21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2017 — 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 parts of Question VI (grain-size strengthening) touch mechanical properties directly; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic planes, solid solubility, XRD, diffusion 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.
Ionic bonding is the electrostatic attraction between oppositely-charged ions formed by electron transfer (e.g. Na⁺ and Cl⁻); it is strong, non-directional, and gives hard, brittle, electrically-insulating solids — the dominant bond type in ceramics (and in ionic compounds generally). Covalent bonding is the sharing of electron pairs between specific neighbouring atoms, forming strong, highly directional bonds; it dominates in covalent ceramics and semiconductors (e.g. Si, GaAs, SiC, diamond). Metallic bonding is the delocalized "sea" of valence electrons shared non-directionally among a lattice of positive ion cores; it is the bond type of metals, and its non-directionality and free electrons explain metallic ductility and electrical/thermal conductivity. Secondary (van der Waals/hydrogen) bonding is a comparatively weak dipole–dipole attraction between already-bonded molecules or chains; it is the bond that holds polymer chains to one another between the strong covalent bonds along each chain backbone (and is also the interlayer bond in graphite, Question II.2).
Both diamond and graphite are pure carbon, so the dramatic hardness difference is purely a consequence of how the carbon atoms are bonded, not what they are made of. In diamond, every carbon atom is $sp^3$-hybridised and covalently bonded to four neighbours in a rigid, three-dimensional tetrahedral network (the diamond cubic structure); because strong covalent bonds extend in every direction through the crystal, there is no easy internal plane along which the structure can shear, making diamond the hardest known natural material. In graphite, each carbon atom is $sp^2$-hybridised and covalently bonded to only three neighbours within a flat hexagonal sheet (strong, in-plane covalent bonds, the same strength as diamond's), but adjacent sheets are held together only by weak secondary (van der Waals) bonding. The sheets readily slide over one another under a comparatively small shear stress, which is why graphite is soft and is used as a solid lubricant — the bonding is strong within a plane and weak between planes, exactly the opposite of an isotropic solid like diamond.
Both compounds form by complete transfer of valence electrons from the metal to the non-metal, so each resulting ion attains the electron configuration of the nearest noble gas.
| Ion | Parent atom configuration | Ion configuration | Isoelectronic with |
|---|---|---|---|
| Na⁺ (in NaCl) | Na: $1s^2 2s^2 2p^6 3s^1$ | $1s^2 2s^2 2p^6$ | Ne (10 e⁻) |
| Cl⁻ (in NaCl) | Cl: $1s^2 2s^2 2p^6 3s^2 3p^5$ | $1s^2 2s^2 2p^6 3s^2 3p^6$ | Ar (18 e⁻) |
| Mg²⁺ (in MgO) | Mg: $1s^2 2s^2 2p^6 3s^2$ | $1s^2 2s^2 2p^6$ | Ne (10 e⁻) |
| O²⁻ (in MgO) | O: $1s^2 2s^2 2p^4$ | $1s^2 2s^2 2p^6$ | Ne (10 e⁻) |
Na loses its single $3s$ electron to reach the neon configuration, while Cl gains one electron into its $3p$ sub-level to reach the argon configuration — the ionic bond in NaCl is the electrostatic attraction of the resulting Na⁺/Cl⁻ pair. Mg loses both $3s$ electrons to reach the neon configuration, and O gains two electrons into its $2p$ sub-level to reach that same neon configuration, so both ions in MgO are isoelectronic with neon; the doubled ionic charges ($+2$/$-2$, vs. $+1$/$-1$ in NaCl) give MgO a substantially stronger ionic bond, consistent with its much higher melting point (∼2852°C) than NaCl (∼801°C).
