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