A) \[Sn{{F}_{2}}<SnC{{l}_{2}}<Sn{{F}_{4}}<SnC{{l}_{4}}<SiC{{l}_{4}}\]
B) \[Sn{{F}_{2}}<SnC{{l}_{2}}<Sn{{F}_{4}}<SiC{{l}_{4}}<SnC{{l}_{4}}\]
C) \[SiC{{l}_{4}}<SnC{{l}_{4}}<Sn{{F}_{4}}<SnC{{l}_{2}}<Sn{{F}_{2}}\]
D) \[SnC{{l}_{4}}<Sn{{F}_{4}}<SnC{{l}_{2}}<Sn{{F}_{2}}<SiC{{l}_{4}}\]
Correct Answer: C
Solution :
[c] As the size of the cations increases in the order, \[S{{i}^{4+}}<S{{n}^{4+}}<S{{n}^{2+}}\] and for size of anions, \[{{F}^{-}}<C{{l}^{-}}\] so the order of increasing ionic character is, \[SiC{{l}_{4}}<SnC{{l}_{4}}<Sn{{F}_{4}}<SnC{{l}_{2}}<Sn{{F}_{2}}\]
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