As we know that the combined effect of attractive and repulsive force on valence electrons results as an attraction of the valence electrons by the nucleus which by the way is obvious

  • But i have a difficulty understanding it happens as electrostatic force >>>>> gravitational force and electrostatic force is inversely proportional to the square of the distance between the charges ,therefore as the inner electrons is close to the valence electrons they should repel it with much greater force than the nucleus attracting it (as proton and electron have same magnitude of charge).

    -It's the problem i have.Lemme know if it has something to do with vector addition of forces or something else.


1 Answer 1


Though the distance between the valence electrons and the inner shell electrons is much smaller than the distance between the nucleus and the valence electron, the electrons individually have little charge as compared to the nucleus.

Moreover all the electrons don't usually come very close to each other due to mutual repulsion among them. They also never line up. There will be a lot of electrons on the other side of the nucleus rather than crowding up at a place hence their distance from the concerned valence electron will be more than that from the nucleus. Also, the distance between a valence electron and the nucleus more or less remains the same as long as the electron continues to be in the same orbital.

However, a lot of shielding is done by the inner electrons which makes removing an electron from the valence shell very easy.

I forgot to mention this point earlier that when an element has a half-filled or full-filled (valence) sub-shell, the energy required to remove an electron from there suddenly increases due to stability reasons, due to symmetry and exchange energy in case of half-filled sub-shells or because of spin-pairing in the case of full-filled sub-shells. The effect is more pronounced in case of full-filled sub-shells like in group $12$ (fully filled d subshell) and group $18$ (fully filled p subshell) elements.

  • 1
    $\begingroup$ However, a lot of shielding is done by the inner electrons which makes removing an electron from the valence shell very easy. Not so in the case of noble gases (formerly Group VIII) $\endgroup$
    – Gert
    Oct 24, 2019 at 9:12
  • $\begingroup$ @Gert Sorry, I forgot to write that before. Thanks for pointing out. $\endgroup$ Oct 24, 2019 at 10:28
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    $\begingroup$ You can include halogens in your list of exceptions. Maybe more. Come to think of it, what is your criterion for "easy"? $\endgroup$
    – my2cts
    Oct 24, 2019 at 11:36
  • $\begingroup$ @my2cts Yeah...thanks for telling. We all know it's a general trend that the ionization energy increases from left to right. The halogens have high electronegativities. I just answered the OP's question without taking these into account. :-) $\endgroup$ Oct 24, 2019 at 12:23

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