The order of successive ionization enthalpies of group 15 are as follows :

$$\Delta H_1<\Delta H_2<\Delta H_3$$

But as all the fifteen group elements have a very stable electronic configuration with half filled p orbital so it would require more energy to remove an electron from an extraordinarily stable configuration so the order according to me should be $\Delta H_1>\Delta H_2>\Delta H_3$.

please guide me if I am wrong.

  • $\begingroup$ @Jan: the answer states, "Second ionization energies are almost always higher". In which cases for the second, third, or any of the further ionisations does the "successive ionisations cost increasing amounts of energy" rule not work? The answerer hasn't been online for a while, which is why I pinged you. $\endgroup$ – harry May 22 at 1:16
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    $\begingroup$ @harry The ping isn’t really appreciated. I am not aware of any case. The answerer may have chosen a phrasing that allows wiggle room in case there is indeed one case somewhere so that the answer remains correct. I don’t know, I can’t look into their brain. $\endgroup$ – Jan May 24 at 17:04
  • $\begingroup$ @Jan: Okay, sorry and thanks. $\endgroup$ – harry May 25 at 1:03

While you are correct that the first ionization energy should have a larger value for all the group 15 elements, this has a minuscule effect on the ionization energy compared to the energy that is required to separate charges. Second ionization energies are almost always higher than first ionization energies because the nucleus already has a positive charge, creating a larger coulomb force that needs to be overcome as part of the ionization process. In addition, additional ionization increases the charge on the nucleus of the ion, which progressively makes the ion less and less stable.


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