Slater's rules are a simple way to estimate the effective nuclear charge
experienced by an electron. In this approach, the "shielding constant,"
\(\mathrm{S}\), is calculated. The effective nuclear charge is then the
difference between S and the atomic number, \(Z\). (Note that the results in
Table 7.2 and Figure 7.2 were calculated in a slightly different way.)
$$
Z^{*}=Z-\mathrm{S}
$$
The shielding constant, \(\mathrm{S}\), is calculated using the following rules:
1\. The electrons of an atom are grouped as follows:
(1s) \((2 s, 2 p)(3 s, 3 p)\) (3d) \((4 s, 4 p)\) (4d), and so on.
2\. Electrons in higher groups (to the right) do not shield those in the lower
groups.
3\. For \(n s\) and \(n p\) valence electrons
(a) Electrons in the same \(n s, n p\) group contribute 0.35 (for \(1 s 0.30\)
works better).
(b) Electrons in the \(n-1\) group contribute 0.85
(c) Electrons in the \(n-2\) group (and lower) contribute 1.00
4\. For \(n d\) and \(n f\) electrons, electrons in the same \(n d\) or \(n f\) group
contribute \(0.35,\) and those in groups to the left contribute 1.00
As an example, let us calculate \(Z^{*}\) for the outermost electron of oxygen:
$$
\begin{array}{c}
\mathrm{S}=(2 \times 0.85)+(5 \times 0.35)=3.45 \\
Z^{*}=8-3.45=4.55
\end{array}
$$
Here is a calculation for a \(d\) electron in \(\mathrm{Ni}\) :
$$
Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55
$$
and for an s electron in \(\mathrm{Ni}\) :
$$
\begin{aligned}
Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] & \\
-[1 \times 0.35] &=4.05
\end{aligned}
$$
(Here \(3 s, 3 p,\) and \(3 d\) electrons are in the \((n-1)\) groups.)
(a) Calculate \(Z^{*}\) for \(F\) and \(N e\). Relate the \(Z^{*}\) values for
O, F, and Ne to their relative atomic radii and ionization energies.
(b) Calculate \(Z^{* *}\) for one of the \(3 d\) electrons of \(\mathrm{Mn}\), and
compare this with \(Z^{*}\) for one of the \(4 s\) electrons of the element. Do
the \(Z^{*}\) values give us some insight into the ionization of Mn to give the
cation?
