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Anomalous Behaviour of Fluorine

The halogen t...

Question

The halogen that is most easily reduced is:

F_{2}

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C l_{2}

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B r_{2}

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I_{2}

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Solution

The correct option is B $F_{2}$
$F_{2}$ is easily reduced as $F^{- 1}$ is the strongest oxidising agent. Oxidising power can be explained by the fact that the ease with which a halogen $[X]$ forms the halide ion $H^{-}$ decreases down the group.

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Similar questions

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

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

x

y

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

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

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

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

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

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