An Oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:

$$\ce{F2 + 2 X- → 2F^- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$

We also know that Fluorine has very high charge/mass ratio as it is very small in size,

its electron gain enthalpy is very high.

Also, many sources state that fluorine

1.florine has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.

2.it is achieving octect so electron gain is favorable

My questions

1.What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power? (As we know that lithium is the best reducing agent , and also has high hydration enthalpy , we can’t say it is a oxidising agent)

2.What am I missing here? What is the reason for such high standard electrode potential for fluorine ?

3.Why wouldn't the electron cloud of fluorine atom repel the electron cloud of substance being oxidized? It seems paradoxical the fluorine pulls electron cloud towards itself so easily

Even in this post it is mentioned that electron cloud is repelled.

is there any better explanation other than "octect is being achieved" , as mentioned in 2nd statement

  • $\begingroup$ look up Born-Haber cycle, hydration energy effects will make sense. $\endgroup$ – William R. Ebenezer Mar 10 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.