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: 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 \[NaCl<LiCl<BaC{{l}_{2}}\] |
| (The order of size of cation\[N{{a}^{+}}>L{{i}^{+}}>B{{e}^{2+}}\]) |
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