The electron affinity of chlorine is 3.7 eV. 1 gram of chlorine is completely converted to $$\text{Cl}^-$$ ion in a gaseous state. ($$1\ \text{eV} = 23.06\ \text{kcal mol}^{-1}$$). Energy released in the process is

Electron affinity (E.A.) of chlorine is given per atom: $$\text{E.A.}=3.7\ \text{eV atom}^{-1}$$.

Step 1 - Convert electron affinity from eV atom⁻¹ to kcal mol⁻¹.
Given $$1\ \text{eV}=23.06\ \text{kcal mol}^{-1}$$, therefore

$$\text{E.A.}=3.7 \times 23.06 = 85.322\ \text{kcal mol}^{-1}\,.$$

Step 2 - Calculate moles of chlorine atoms present in 1 g of Cl.
Atomic mass of Cl ≈ $$35.5\ \text{g mol}^{-1}$$, so

$$n = \frac{1\ \text{g}}{35.5\ \text{g mol}^{-1}} = 0.02817\ \text{mol}\,.$$

Step 3 - Energy released when these moles gain electrons to form $$\text{Cl}^-$$ (g).
Energy released $$= n \times \text{E.A.}$$

$$\text{Energy} = 0.02817\ \text{mol} \times 85.322\ \text{kcal mol}^{-1} \approx 2.40\ \text{kcal}\,.$$

Thus, the energy liberated when 1 g of chlorine atoms is converted completely into gaseous $$\text{Cl}^-$$ ions is about $$2.4\ \text{kcal}$$.

Option D which is: 2.4 kcal