I want to address two exceptions in the trend of ionization energy across the period, that are causing me problems:
Taking the second period as an example :
The two exceptions from the general trend are the ionization energies of B lesser than Be and that of O less than N. My teacher told me the reason to both was that half filled and fully filled orbitals of N and Be are more stable and hence require more energy to pull off an electron. He also said that this affects only the next element in the period for a change in trend( he never said why though). This place and my textbook tells me that the reasons are:
1) Be,B :
The electron is to be removed from the 2p orbital in B and the 2s orbital in Be. The 2s orbital penetrates more into the nucleus and hence it is more difficult to remove an electron form Be than from B.
Question: In that case why doesn't this apply to C? Shouldn't it be easier to remove one of the electrons from C's 2p orbital than Be's 2s orbital just like the B's 2p orbital? And so shouldn't the ionization energy of Be must be more than C too?
N 1s2 2s2 2px1 2py1 2pz1
O 1s2 2s2 2px2 2py1 2pz1
The reason I have understood here is that there are two electrons in the 2px orbital of O and because of electron-electron repulsion, it is easier to remove an electron from here as opposed to the 2px orbital of N which had only one electron.
Question: Why doesn't this apply to F?
F 1s2 2s2 2px2 2py2 2pz1
F has paired electrons too like O in the px and py subshells. Wouldn't there be repulsion here too? And hence why isn't it easier to remove electrons from F than is it to from N which doesn't have any paired electrons at all? Why is the ionization energy of F more than N instead?
The same problems can be seen in the next period.I hope they can be sorted out with the same logic that i am missing for this period.
Finally, was what my teacher said just a simplification of everything I found in the link and the textbook? Was he trying to convey the same thing? Or are they two complete different explanations?