Shouldn't the first ionization energy of fluorine be lower than oxygen?

Firstly, ionisation energy is the amount of energy required to remove an electron from the valence shell in the gaseous state of an atom. As we move across the period, electrons are added to the same shell. Hence, effective nuclear charge increases. This translates to the fact that the electrons are held more tightly on moving across the period. Hence, fluorine has more I.E compared to oxygen. The fact that on losing an electron oxygen will attain the half filled configuration is more than compensated by the high electronegativity of fluorine. Hence, the following factors may be considered: 1) The increase in ENC across the period. 2) The increase in electronegativity across the period. 3) Half filled configuration attained by oxygen. 1) and 2) overpower 3) and hence fluorine has higher I.E than oxygen.

The atomic number of O is 8. Atomic number of F is 9. As you yourself correctly show the pic of electronic configuration of F, you will observe that if F gets 1 electron, i.e, F-, then it has a completely filled orbit. Hence, it is highly stable. Whereas, O- will now have 9 electrons and still O needs 1 more electron to be completely filled. Hence, for first ionization energy of Flourine is >> as compared to Oxygen . Completely filled orbits are highly stable - this is your answer Edit :  After Michael Flynn correctly pointed out my mistake, god bless him, i wanna correct my answer. The above answer applies to electronegativity. As for ionization energy, it is because of the increased nucleus attraction towards the same orbital, hence, more difficult to remove that electron...