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Why is the second ionization energy always greater than the first? By shielding effect, it shouldn't have been true. Let's take $\ce{Mg}$ and $\ce{Mg^+}$ for example. Effective nuclear charge(ENC) for $\ce{Mg},Z_{\mathrm{eff}}=Z-$ no. of core electrons $=12-10=2$. For $\ce{Mg^+}$, it's also $12-10=2$, since there are still $2$ shells before valence shell like in $\ce{Mg^+}$. The result is giving equal number of attraction in this case. So energy needed to extrapolate one electron should be the same in both cases. Moreover, according to ENC, the size of the cation and anion are equal here, but we know cation<atom. I will be very grateful if someone sheds light on this atomic radius and attraction property in light of shielding and screening effect since that's how atomic and ionic radii are compared.

Pallas
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madness
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    Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions /
    upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.     – Poutnik Jun 08 '22 at 07:05
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    The electron energy is not determined by the effective nuclear charge. which is just a contributing factor. BTW, the ENC is better computed by Slater's rules. The integer counting you have used would have been applicable if orbitals had been non-overlaping regions with spherical symmetry of probability density. – Poutnik Jun 08 '22 at 07:07
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    Stealing one electron from a magnesium atom is hard and results in a positively charged magnesium ion. Then stealing another electron, which is negatively charged, from the positively charged magnesium ion, is harder still. Each subsequent theft gets harder. – Ed V Jun 08 '22 at 13:28
  • @ed v you may be aware that magnesium can actually be stopped at one ionization, though it usually must dimerize to do so. See https://en.wikipedia.org/wiki/Low_valent_magnesium_compounds. – Oscar Lanzi Jun 08 '22 at 20:50
  • @OscarLanzi I have only heard very little about low valent magnesium species, so thank you for the link! I was just trying a simple rationalization of the typical trend. I used to tell my students in first year chemistry that if a rich guy had multiple wallets and a mugger stole one, it would be harder to mug the guy for another wallet. So, just the usual trend, not pesky details about energetics, etc.! – Ed V Jun 08 '22 at 21:03
  • A good mugger renders the victim at least senseless and takes both wallets. It is basic high school physics the first electron is removed from an effective nuclear charge of +2 each successive electron Has one more charge hence needs more energy. Compound formation is the effective mugger It returns the energy in much higher lattice and or solvation energies – jimchmst Jun 27 '22 at 21:34

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One way to think about this is to consider the integral of the force required to bring the electron from its radius to infinity.

When you remove the first electron from a neutral species like $\ce{Mg}$, the electron leaves behind a positive ion of charge $1$. When you remove the second electron, it leaves behind a positive ion of charge $2$. Therefore, the second ionization energy will generally be about twice as large as the first ionization energy.

In addition, there can be a significant radius effect if the electrons belong to different subshells, but that is not the case for $\ce{Mg}$, as you are removing two electrons from the $\ce{3s}$ shell. As a matter of fact, for $\ce{Mg}$ the ratio of 1st and 2nd ionization energies is very close to 1:2.

Pallas
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    Exchange energy effects, which favor half-filled subshell configurations, can also enter. They cause phosphorus and chlorine to each have IE2 < 2 × IE1. – Oscar Lanzi Jun 25 '22 at 11:05