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 ΔᵢH than B (ii) O has lower ΔᵢH than N and F?
In beryllium, the electron removed during ionization is a -electron, whereas in boron, it is a -electron. The penetration of a -electron to the nucleus is more than that of a -electron. Hence, the electron of boron is more shielded from the nucleus by the inner core of electrons than the electrons of beryllium. Therefore, it is easier to remove the -electron from boron compared to the removal of a -electron from beryllium.
In the nitrogen atom, three -electrons reside in different atomic orbitals (Hund’s rule) whereas in the oxygen atom, two of the four -electrons must occupy the same -orbital resulting in an increased electron-electron repulsion. Consequently, it is easier to remove the fourth -electron from oxygen than it is to remove one of the three -electrons from nitrogen. Further, the first ionization enthalpy generally increases as we go across a period, so F has a higher ionization enthalpy than O.
Explanation
The answer relies on the concepts of penetration effect and shielding for part (i), and Hund's rule along with electron-electron repulsion for part (ii). For Be vs B, the 2s electron is more penetrating and less shielded than the 2p electron, making it harder to remove. For O vs N, pairing of electrons in the 2p orbital of oxygen causes repulsion, making its ionization enthalpy lower than nitrogen. For O vs F, the general trend of increasing ionization enthalpy across a period due to increasing nuclear charge applies.