A) \[LiCl<NaCl<BeC{{l}_{2}}\]
B) \[BeC{{l}_{2}}<NaCl<LiCl\]
C) \[NaCl<LiCI<BeC{{l}_{2}}\]
D) \[BeC{{l}_{2}}<LiCl<NaCl\]
Correct Answer: C
Solution :
On the basis of Fajan's rule, lower the size of cation higher will be its- polarising power and higher will be covalent character. \[\therefore \]Polarising power\[\propto \frac{1}{size\,\,of\,\,cation}\] Covalent character \[\propto \] Polarising power So the correct order is\[NaCl<LiCl<BeC{{l}_{2}}\] (The order of size of cation\[N{{a}^{+}}>L{{i}^{2+}}>B{{e}^{2+}})\]
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