Amongst halogens,fluorine is the most powerfu | vedclass

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(D) The oxidizing power of a halogen is determined by the standard electrode potential $(E^\circ)$,which depends on three factors: enthalpy of dissoci

Vedclass · Accepted Answer

All of the above are correct

Vedclass · Answer

(D) The oxidizing power of a halogen is determined by the standard electrode potential $(E^\circ)$,which depends on three factors: enthalpy of dissociation,enthalpy of hydration,and enthalpy of sublimation.$1$. Fluorine has the lowest enthalpy of dissociation ($F-F$ bond is weak).$2$. Fluorine has the highest enthalpy of hydration due to its small size.$3$. These factors combined make the reduction potential of fluorine the highest,making it the strongest oxidizing agent.Therefore,the correct answer is $D$.

Amongst halogens,fluorine is the most powerful oxidizing agent because:

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