According to Wikipedia, the shielding effect only happen in atoms which have more than 1 electron shells. The core electrons repel the electron in the valence shell. However, I have heard some lectures mention the shielding effect on helium atom when they explained why the first ionization energy of $\ce{He}$ is not exactly 2 times more than $\ce{H}$.

Why is this? According to the definition of shielding effect, there shouldn't be one in $\ce{He}$ atom.

  • $\begingroup$ In $\ce{He}$ you have 2 electrons instead of 1 for $\ce{He^{+}}$ (which would have two times the ionization potential of $\ce{H}$). Loosely speaking, the electrons repel each other, thus raising the energy of the electron state compared to that in $\ce{He^{+}}$. Thus, the ionization energy is smaller than 2 times that of $\ce{H}$. I'm not sure whether one should call that shielding but it has a similar effect - only that the effect is weaker than with "traditional" shielding by core electron shells. $\endgroup$ – Philipp Feb 26 '14 at 2:15
  • $\begingroup$ I'm pretty sure it still counts as shielding. Even the simplified Slater rules take same-subshell electron shielding into effect. More elaborate schemes like Clementi-Raimondi shielding actually incorporate shielding of inner electrons by outer shells, too! $\endgroup$ – Nicolau Saker Neto Feb 26 '14 at 2:24
  • $\begingroup$ @NicolauSakerNeto I was a bit unsure whether this is in general referred to as shielding. Since in my courses in general chemistry the term shielding was introduced only in the context of core electron shielding I thought there might be a different term for same-subshell shielding in the general chemistry community. $\endgroup$ – Philipp Feb 26 '14 at 2:47

The premise is problematic as there is no 'shielding' effects in an isolated Helium atom.

However, Phillipp was correct in explaining why the ionization energy of Helium is not exactly twice that of Hydrogen. This is because the energies of each electron are higher due to their correlated nature (i.e. they repel each other into higher, non-ground-state-like energies). Therefore, it takes LESS energy to rip one of those electrons out of the system.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.