MCQ
Arrange the halogens $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2, \mathrm{I}_2$, in order of their increasing reactivity with alkanes :
  • $ \mathrm{I}_2 < \mathrm{Br}_2 < \mathrm{Cl}_2 < \mathrm{F}_2 $
  • B
    $ \mathrm{Br}_2 < \mathrm{Cl}_2 < \mathrm{F}_2 < \mathrm{I}_2 $
  • C
    $ \mathrm{~F}_2 < \mathrm{Cl}_2 < \mathrm{Br}_2 < \mathrm{I}_2 $
  • D
    $ \mathrm{Br}_2 < \mathrm{I}_2 < \mathrm{Cl}_2 < \mathrm{F}_2 $

Answer

Correct option: A.
$ \mathrm{I}_2 < \mathrm{Br}_2 < \mathrm{Cl}_2 < \mathrm{F}_2 $
Fluorine is highly electronegative element. Electronegativity of halogens decreases down the group. As the electronegativity of halogens decreases, reactivity with alkanes, decreases. Therefore, $F_2$ reacts vigorously and $I_2$ reacts too slow.

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