Giving the electronic configurations for each element, predict the trend in 1st ionisation energies going across period 2 from Lithium to Neon.

First start by giving the definition of first ionisation energy - the energy required to remove one mole of electrons from one mole of gaseous atoms to produce one mole of gaseous ions each with a +1 charge, so we're removing the outermost electron from each of our atoms.

Then have a go at writing down the electronic configurations of each element, it will help to have your periodic table in front of you for this, it is also useful to split the 2p sub-shell into its individual orbitals i.e p_x, p_y and p_z as well.
Li - 1s²2s¹
Be - 1s²2s²
B - 1s²2s² 2p_x¹
C - 1s²2s² 2p_x¹2p_y¹
N - 1s²2s² 2p_x¹2p_y¹2p_z¹
O - 1s²2s² 2p_x²2p_y¹2p_z¹
F - 1s²2s² 2p_x²2p_y²2p_z¹
Ne - 1s²2s² 2p_x²2p_y²2p_z²

We expect to see that there is a general increase in 1^st ionisation energy as we go from Li to Ne and this is because the nuclear charge increases as you go across a period (more protons in the nucleus). However there will be a dip at Boron and Oxygen, they will have a lower actual 1st ionisation energy then we predict from the general increase. The dip at Boron is because the outermost electron is now in a p orbital instead of an s orbital as in beryllium, this p orbital is further from the nucleus than the s orbital so feels a lower nuclear charge meaning the electron is easier to remove. The second dip at oxygen is because the electrons in the px orbital are now paired, when two electrons share an orbital they repel each other so it is slightly easier to remove this electron.

Why do you think Be doesn’t show the same dip as oxygen even though it has two paired electrons in the s orbital?

You might find it helpful also to write out the electron configurations of the +1 ions produced from the ionisation.