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Problem Which step(s) is/are endothermic in the Born–Haber cycle for the formation of LiCl?
a. 1 2 Cl 2 ( g ) → Cl ( g ) \frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g}) 2 1 Cl 2 ( g ) → Cl ( g ) and Li ( s ) → Li ( g ) \text{Li}(\text{s}) \rightarrow \text{Li}(\text{g}) Li ( s ) → Li ( g )
b. Cl ( g ) + e − → Cl − ( g ) \text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g}) Cl ( g ) + e − → Cl − ( g ) and Li ( g ) → Li + ( g ) + e − \text{Li}(\text{g}) \rightarrow \text{Li}^+(\text{g}) + \text{e}^- Li ( g ) → Li + ( g ) + e −
c. Li + ( g ) + Cl − ( g ) → LiCl ( s ) \text{Li}^+(\text{g}) + \text{Cl}^-(\text{g}) \rightarrow \text{LiCl}(\text{s}) Li + ( g ) + Cl − ( g ) → LiCl ( s )
d. 1 2 Cl 2 ( g ) → Cl ( g ) \frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g}) 2 1 Cl 2 ( g ) → Cl ( g ) and Cl ( g ) + e − → Cl − ( g ) \text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g}) Cl ( g ) + e − → Cl − ( g )
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