Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne.

Explain why

(i) Be has higher than B

(ii) O has lower than N and F?

(i) Beryllium has higher ionization enthalpy as compared to boron which can easily explained using the concept of symmetry factor. Beryllium has both the orbitals filled fully and boron has one half filled orbital in 2p subshell because of which beryllium has more symmetric nature as compared to boron. The electron of beryllium is places in 2s subshell and the 2s subshell is more tightly held to the nucleus as compared to 2p subshell of boron. Thus, during ionization more energy is required to remove an electron from beryllium atom as compared to boron atom. Hence, beryllium has higher ∆_i H than boron.

(ii) In nitrogen atom all three orbitals of the 2p subshell are fully filled and in case of oxygen one orbital is fully filled and the two orbitals are partially filled of 2p subshell. So, nitrogen has greater symmetricity as compared to oxygen and hence it is difficult to remove an electron from nitrogen atom as compared to oxygen atom. Thus, nitrogen has higher value of ionization enthalpy as compared to oxygen.

Fluorine has one electron and proton more as compared to oxygen atom and hence in case of fluorine atom the electron is more tightly held as compared to oxygen atom. Thus, it is difficult to remove an electron from fluorine atom as compared to oxygen atom. Thus fluorine has higher value of ionization enthalpy as compared to oxygen.