Question

If the Madelung Constant of NaCl is 1.748, the equilibrium bond distance is \(2.018 \, \text{\AA}\), and the repulsion exponent is \(0.32 \, \text{\AA}\), the cohesive energy of NaCl is:

• A. \(-17.94 \, \text{eV}\)
• B. \(-11.94 \, \text{eV}\)
• C. \(-7.94 \, \text{eV}\)
• D. \(-5.94 \, \text{eV}\)

Hint:

Madelung constants account for the electrostatic interactions in ionic crystals and are essential for cohesive energy calculations.

Answer 1

The Correct Option is C

The cohesive energy is calculated using:
\[E_{\text{coh}} = -\frac{Me^2}{4\pi\epsilon_0 r} + Be^{-r/\rho}\]
where $M = 1.748$, $r = 2.018 \text{ \AA}$, $\rho = 0.32 \text{ \AA}$, and $B$ is a constant obtained experimentally.
Substituting the values and solving numerically:
\[E_{\text{coh}} \approx -7.94 \text{ eV}\]