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Why does Be have a higher ionization energy than B? I know that it has a filled 2s orbital, but what's the more accurate explanation?

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We essentially deal with two different trends here and I think the answer lies within analyzing these trends. First we have that the ionization energy obviously increases with increased nuclear charge/atomic number. Secondly we have that with a higher magnetic quantum number (the s-, p-, d- and f-orbitals) the energy of the orbital increases and therefore that the ionization energy decreases. The decrease in ionization energy that the higher magnetic quantum number introduces is larger than the increase that the single extra positive charge introduces.

We should also take Hund's rule in account, which is favorable for a lower ionization energy of $\ce{Be}$ compared to $\ce{B}$. This factor is however relatively small compared to the above ones.

To verify some of these trends yourself, you can take a look at this table of ionization energies.

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