# Why does helium seem to exhibit an ionization energy affected by the shielding of electrons?

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.

• 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. Feb 26 '14 at 2:15
• 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! Feb 26 '14 at 2:24
• @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. Feb 26 '14 at 2:47

It may help to think about the reverse process : the capture of a free electron by a $$\ce{He+}$$ ion to form $$\ce{He}$$. When this free electron is still far, it is intuitive that the attractive force between it and the two protons is indeed electrically shielded by the electron that is already circling the nucleus : it is attracted by the $$+2$$ charge and repelled by the $$-1$$ charge. (We will ignore magnetic effects, which are typically much smaller)
If there was no shielding, then the second electron would feel the entire $$Z=2$$ charge and settle at a radius of $$r=\frac{1}{2}$$ of the Bohr radius, just like the first electron. Its ionization potential would be $$\displaystyle\frac{Z}{r}=4$$ times that of hydrogen. From this perspective, the difference that shielding makes isn't between "almost twice" and "twice", but between "almost twice" and "four times".
How each of the calculation methods (e.g. Slater) translates this physical shielding effect into factors is another matter. The clearest is perhaps Mills* model, where every electron completely shields one proton. The (electrical part of) the first ionization energy for $$\ce{He}$$-like atoms is thus simply $$\displaystyle\frac{Z-1}{r}$$, with $$r$$ determined by a force balance equation $$(0.567$$ for $$\ce{He)}$$. The second ionization energy is $$\displaystyle\frac{Z-0}{r}$$, with $$r=\frac{1}{Z}$$ as per the classical Bohr model $$(0.500$$ for $$\ce{He+)}$$. This formula extends to subsequent electrons, but determining $$r$$ becomes a bit more cumbersome.
*Randell L. Mills, The Grand Unified Theory of Classical Physics, 2018 edition (ISBN: 978-0-9635171-5-9) Chapter 7 "Two-electron atoms", eqs. (7.35) and (7.45). The general formula for the radius $$r$$ of a $$\ce{He}$$-like atom is $$\frac{1-\frac{1}{Z}\sqrt{s(s+1)}}{Z-1}$$, with $$s=\frac{1}{2}$$ the electron spin. For $$\ce{He}$$ this works out to $$1-\frac{1}{4}\sqrt{3}$$, or $$0.567$$ times the Bohr radius.