A) \[{{F}_{2}}\]
B) \[C{{l}_{2}}\]
C) \[{{I}_{2}}\]
D) \[B{{r}_{2}}\]
Correct Answer: A
Solution :
\[n{{F}_{2}}C=C{{F}_{2}}\xrightarrow{Polymerisation}-{{[{{F}_{2}}C-C{{F}_{2}}]}_{\pi }}-\]tetra fluoroethene Hence,\[{{F}_{2}}\]is present in teflon.
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