A) \[LiCl>KCl>NaCl>CsCl\]
B) \[CsCl>KCl>NaCl>LiCl\]
C) \[NaCl>KCl>LiCl>CsCl\]
D) \[KCl>CsCl>NaCl>LiCl\]
Correct Answer: D
Solution :
Stability of a compound depends upon its enthalpy of formation\[\Delta {{H}_{f}}\]. The more negative value of shows more stability of a compound. Thus, \[KCl\] is more stable and \[\Delta {{H}_{f}}\] is least stable. \[\Delta {{H}_{f}}\]for\[LiCl=-408.8\,\,kJ\,\,mo{{l}^{-1}}\] \[\Delta {{H}_{f}}\]for\[NaCl=-412.5\,\,kJ\,\,mo{{l}^{-1}}\] \[\Delta {{H}_{f}}\]for\[CsCl=-433\,\,kJ\,\,mo{{l}^{-1}}\] \[\Delta {{H}_{f}}\]for\[KCl=-436\,\,kJ\,\,mo{{l}^{-1}}\]
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