How much is the effecting shielding of the 1s electrons from the 2s electrons? . Because the 2s subshell overlaps with the 1s subshell this will cause the 1s electrons to 'feel' less charge than the nucleus charge.

However due to the fact that the 2s subshell is much bigger than the 1s subshell the effect of shielding will be small correct?


1 Answer 1


The effect of "reverse shielding", as I call it, is on the order of one volt of potential difference in light elements where the $n=2$ electrons are the valence electrons.

Let's look at lithium. From this data page the effective nuclear charge on the $1s$ electrons in the neutral atom is 2.691. If we assume that the effective nuclear charge is related to the electronic energy level by

Energy$=(-13.6 eV)(Z_{eff}^2)$

then we infer that the orbital energy level is -98.5 eV for each electron.

Now what is that energy level if we take away the 2s electron? For this we need the electronic energy level of the $1s$ orbital in the lithium ion, which is obtained by averaging the second and third ionization energies of lithium given here. That leads to -99.1 eV per electron. The 2s electron has effectively increased the energy level of each 1s electron by 0.6 eV.

A similar calculation for beryllium gives a difference of about 1.1 eV per $1s$ electron (-184.7 eV from the effective charge on $1s$ in neutral Be vs -185.8 eV from the third and fourth ionization energies after forming the di-cation). The second 2s electron has basically doubled the shielding.

The electron volt differences determined above have some impact on chemical reactions, whose energies are on the order of one electron volt per electron transferred in a redox reaction.

  • $\begingroup$ Can we apply this to molecular orbitals as well? $\endgroup$
    – McCarter
    Jun 6, 2020 at 3:46

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