The enthalpy changes for the following processes are listed below
$\begin{array}{ll} {Cl}_{2} (g) = 2 Cl (g); 242.3 kJ {mol}^{- 1} \\ I_{2} (g) = 2 I (g); 151.0 kJ {mol}^{- 1} \\ ICl (g) = I (g) + Cl (g); 211.3 kJ {mol}^{- 1} \\ I_{2} (g) = I_{2} (g); 62.76 kJ {mol}^{- 1} \end{array}$
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 :

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$+ 16.8 kJ {mol}^{- 1}$

$- 14.6 kJ {mol}^{- 1}$

$- 16.8 kJ {mol}^{- 1}$

$+ 244.8 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} I_{2} \to \frac{1}{2} {Cl}_{2} (g) \to ICl (g) \\ {ΔH}_{1 Cl} = - [\frac{1}{2} Δ {HI}_{2} (s) \to I_{2} (g) + \frac{1}{2} {ΔH}_{T - I} + \frac{1}{2} {ΔH}_{Cl} - cl] \\ - [{ΔH}_{Y} - Cl] \\ = - [\frac{1}{2} \times 62.76 + \frac{1}{2} \times 1.51.0 + \frac{1}{2} \times 242.3] - [2.11.3] \\ = - 16.83 KJ / mol \end{array}$