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The effective nuclear charge is defined as the net positive charge experienced by an electron in a polyelectronic atom. It can be calculated using the well-known Stars Rule.

Once I have calculated it for different chemical elements, I would like to know if there is (if there is any justification) any dependence between its value and the periodic property.

I remember hearing, for example, that the greater the value of the effective nuclear charge, the smaller the atomic radius (half the internuclear distance). And that the greater the effective nuclear charge, the greater the ionization potential.

However, I cannot justify these assumptions. Nor do I know whether this relationship exists in electronic affinity or electronegativity.

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You can think of it in terms of electrostatic attraction.

If you model an atom as a positive point charge surrounded by orbiting electrons, the radius of each orbit is defined by the equiilibrium between the attraction from the nucleus and the repulsion from the other electrons. The effective nuclear charge would be the net force resulting from this competition.

As nuclear charge increases, the attractive force increases and the radius of the orbit decreases; hence the atomic radius decreases.

Because the attraction is higher, more energy is required to remove an electron from that low-energy state, which means the ionization potential is higher. In the same fashion, a new electron added to a neutral atom will have lower energy with increased nuclear charge, which makes it easier to overcome the repulsive forces that the other valence-shell electrons exert on it. Thus the electron affinity is also higher.

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  • $\begingroup$ In summary, one could say that: the higher the effective nuclear charge, the greater the ionization energy, the greater the electronic affinity and the smaller the radius, right? @F Bat $\endgroup$ – user218559 Jan 28 at 16:58
  • $\begingroup$ @user218559 Exactly. All those can be justified by this simple principle. Note though that I'm talking exclusively about the first electron affinity/ionization energy. Adding or removing many electrons is often more difficult and there are other considerations. $\endgroup$ – chemicalromance Jan 28 at 17:27

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