Reason of being 2nd ionization energy

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.




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.


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