Explain the trends in ionisation energies across the 2nd period of the periodic table?

The general trend observed is that ionisation energies increase from left to right across the periodic table. There are 3 primary factors in determining ionisation energies: 1.) Atomic radius- the smaller the atom, the closer the outermost electrons will be to the positively charged nucleus, so the strength of attraction will be larger. 2.) Nuclear charge- the more protons an atom has, the more positive the nucleus will be, so electrons will experience greater attraction. 3.) Nuclear shielding- the negative charges of other electrons "shield" the positive charge of the nucleus, which decreases the attraction experienced by the outermost electrons. The more electrons, the greater the shielding effect.Across a period, the number of protons in an atom increases, so the nuclear charge increases. The atomic radius also decreases, because of the greater nuclear charge attracting the electrons more strongly. Finally, the amount of electron shielding also increases, but the first two effects dominate and so the ionisation energy increases across the period. There are two exceptions in the second period: 1.) Beryllium to boron- the ionisation energy decreases, because the highest energy electron in boron occupies an s orbital, while in boron, it occupies a p orbital. As the p orbital is higher in energy, it is easier to remove an electron from the latter and so the ionisation energy decreases. 2.) Nitrogen to oxygen- the ionisation energy decreases. Nitrogen has 3 2p electrons, each held in a p_x, p_y and a p_z orbital. Oxygen however has 4 2p electrons, meaning 1 set of electrons must pair up, with opposite spins. The repulsion between the paired electrons means they are higher in energy, so they are easier to remove and so the ionisation energy decreases.