Answer 1

In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three 2p-electrons from nitrogen. Hence, oxygen has lower Δ_iH than nitrogen.

Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from fluorine atom than that required to remove an electron from oxygen atom. Hence, oxygen has lower Δ_iH than fluorine.

Answer 2

The electronic configuration of

N₇ = 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ 

O₈ =1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹

We can see that in case of nitrogen 2p-orbitals are exactly half filled. Therefore, it is difficult to remove an electron from N than from O. As a result  ∆_iH₁ of N is higher than that of O