# Order of second iozination energies of lithium,berillium,neon,carbon and boron?

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?

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.