NivaarExam PrepOfficial exam papers ↗

21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2016

Question 2 of 7: Bonding (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2016 — 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 (dislocation theory) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic directions/planes, solid solubility, XRD, microscopy 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 II — 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.

II.1 — Force–spacing relation, equilibrium spacing, maximum binding energy

Given. $E(r)=-\dfrac{A}{r^m}+\dfrac{B}{r^n}$, with $n\gt m$ (the repulsive term must fall off faster than the attractive term, or no stable minimum exists).

Find. $F(r)$; the equilibrium spacing $r_0$; the maximum binding energy $E_0=E(r_0)$.

Approach. Force is $F=-dE/dr$. The equilibrium spacing is where the net force is zero (the minimum of $E$); substituting $r_0$ back into $E(r)$ gives the bond (binding) energy.

  1. Differentiate for the force. $$\frac{dE}{dr}=\frac{mA}{r^{m+1}}-\frac{nB}{r^{n+1}} \quad\Rightarrow\quad F(r)=-\frac{dE}{dr}=\boxed{\dfrac{nB}{r^{n+1}}-\dfrac{mA}{r^{m+1}}}$$ For $r\lt r_0$ the second (repulsive) term dominates, $F\gt0$ (net repulsion, pushing the atoms apart); for $r\gt r_0$ the first term dominates, $F\lt0$ (net attraction).
  2. Equilibrium spacing. Set $F(r_0)=0$: $$\frac{nB}{r_0^{n+1}}=\frac{mA}{r_0^{m+1}} \;\Rightarrow\; r_0^{\,n-m}=\frac{nB}{mA} \;\Rightarrow\; \boxed{r_0=\left(\frac{nB}{mA}\right)^{1/(n-m)}}$$
  3. Maximum binding energy. From the equilibrium condition, $mA/r_0^m=nB/r_0^n$, so $B/r_0^n=(m/n)(A/r_0^m)$. Substituting into $E(r_0)$: $$E_0=-\frac{A}{r_0^{m}}+\frac{B}{r_0^{n}}=-\frac{A}{r_0^m}+\frac{m}{n}\frac{A}{r_0^m}=\boxed{-\dfrac{A}{r_0^{m}}\left(\dfrac{n-m}{n}\right)}$$ The negative sign confirms $E_0$ is a true minimum (a bound state); its magnitude is the maximum binding (bond) energy.
r E r₀ E₀ (max binding energy) repulsive (r<r₀) attractive (r>r₀) (a) Potential energy E vs. r r F r₀ repulsive, F>0 attractive, F<0 (b) Force F vs. r
Qualitative $E$ vs. $r$ (a) and $F$ vs. $r$ (b) for a generic attractive+repulsive pair potential. $r_0$ marks the energy minimum / zero-force crossing; the shape shown uses illustrative exponents and is not to the scale of any specific $m,n$.

The two curves are linked point-for-point: wherever $E(r)$ has zero slope (its minimum, at $r_0$), $F(r)=-dE/dr$ crosses zero: $F\gt0$ (repulsive) for $r\lt r_0$, where $E$ is falling steeply toward the minimum from a very high value, and $F\lt0$ (attractive, net pull-together) for $r\gt r_0$, where $E$ is rising slowly back toward zero at large separation.

II.2 — Dominant bonding type

Dominant bond type by material
MaterialDominant bondWhy
(a) SiCovalentGroup-IVA element; each Si shares 4 electrons with 4 neighbours in a diamond-cubic network (sp³ hybridised).
(b) GraphiteCovalent (in-plane) + secondary (between layers)Strong covalent sp² bonds within each hexagonal carbon sheet; the sheets themselves are held together only by weak van der Waals forces, which is why graphite cleaves easily along the basal planes.
(c) NaClIonicLarge electronegativity difference between Na ($\chi\approx0.9$) and Cl ($\chi\approx3.0$) transfers an electron, giving Coulombic attraction between Na⁺ and Cl⁻.
(d) SiO₂Covalent (with partial ionic character)The Si–O bond has a large electronegativity difference ($\chi_O-\chi_{Si}\approx1.7$), giving substantial ionic character, but the tetrahedral SiO₄ network is conventionally classed as (mixed ionic–covalent, predominantly covalent) network bonding, the same family as Si and diamond.
(e) ZrMetallicA transition metal; delocalised valence electrons form an electron "sea" around the positive ion cores, the hallmark of metallic bonding.

II.3 — Four quantum numbers; valence electron of Br⁻¹ (Z=35)

The four quantum numbers. $n$ (principal — energy level), $l$ (azimuthal/angular-momentum — orbital sub-level shape, $0,1,\ldots,n-1$, labelled $s,p,d,f,\ldots$), $m_l$ (magnetic — orbital orientation within that sub-level, $-l,\ldots,+l$), and $m_s$ (spin, $\pm\tfrac12$). Together they uniquely specify a single electron state (Pauli's exclusion principle, I.1b).

Br⁻¹ configuration. Neutral bromine ($Z=35$) is $[\text{Ar}]3d^{10}4s^24p^5$ (35 electrons). The Br⁻¹ anion has gained one electron, filling the $4p$ sub-level to $4p^6$ — a total of 36 electrons, isoelectronic with krypton: $[\text{Ar}]3d^{10}4s^24p^6$. Every sub-level is now completely full, so any one of the six $4p$ electrons is an equally valid "valence" electron to quote. One valid combination, for the electron completing the last $4p$ orbital: $$\boxed{n=4,\ \ l=1,\ \ m_l=+1,\ \ m_s=+\tfrac12}$$