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Some textbook write that effective nuclear charge is constant when gping down the group,while some write that the effective nuclear charge decrease down the group

So which statement is true? I try to dtermined it by using slater's rule.However,it came out a trend that $Z_{eff}$ is increasing,which mean effective nuclear charge is increase down the group instead of the 2 other trend i stated above.

I am very confuse about itenter image description here

Pck Tsp
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  • Consider the major feature for the Slater's rules: Weak mutual screening of neighbour electrons within the same ns or np orbital, versus strong screening by fully filled orbitals of the "lower" subshells. Applied to the chart, it should be clear. – Poutnik Jun 30 '22 at 08:15

2 Answers2

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It would probably be much easier to think about this question if you clearly defined what you mean by effective nuclear charge.

According to my understanding, the effective nuclear charge $Z_\mathrm{eff}$ is a fitting parameter appearing when you try to model your valence electrons by a hydrogen orbital.

In this sense, there's two effects in play: $Z_\mathrm{eff}$ will increase as the total charge of the nucleus increases, but $Z_\mathrm{eff}$ will decrease from the screening effects of the non-valence electrons (there's also some weak screening due to other valence electrons but that's a much smaller effect).

If you consider let's say the valence electrons of Na (well, the single one of them), then it will feel the positive charge of $+11$ from the nucleus, shielded by the negative charge $-10$ of the core electrons, leading to an effective nuclear charge somewhat larger than $+1$ because the screening is not perfect (in fact, $Z_\mathrm{eff}$ is about 2 for the valence shell of Na).

Then, consider Cl in the same row. The valence electrons will feel $+17$ charge from the nucleus, shielded by $-10$ valence electrons, meaning $Z_\mathrm{eff}$ will be in the order of magnitude of 7.

So now that we have established how to think about it, we can easily answer your question.

Compare Na to K. Both of these atoms will have one single valence electrons, so the effective nuclear charge is about $+1$, with the differences accounting for the efficiency of screening.

In general, the closer to the core an electron is, the more effectively it screens the nucleus. Meaning that a $2p$ electron will contribute more to screening than a $3p$ electron. This really all comes down to the radius of the orbital.

The core electrons in K are more diffuse than the ones in Na, so they don't screen as efficiently.

Therefore $Z_\mathrm{eff}$ is larger for the valence electron of K than for the valence electron of Na $\rightarrow$ $Z_\mathrm{eff}$ increases as you go down a group.

Szgoger
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  • that means effective nuclear charge increase down the group, right ?If this the fact,then how would u explain the trend of atomic radius increasing going down the group as effective nuclear charge is increasing? – Pck Tsp Jun 30 '22 at 11:57
  • It is more or less the same explanation: the orbitals of the valence electrons are getting larger as you increase the principal quantum number. It is true that the effective nuclear charge is increasing, so the radius is smaller than it would be with a hypothetical constant effective nuclear charge, but it is not enough to offset the fact that the radius of the orbital is so much larger. – Szgoger Jun 30 '22 at 12:06
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I am not sure which texts you saw which said that $Z_{eff}$ (effective nuclear charge) decreases down the group. Many will say that it is constant (an approximation, but it does simplify the explanation of trends) and a more in-depth analysis will show (using Slater's Rules and bearing in mind the fact that more diffuse inner electrons do a poorer job of shielding protons completely) that $Z_{eff}$ increases down the group - which is what your graph shows.

The simple explaination (constant $Z_{eff}$ down the group) assumes that all inner electrons exactly cancel the effect of protons in the nucleus. In a neutral atom, the number of protons not cancelled by inner electrons is equal to the number of outer shell electrons. As this is constant down a group, so is the $Z_{eff}$. Adding Slater's Rules simply shows that the inner electrons are not 100% effective at cancelling the charge of their corresponding protons, and the effectiveness of the inner electrons decreases as the orbitals get more diffuse, so $Z_{eff}$ increases down the group.

Where you may be getting confused is that although the $Z_{eff}$ is increasing down the group (or remaining constant in the simple model), the force of the attraction between the valence electrons and the nucleus is decreasing, as the distance between them is increasing. This is due to Coulomb's Law.

$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$

Many people confuse the meaning of $Z_{eff}$: it does not correspond to the force felt by the outer electrons, it is only the apparent charge of the nucleus, from the outer electrons' point of view. In Coulomb's Law, $Z_{eff}$ corresponds to one of the q charges on the numerator of the quotient.

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