Does effective nuclear charge decrease down the group?

I've read that the effective nuclear charge increases down the group.


[![enter image description here](https://i.stack.imgur.com/Oa9awm.png)](https://i.stack.imgur.com/Oa9awm.png)


This seems incorrect. As we go down the group the number of protons increases and the shielding constant also increases. We can approximate the shielding constant with the number of electrons between the nucleus and the valence electrons. Hence it seems that the effective nuclear charge is the same within a given group.


The effective nuclear charge doesn't depend on the distance ( Shown through the formula above ). So the valence electrons are further away from the nucleus and experience the same effective nuclear charge down the group. Hence they have less attraction to the nucleus and the atomic radius increases down the group.


Is my intuition correct? Or does the effective nuclear charge increase down the group?




The effective nuclear charge does not decrease along the groups.


It rather increases with the increase converging to zero. What usually decreases is the first ionisation energy, which depends also on the atomic radius:


$$E\_\text{ion} \propto \frac{Z\_\text{eff}}{r\_\text{eff}}$$




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Shielding constants $S$ in


$$Z\_\text{eff}= Z - S$$


cannot be approximated by the number of electrons.


Electrons shield the nucleus just partially and differently, depending on orbital type and relation to the valence orbital.


For electrons much closer to nucleus, the constant for that 1 electron shielding contribution just converges to 1.


For more, see [Slater's rules](https://en.wikipedia.org/wiki/Slater%27s_rules).




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Imagine lithium atom with configuration $\mathrm{1s^2 2s^1}$. If $\mathrm{1s}$ and $\mathrm{2s}$ orbitals had not overlapped, there would have been valid $Z\_\text{eff} =Z-2=1$, due the spherical symmetry of these orbitals and the Gauss law of electrostatics.


But as they do overlap, part of 1s electron shielding is close to zero or even **negative**, being behind the shielded 2s electron, effectively acting as additional nucleus charge.


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The effective nuclear charge **can increase or decrease** down a group. As the nuclear charge is well known, the effective charge depends significantly on the shielding factor S which contains the effects of the **electronic distribution** (and not the *electronic population* : number of electrons).


For most transition metals groups, as the shielding of the d orbital is low (due to their symmetry) the effective nuclear charge tend to increase down the group compared to p or s block elements where the shielding is significant, the effective nuclear charge decreases down the group.


The shielding parameter is not trivial because it contains major electronic effects. Beyond this approximation, the shielding depends on the distance.


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