Question

Flourine is the best oxidising agent because it has:

• A. highest electron affinity
• B. highest $E_{red}^{\circ }$
• C. highest $E_{oxid}^{\circ }$
• D. lowest electron affinity

Answer 1

The correct option is A highest electron affinity

Oxidizing agent means element, which oxidizes others by itself getting reduced.

Oxidation occurs by loss of $e^{-}$

For eg:-

$\stackrel{0}{H}⟶\stackrel{+1}{H^{+}}+e^{-}$

loss of $e^{-}$

$F+e^{-}⟶F^{-}$

Here $F$ oxidizes $H$ and itself gets reduced.

(Refer to Image)

$\because F$ has high tendency to gain $e^{-}$ and to stabilize itself, i.e. have high $EGE$ or $e^{-}$ affinity. Hence accepts $e^{-}$ from $H$ and oxidized $H$ and act as good oxidizing agent.