I recently came across a question about the stability of radicals and this option was given as correct.

  • $\ce {F2C^.H}>\ce {C^.H3} $

And the solution was given that $\ce{FC^.H2}$ and $\ce{F2CH^.}$ are more stable than $\ce {C^.H3}$ but $\ce{C^.F3}$ is less stable. Can anyone explain why?
If it has got something to do with inductive effect then all three $\ce {F2C^.H},\ce{FC^.H2},\ce{C^.F3}$ should be less stable than $\ce {C^.H3} $. Right?

  • $\begingroup$ you might also want to consider lone pair repulsion between F atoms. $\endgroup$ – Safdar Faisal Jul 15 '20 at 13:15
  • $\begingroup$ chemistry.stackexchange.com/questions/49754/… $\endgroup$ – Mithoron Jul 15 '20 at 15:36
  • 1
    $\begingroup$ @RahulVerma Do you have anything to back up that claim? Normal radicals are electron-deficient. $\endgroup$ – Zhe Jul 16 '20 at 17:11
  • $\begingroup$ @Zhe: No. I don't have. As the claim was wrong. $\endgroup$ – Rahul Verma Jul 17 '20 at 4:32

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