Arrange the halogens F_{2}, Cl_{2}, Br_{2}, I_{2}, in order of their increasing reactivity with alkanes.
(i) I_{2}< Br_{2} < Cl_{2} < F_{2}
(ii) Br_{2} < Cl_{2} < F_{2} < I_{2}
(iii) F_{2} < Cl_{2} < Br_{2} < I_{2}
(iv) Br_{2} < I_{2} < Cl_{2} < F_{2}

Answers (1)

The answer is the option (i) I_{2}< Br_{2} < Cl_{2} < F_{2}

Explanation:  Since electronegativity of halogens decreases down the group, Fluorine is the most electronegative. As electronegativity of Fluorine decreases their reactivity with alkanes also decreases. Thus, F_{2} is highly reactive, whereas I_{2} is the least reactive.

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