(i) During the process of ionization, the electron to be removed from beryllium atom is a

2s-electron, whereas the electron to be removed from boron atom is a 2p-electron.

Now, 2s-electrons are more strongly attached to the nucleus than 2p-electrons. Therefore,

more energy is required to remove a 2s-electron of beryllium than that required to remove

a 2p-electron of boron. Hence, beryllium has higher ∆iH than boron.

(ii)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 ∆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 ∆iH than fluorine.