Question 1.3: IONIC BONDING AND LATTICE ENERGY The potential energy E per ...
IONIC BONDING AND LATTICE ENERGY The potential energy E per Na^{+} –Cl^{−} pair within the NaCl crystal depends on the interionic separation r as
E(r)=-\frac{e^{2} M}{4 \pi \varepsilon_{o} r}+\frac{B}{r^{m}} [1.4] Energy per ion pair in an ionic crystal
where the first term is the attractive and the second term is the repulsive potential energy, and M, B, and m are constants explained in the following. If we were to consider the potential energy PE of one ion pair in isolation from all others, the first term would be a simple Coulombic interaction energy for the Na^{+}–Cl^{−} pair, and M would be 1. Within the NaCl crystal, however, a given ion, such as Na^{+}, interacts not only with its nearest six Cl^{−} neighbors (Figure 1.9b), but also with its twelve second neighbors (Na^{+}), eight third neighbors (Cl^{−}), and so on, so the total or effective PE has a factor M, called the Madelung constant, that takes into account all these different Coulombic interactions. M depends only on the geometrical arrangement of ions in the crystal, and hence on the particular crystal structure; for the FCC crystal structure, M = 1.748. The Na^{+} –Cl^{−} ion pair also has a repulsive PE that is due to the repulsion between the electrons in filled electronic subshells of the ions. If the ions are pushed toward each other, the filled subshells begin to overlap, which results in a strong repulsion. The repulsive PE decays rapidly with distance and can be modeled by a short-range PE of the form B ∕ r^{m} as in the second term in Equation 1.4 where for Na^{+} –Cl^{−}, m = 8 and B = 6.972 \times 10^{−96} J m^{8}. Find the equilibrium separation (r_{o}) of the ions in the crystal and the ionic bonding energy, defined as −E(r_{o}). Given the ionization energy of Na (the energy to remove an electron) is 5.14 eV and the electron affinity of Cl (energy released when an electron is added) is 3.61 eV, calculate the atomic cohesive energy of the NaCl crystal as joules per mole.
Bonding occurs when the potential energy E(r) is a minimum at r = r_{o} corresponding to the equilibrium separation between the Na^{+} and Cl^{−} ions. We differentiate E(r) and set it to zero at r = r_{o,}
\frac{d E(r)}{d r}=\frac{e^{2} M}{4 \pi \varepsilon_{o} r^{2}}-\frac{m B}{r^{m+1}}=0 \quad \text { at } r=r_{o}
Solving for r_{o,}
r_{o}=\left[\frac{4 \pi \varepsilon_{o} B m}{e^{2} M}\right]^{1 /(m-1)} [1.5] Equilibrium ionic separation
Thus,
\begin{aligned} r_{o} &=\left[\frac{4 \pi\left(8.85 \times 10^{-12} Fm ^{-1}\right)\left(6.972 \times 10^{-96} J m ^{8}\right)(8)}{\left(1.6 \times 10^{-19} C \right)^{2}(1.748)}\right]^{1 /(8-1)} \\ &=0.281 \times 10^{-9} m \quad \text { or } \quad 0.28 nm \end{aligned}
The minimum energy E_{\min} per ion pair is E(r_{o}) and can be simplified further by substituting for B in terms of r_{o}:
E_{\min }=-\frac{e^{2} M}{4 \pi \varepsilon_{o} r_{o}}+\frac{B}{r_{o}^{m}}=-\frac{e^{2} M}{4 \pi \varepsilon_{o} r_{o}}\left(1-\frac{1}{m}\right) [1.6] Minimum PE at bonding
Thus,
\begin{aligned} E_{\min } &=-\frac{\left(1.6 \times 10^{-19} C \right)^{2}(1.748)}{4 \pi\left(8.85 \times 10^{-12} Fm ^{-1}\right)\left(2.81 \times 10^{-10} m \right)}\left(1-\frac{1}{8}\right) \\ &=-1.256 \times 10^{-18} J \quad \text { or } \quad-7.84 eV \end{aligned}
This is the energy with respect to two isolated Na^{+} and Cl^{−} ions. We therefore need 7.84 eV to break up a NaCl crystal into isolated Na^{+} and Cl^{−} ions, which represents the ionic cohesive energy. Some authors call this ionic cohesive energy simply the lattice energy. To take the crystal apart into its neutral atoms, we have to transfer the electron from the Cl^{−} ion to the Na^{+} ion to obtain neutral Na and Cl atoms. It takes 3.61 eV to remove the electron from the Cl^{−} ion, but 5.14 eV is released when it is put into the Na^{+} ion. Thus, we need 7.84 eV + 3.61 eV but get back 5.14 eV.
Bond energy per Na–Cl pair = 7.84 eV + 3.61 eV − 5.14 eV = 6.31 eV
The atomic cohesive energy is 3.1 eV/atom. In terms of joules per mole of NaCl, this is
E_{\text {cohesive }}=(6.31 eV )\left(1.6022 \times 10^{-19} J / eV \right)\left(6.022 \times 10^{23} mol ^{-1}\right)=608 kJ mol ^{-1}