1. Write the required formation reaction

The standard enthalpy of formation of $ICl(g)$ is for:

$\frac{1}{2}I_2(s) + \frac{1}{2}Cl_2(g) \rightarrow ICl(g)$

We need $\Delta H_f^\circ$ for this reaction.

2. Use the given bond dissociation / phase-change data

Given:

$Cl_2(g) \rightarrow 2Cl(g), \quad \Delta H = 242.3\,\text{kJ mol}^{-1}$ $I_2(g) \rightarrow 2I(g), \quad \Delta H = 151.0\,\text{kJ mol}^{-1}$ $ICl(g) \rightarrow I(g)+Cl(g), \quad \Delta H = 211.3\,\text{kJ mol}^{-1}$ $I_2(s) \rightarrow I_2(g), \quad \Delta H = 62.76\,\text{kJ mol}^{-1}$

3. Convert reactants in standard states into atoms

From $\frac12 I_2(s)$ to $I(g)$:

• Sublimation of iodine: $\frac12 I_2(s) \rightarrow \frac12 I_2(g), \quad \Delta H = \frac{62.76}{2}=31.38$

• Dissociation of gaseous iodine: $\frac12 I_2(g) \rightarrow I(g), \quad \Delta H = \frac{151.0}{2}=75.5$

So,

$\frac12 I_2(s) \rightarrow I(g), \quad \Delta H = 31.38+75.5=106.88\,\text{kJ mol}^{-1}$

From $\frac12 Cl_2(g)$ to $Cl(g)$:

$\frac12 Cl_2(g) \rightarrow Cl(g), \quad \Delta H = \frac{242.3}{2}=121.15\,\text{kJ mol}^{-1}$

Thus,

$\frac12 I_2(s)+\frac12 Cl_2(g) \rightarrow I(g)+Cl(g)$

has

$\Delta H = 106.88+121.15=228.03\,\text{kJ mol}^{-1}$

4. Form $ICl(g)$ from atoms

Given dissociation:

$ICl(g) \rightarrow I(g)+Cl(g), \quad \Delta H = 211.3\,\text{kJ mol}^{-1}$

So the reverse reaction is:

$I(g)+Cl(g) \rightarrow ICl(g), \quad \Delta H = -211.3\,\text{kJ mol}^{-1}$

5. Apply Hess's law

Add the two steps:

$\frac12 I_2(s)+\frac12 Cl_2(g) \rightarrow I(g)+Cl(g) \quad (228.03)$ $I(g)+Cl(g) \rightarrow ICl(g) \quad (-211.3)$

Net:

$\frac12 I_2(s)+\frac12 Cl_2(g) \rightarrow ICl(g)$

Therefore,

$\Delta H_f^\circ = 228.03-211.3=16.73\,\text{kJ mol}^{-1}$

So,

$\boxed{\Delta H_f^\circ\big(ICl(g)\big) \approx +16.8\,\text{kJ mol}^{-1}}$

6. Check options

• A: $-14.6$ ❌
• B: $-16.8$ ❌
• C: $+16.8$ ✅
• D: $+244.8$ ❌

Hence the correct option is C.