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The correct arrange of them is like this:

Be < B < Li ... But why?

Li will have the greatest IE2 (second ionization energy) because that will involve removing a core electron, but what I am confused about is about Be and B.

I think Be will have greater IE2 than B, because B has 2 electrons in the same orbital, so the repulsive force between them will make it easier to remove one of them, but in Be there is only one electron in the orbital and there is no repulsive force in the same orbital, therefore
$$\ce{ B < Be}$$

What do you think?

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  • $\begingroup$ The key factors are the effective nuclear charge and the principal quantum number n Read my answer below $\endgroup$
    – jimchmst
    Commented Aug 28, 2023 at 22:18

2 Answers 2

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To reach the second ionization potential we remove a second electron from an already ionized element $$\ce{M^{+1} -> M^{+2}}$$ Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol) \begin{aligned} \ce{& Li^{+1} & 1s^2~~ &~~ 7297}\\ \ce{& Be^{+1} & 1s^2~ 2s^1~~&~~ 1757}\\ \ce{& B^{+1} & 1s^2~ 2s^2~~&~~ 2426}\\ \end{aligned} We see that both $\ce{Li^{+1}}$ and $\ce{B^{+1}}$ have completely filled shells, the stable "inert gas" configuration. We would expect it to be difficult to remove one electron from these two ions and destroy this stable configuration. Further, we would expect removal of one electron from $\ce{Li^{+1}}$ to be more difficult than from $\ce{B^{+1}}$ because in the former case we are removing a 1s electron which is closer to the nucleus than the 2s electron in the case of $\ce{B^{+1}}$. On the other hand, if $\ce{Be^{+1}}$ loses one electron it will have a $\ce{1s^2}$ electron configuration and achieve a stable "inert gas" configuration. This should make it much easier to remove one electron from $\ce{Be^{+1}}$ than from the other two ions.

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Li+ is fairly obvious, the 1s electrons do not shield each other well so each is attracted to an effective nuclear charge approaching 3 resulting in a very high IP. Be+ is different, the 1s2 2s1 configuration, isoelectronic with Li atom, has a well shielded 2s1 electron. The effective nuclear charge approaches 2 and the distance is greater. The result is a large second IP but less that of Li+ [and much greater than Li atom]. B+ is similar to Li+ in that the 2s2 electrons do not effectively shield each other, but they are shielded strongly by the inner 1s2 electrons resulting in an effective nuclear charge approaching 3 at a greater distance than Li+. This results in an IP less than Li+ but greater than Be+.

The reason that inert gas configurations or filled orbitals are of lower energy is not magic. Electrons added to the same orbital or to orbitals of the same [or very close] energy do not effectively shield each other from the nucleus. The increased nuclear charge means each electron is more strongly attracted to the nucleus, is of lower energy. These effects are very evident in the first and second periods and become more subtle at higher atomic numbers.

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