# Do Slater's rules apply to neutral lithium (at least approximately)?

For lithium, the electronic configuration is $\ce{(1s)^2(2s)^1}$. I tried to calculate $Z_{eff}$ for the $\ce{2s}$ orbital of the neutral atom as follows:

$$Z_{eff} = Z-\sigma$$ $$Z_{eff} = 3-(2\times0.3)$$ $$Z_{eff} = 2.4$$

But the Wikipedia article's table of effective nuclear charges (linked below) states that $Z_{eff}$ is 1.279 for the $\ce{2s}$ lithium orbital, which is not even close to my answer (above). So do Slater's rules not apply to lithium or I have done something wrong?

Wikipedia article: https://en.wikipedia.org/wiki/Effective_nuclear_charge

You need to use the factor of 0.85 for each of the $\ce{1s}$ electrons: $$Z_{eff} = 3 - (2\times 0.85) = 1.3$$ as per rule #2 in the Wikipedia article:

1. If the group is of the [s p] type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

The $\ce{1s}$ electrons both qualify as having "a principal quantum number $n$ one less than that of the group" (here, the principal quantum number of the group of the $\ce{2s}$ electron is 2). Here is a nicely worked example for further edification.

• So the screening of 1s orbital being 0.3 only applies to Helium? – Abhishek Mhatre Aug 3 '15 at 16:22
• The screening of a 1s electron by another 1s electron, as in the case of helium can also apply to other atoms that are missing electrons that result in the $\ce{(1s)^2}$ electronic configuration such as $\ce{Li+}$, where you calculate $Z_{eff} = 3 - (1\times 0.3) = 2.7$ (compare to 2.691 on Wikipedia's page on effective nuclear charge, linked in the main thread here). – Todd Minehardt Aug 3 '15 at 16:35