Among the second period elements the actual ionization enthalpies are in the
Question:

Among the second period elements the actual ionization enthalpies are in the

order $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne} .$

Explain why

(i) Be has higher $\Delta_{i} H$ than B

(ii) $\mathrm{O}$ has lower $\Delta_{i} H$ than $\mathrm{N}$ and $\mathrm{F}$ ?

Solution:

(i) During the process of ionization, the electron to be removed from beryllium atom is a $2 s$-electron, whereas the electron to be removed from boron atom is a $2 p$-electron. Now, $2 s$-electrons are more strongly attached to the nucleus than $2 p$-electrons. Therefore, more energy is required to remove a $2 s$-electron of beryllium than that required to remove a $2 p$-electron of boron. Hence, beryllium has higher $\Delta_{i} H$ than boron.

(ii) In nitrogen, the three $2 p$-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four $2 p$-electrons of oxygen occupy the same $2 p$-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth $2 p$-electron from oxygen is less as compared to the energy required to remove one of the three $2 p$-electrons from nitrogen. Hence, oxygen has lower $\Delta_{j} H$ 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 $\Delta_{i} H$ than fluorine.