $r_0$ | $0.236$ nm |
