Mathrm-F-2- is better oxidising agent than mathrm-B-mathrm-r-2- It is due to-small size of flourine atommore electron repulsions in flourinehigher electronegativity of flourinenon-metallic nature of fluorine

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Toppr Admin · Accepted Answer

F is more electronegative than-xA0-Br- Therefore it can easily-xA0-accept e-x2212- and oxidise others-Therefore F2 is better oxidising agent than Br2

$F_{2}$ is better oxidising agent than $B r_{2}$ . It is due to:
small size of flourine atom
more electron repulsions in flourine
higher electronegativity of flourine
non-metallic nature of fluorine

F is more electronegative than Br. Therefore it can easily accept $e^{-}$ and oxidise others.
Therefore $F_{2}$ is better oxidising agent than $B r_{2}$

F2 is better oxidising agent than Br2. It is due to:small size of flourine atommore electron repulsions in flourinehigher electronegativity of flourinenon-metallic nature of fluorine

F is more electronegative than Br. Therefore it can easily accept e− and oxidise others.Therefore F2 is better oxidising agent than Br2

$F_{2}$ is better oxidising agent than $B r_{2}$ . It is due to:
small size of flourine atom
more electron repulsions in flourine
higher electronegativity of flourine
non-metallic nature of fluorine

F is more electronegative than Br. Therefore it can easily accept $e^{-}$ and oxidise others.
Therefore $F_{2}$ is better oxidising agent than $B r_{2}$