According to most textbook and the Wikipedia page of lanthanide contraction, the phenomenon is due to poor shielding of nuclear charge by 4f electrons. However, the σ value of 4f electrons to 6s is 1.00 in Slater’s rule, meaning that if we go through the lanthanide elements, there should be no net change in the effective nuclear charge of 6s electrons. So are they contradictory? Is it that the σ value of 4f electrons should be smaller than 1.00?

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    $\begingroup$ Slater’s rules are an approximation. You’re reading too much into them. $\endgroup$ Aug 5, 2019 at 11:45

1 Answer 1


Take the example of group-12 chlorides $\ce{ZnCl2}$, $\ce{CdCl2}$ and $\ce{HgCl2}$.

The ionic radii (VI coordination) increases from $\ce{Zn++}(\pu{0.74~\overset \circ A)}$ to $\ce{Cd++}(\pu{0.95~\overset \circ A)}$ to $\ce{Hg++}(\pu{1.02~\overset \circ A)}$. Due to lanthanide contraction or poor shielding of $\ce{4f^14 e-}$ in $\ce{Hg}$, the difference of atomic and ionic radii between $\ce{Cd}$ and $\ce{Hg}$ is very less. $$\ce{Zn: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2}$$ $$\ce{Cd: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2 4p^6 4d^10, 5s^2}$$ $$\ce{Hg: 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^10, 4s^2 4p^6 4d^10 4f^14, 5s^2 5p^6 5d^10, 6s^2}$$ Applying Slater's rule to the valence shell $n\mathrm{s^2}$ $\ce{e-}$, the calculation of screening constant $s$:

$s[\ce{Zn(4s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times10=25.65}$

$s[\ce{Cd(5s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times28=43.65}$

$s[\ce{Hg(6s)]}$ $\mathrm{=0.35\times1 + 0.85\times18 + 1\times60=75.65}$

Calculation of effective nuclear charge $Z_{eff}=Z-s$




We can see that effective nuclear charge is same in all three cases. But the covalent character in these compounds increases gradually as:

$\ce{ZnCl2 < CdCl2 < HgCl2}$

This might be due to gradual increase in effective nuclear charge in cation.

Therefore, Slater's rules are an approximation to give us the idea about effective nuclear charge and it might be contradictory to lanthanide contraction in some cases.


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