A) \[S{{F}_{4}}\]is polar and nonreactive
B) \[S{{F}_{6}}\]is nonpolar and very reactive
C) \[S{{F}_{6}}\]is a strong fluorinating agent
D) \[S{{F}_{4}}\]is prepared by fluorinating\[SC{{l}_{2}}\]with\[NaF\]
Correct Answer: D
Solution :
:\[3SC{{l}_{2}}+4NaF\xrightarrow[{}]{75{}^\circ C}S{{F}_{4}}+{{S}_{2}}C{{l}_{2}}+4NaCl\].
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