Question 1.4: ENERGY OF SECONDARY BONDING Consider the van der Waals bondi...
ENERGY OF SECONDARY BONDING Consider the van der Waals bonding in solid argon. The potential energy as a function of interatomic separation can generally be modeled by the Lennard–Jones 6–12 potential energy curve, that is,
E(r) = −Ar^{−6} + Br^{−12}
where A and B are constants. Given that A = 8.0 × 10^{−77} J m^{6} and B = 1.12 × 10^{−133} J m^{12}, calculate the bond length and bond energy (in eV) for solid argon.
Bonding occurs when the potential energy is at a minimum. We therefore differentiate the Lennard–Jones potential E(r) and set it to zero at r = r_{o^{,}} the interatomic equilibrium separation or
\frac{d E}{d r}=6 A r^{-7}-12 B r^{-13}=0 at r=r_{o}
that is,
r_{o}^{6}=\frac{2 B}{A}
or
r_{o}=\left[\frac{2 B}{A}\right]^{1 / 6}
Substituting A = 8.0 × 10^{−77} and B = 1.12 × 10^{−133} and solving for r_{o}, we find
r_{o} = 3.75 × 10^{−10} m \quad or \quad 0.375 nm
When r = r_{o} = 3.75 × 10^{−10} m, the potential energy is at a minimum, and the magnitude E_{min} is the bonding energy E_{bond^{,}} so
E_{bond}=\left|-A r_{o}^{-6}+B r_{o}^{-12}\right|=\left|-\frac{8.0 \times 10^{-77}}{\left(3.75 \times 10^{-10}\right)^{6}}+\frac{1.12 \times 10^{-133}}{\left(3.75 \times 10^{-10}\right)^{12}}\right|
that is,
E_{bond} = 1.43 × 10^{−20 } J or 0.089 eV
Notice how small this energy is compared to primary bonding.