Question 2 of 7: Bonding

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.

Part 1 — Given. $E(r)=-\dfrac{A}{r^m}+\dfrac{B}{r^n}$, with $n>m>0$ (the $1/r^m$ term is the long-range attraction, $1/r^n$ the short-range repulsion).

Find. $F(r)$, the equilibrium spacing $r_0$, the maximum binding energy $E(r_0)$, and qualitative $E$–$r$/$F$–$r$ sketches.

Approach. Force is the negative gradient of potential energy, $F=-dE/dr$; equilibrium is where $F=0$; the binding energy is $E$ evaluated at that equilibrium spacing.

1. Force–spacing relation. $$F(r)=-\frac{dE}{dr}=-\left(\frac{mA}{r^{m+1}}-\frac{nB}{r^{n+1}}\right) =\boxed{\frac{nB}{r^{n+1}}-\frac{mA}{r^{m+1}}}.$$ The first (repulsive, positive) term dominates at small $r$; the second (attractive, negative) term dominates at large $r$.
2. Equilibrium spacing. Setting $F(r_0)=0$: $\dfrac{nB}{r_0^{n+1}} =\dfrac{mA}{r_0^{m+1}}\ \Rightarrow\ r_0^{\,n-m}=\dfrac{nB}{mA}$, so $$r_0=\boxed{\left(\frac{nB}{mA}\right)^{1/(n-m)}}.$$
3. Maximum binding energy. From the equilibrium condition, $B=\dfrac{mA}{n}\, r_0^{\,n-m}$; substituting into $E(r_0)$ eliminates $B$: $$E(r_0)=-\frac{A}{r_0^m}+\frac{B}{r_0^n}=-\frac{A}{r_0^m}\left(1-\frac{m}{n}\right) =\boxed{-\frac{A}{r_0^{\,m}}\cdot\frac{n-m}{n}}.$$ This is negative (a bound, attractive minimum) since $n>m>0$; its magnitude is the maximum binding energy $E_0$.

Part 2. Dominant bonding type by material:

Part 3 — Given. $\text{Br}^{-1}$, $Z=35$ (neutral Br), so the anion has $35+1=36$ electrons—the same electron count as krypton.

Find. The four quantum numbers, and one valid set for the valence electron of $\text{Br}^{-1}$.

The four quantum numbers that uniquely specify an electron's state are the principal quantum number $n=1,2,3,\dots$ (principal energy level), the orbital (azimuthal) quantum number $\ell=0,1,\dots,n-1$ (orbital-type shape), the magnetic quantum number $m_\ell=-\ell,\dots,+\ell$ (orbital orientation), and the spin quantum number $m_s=\pm1/2$.

Br ($Z=35$) is $[\text{Ar}]\,3d^{10}\,4s^2\,4p^5$; gaining one electron to form $\text{Br}^{-1}$ fills the $4p$ sub-level completely: $[\text{Ar}]\,3d^{10}\,4s^2\,4p^6$ (isoelectronic with Kr). Every valence electron therefore sits in the full $4p^6$ sub-level, $n=4,\ \ell=1$; one valid combination for a specific electron in that sub-level is $$\boxed{n=4,\ \ell=1,\ m_\ell=0,\ m_s=+\tfrac12}$$ (any of the six $(m_\ell,m_s)$ pairs $m_\ell\in\{-1,0,+1\}$, $m_s=\pm1/2$ is an equally valid answer, since all six $4p$ states are occupied).