Question

Determine the energy required to dissociate a Na Cl molecule into a Na and a Cl atom, given that the electron affinity of Cl is 3.70 eV, the first ionization potential of Na is 5.14 eV, and the separation of Na Cl at equilibrium is 0.236 nm.
(a) 6.10 eV
(b) 5.57 eV
(c) 4.66 eV
(d) 3.29
​

Answer 1

Given info : the electron affinity of Cl is 3.70 eV, the first ionization potential of Na is 5.14 eV, and the separation of Na Cl at equilibrium is 0.236 nm.

To find : the energy required to dissociate a NaCl molecule into a Na and a Cl atom is..

solution : energy requied to dissocate NaCl into its constituent atoms = electrical potential energy + electron affinity of Cl + ionization enthalpy of Na.

= $\frac{-Ke^2}{r}+\text{electron affinity}+\text{ionization enthalpy}$

= $-\frac{9\times10^9\times(1.6\times10^{-19})^2}{0.236\times10^{-9}}J+ (-3.7eV)+5.14eV$

= $-97.627\times10^{-20}J +1.44eV$

we know, 1 eV = 1.6 × 10⁻¹⁹ J

so, -97.627 × 10⁻²⁰ J = -6.1 eV

= -6.10 eV + 1.44 eV

= -4.66 eV

therefore the energy required to dissociate NaCl into Na and Cl atoms is 4.66 eV . hence, option (c) is correct.