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Difficulty of Carbon to Form a Stable Ion

Arrange the h...

Question

Arrange the halogens $F_{2}, C l_{2}, B r_{2}, I_{2}$ in order of their increasing reactivity with alkanes.

F_{2} < C l_{2} < B r_{2} < I_{2}

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I_{2} < B r_{2} < C l_{2} < F_{2}

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B r_{2} < I_{2} < C l_{2} < F_{2}

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B r_{2} < C l_{2} < F_{2} < I_{2}

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Solution

The correct option is B $I_{2} < B r_{2} < C l_{2} < F_{2}$

As we move down the group, electronegativity of elements decreases (Since the $F^{-}$ ion formed as HF is most stable, $F_{2}$ will be most reactive). Therefore, for halogens, electronegativity order will be:
$I_{2} < B r_{2} < C l_{2} < F_{2}$
Now, halogens react with alkanes to give halogenation reaction and here reactivity of halogens with alkanes increases as their electronegativity increases.
Thus, increasing order of reactivity of halogens is
$I_{2} < B r_{2} < C l_{2} < F_{2}$

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(x^{2} + y^{2})

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I_{2} < {B r}_{2} < {C l}_{2} < F_{2}

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