The enthalpy changes for the following processes are listed below: $$Cl_2(g) = 2Cl(g),\ 242.3\,kJ\,mol^{-1}$$; $$I_2(g) = 2I(g),\ 151.0\,kJ\,mol^{-1}$$; $$ICl(g) = I(g) + Cl(g),\ 211.3\,kJ\,mol^{-1}$$; $$I_2(s) = I_2(g),\ 62.76\,kJ\,mol^{-1}$$. Given that the standard states for iodine and chlorine are $$I_2(s)$$ and $$Cl_2(g)$$, the standard enthalpy of formation for $$ICl(g)$$ is

The standard enthalpy of formation, $$\Delta_\mathrm fH^\circ$$, is defined for the reaction that produces one mole of the compound from the elements in their standard states.

For iodine monochloride gas this reaction is
$$\tfrac12\,I_2(s) + \tfrac12\,Cl_2(g) \rightarrow ICl(g)$$ $$-(1)$$

We will evaluate the enthalpy change of route (1) by breaking it into simple steps whose enthalpies are given, and then applying Hess’s law.

Step A: Sublimation of solid iodine
$$\tfrac12\,I_2(s) \rightarrow \tfrac12\,I_2(g)$$
Enthalpy change $$= \tfrac12 \times 62.76 = 31.38\;{\rm kJ\,mol^{-1}}$$

Step B: Dissociation of iodine molecule
$$\tfrac12\,I_2(g) \rightarrow I(g)$$
For the full reaction $$I_2(g)\rightarrow 2I(g)$$ the enthalpy change is $$151.0\;{\rm kJ}$$, so per iodine atom
$$\Delta H = \tfrac{151.0}{2}=75.5\;{\rm kJ\,mol^{-1}}$$

Step C: Dissociation of chlorine molecule
$$\tfrac12\,Cl_2(g) \rightarrow Cl(g)$$
For $$Cl_2(g)\rightarrow 2Cl(g)$$, $$\Delta H = 242.3\;{\rm kJ}$$, hence per chlorine atom
$$\Delta H = \tfrac{242.3}{2}=121.15\;{\rm kJ\,mol^{-1}}$$

After steps A-C we have isolated atoms $$I(g)+Cl(g)$$. Their total enthalpy input is
$$\Delta H_{\text{A-C}} = 31.38 + 75.5 + 121.15 = 228.03\;{\rm kJ\,mol^{-1}}$$

Step D: Formation of ICl from its atoms
$$I(g) + Cl(g) \rightarrow ICl(g)$$
The reverse process $$ICl(g)\rightarrow I(g)+Cl(g)$$ has $$\Delta H = 211.3\;{\rm kJ}$$, so step D releases
$$\Delta H = -211.3\;{\rm kJ\,mol^{-1}}$$

Total enthalpy change for reaction (1)
$$\Delta_\mathrm fH^\circ = \Delta H_{\text{A-C}} + \Delta H_{\text{D}}$$
$$= 228.03 - 211.3 = 16.73\;{\rm kJ\,mol^{-1}} \approx 16.8\;{\rm kJ\,mol^{-1}}$$

Therefore, the standard enthalpy of formation of $$ICl(g)$$ is positive and equals about $$+16.8\,{\rm kJ\,mol^{-1}}$$.

Option C which is: $$+16.8\,kJ\,mol^{-1}$$