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According to Slater's rules the effective nuclear charge of Beryllium and Sodium is 1.95 and 2.20 respectively.

That means that the outermost electron of Na feels a stronger attraction from the nucleus than the outermost electron of Be.

But we also know that the atomic radius of Be is smaller than Na and hence Be atom is smaller than Na. That means that the electrons of Be are closer to its nucleus so they must feel a greater attraction than the electrons of Na

The way i think about the above, somewhat contradictory statements, is that the pulling of the outermost electron in Na has to be bigger because there is also a greater shielding effect. Indeed the electron in the 3s orbital of Na feels a greater attraction than the 2s electron of Be but because of the inner electrons pushing it out it can't come closer to the nucleus and so Na ends up being bigger than Be. Is my way of thinking this right?

My question though has to do with the 1st Ionization energy of Be being greater than Na. Why is that? It's the effective nuclear charge that tells us how much attraction the electrons feel from the nucleus.

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    $\begingroup$ $E_{pot} \propto Z_\mathrm{eff}/r$. For Na/ Be is the atomic radius ratio greater than Zeff ratio. $\endgroup$
    – Poutnik
    Jan 15, 2023 at 10:44

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The attraction of alectron to the nucleus also depends on hoq far the electron is from the nucleus. For an outer eletron this is rendered as the atomic radius, which is bigger for sodium at the beginning of the third period versus beryllium in thw second slot of the second period. So sodium has its valence electron fartheraway from the nucleus than beryllium, weakening the nuclear attraction to it.

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