A) \[PbC{{l}_{4}}<PbC{{l}_{2}}<CaC{{l}_{2}}<NaCl\]
B) \[PbC{{l}_{2}}<\text{ }PbC{{l}_{4}}\text{ }<\text{ }CaC{{l}_{2}}\text{ }<\text{ }NaCl\]
C) \[PbC{{l}_{2}}<PbC{{l}_{4}}<NaCl<CaC{{l}_{2}}\]
D) \[PbC{{l}_{4}}<PbC{{l}_{2}}<NaCl<CaC{{l}_{2}}\]
Correct Answer: A
Solution :
\[NaCl\] is the most ionic among these due to largest difference in electro negativities of metal and chlorine. Similarly \[CaC{{l}_{2}}\] is more ionic than other two chlorides. In \[PbC{{l}_{2}}\] and \[PbC{{l}_{4}},\]according to Fazans rule, \[PbC{{l}_{2}}\] is more ionic as \[P{{b}^{4+}}\] cation is smaller than \[P{{b}^{2+}}\] and hence, it will polarise \[C{{l}^{-}}\] ions to greater extent. Hence, the correct increasing order of ionic character will be \[PbC{{l}_{4}}<PbC{{l}_{2}}<CaC{{l}_{2}}<NaCl\]
You need to login to perform this action.
You will be redirected in
3 sec
