A) $ AlCl_3 $
B) $ CI_4 $
C) $ Al_4C_3 $
D) both [a] and [b]
Correct Answer: D
$ AlI_3 $ on reaction with $ CCl_4 $ , gives the $ AlCl_3 $ $ 4AlI_3+3CCl_4\xrightarrow{{}}4AlCl_3+3CI_4 $
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