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?
(i) During the procedure of ionization, the electron to be eliminated from beryllium atom is a 2s-electron, while the electron to be eliminated from boron atom is a 2p-electron. Now, 2s-electrons are greater strongly connected to the nucleus than 2p-electrons. Therefore, extra strength is needed to dispose of a 2s-electron of beryllium than that required to get rid of a 2p-electron of boron. Hence, beryllium has better ΔiH than boron. (ii) In nitrogen, the 3 2p-electrons of nitrogen occupy 3 extraordinary atomic orbitals. However, in oxygen, of the 4 2p-electrons of oxygen occupy the identical 2p-orbital. This outcomes in multiplied electron-electron repulsion in oxygen atom. As a result, the strength required to eliminate the fourth 2p-electron from oxygen is much less in comparison to the strength required to dispose of one of the 3 2p-electrons from nitrogen. Hence, oxygen has decrease ΔiH than nitrogen. Fluorine includes one electron and one proton greater than oxygen. As the electron is being brought to the equal shell, the boom in nuclear attraction (because of the addition of a proton) is extra than the boom in digital repulsion(because of the addition of an electron). Therefore, the valence electrons in fluorine atom enjoy a extra effective nuclear rate than that skilled through the electrons found in oxygen. As a result, extra strength is needed to put off an electron from fluorine atom than that required to get rid of an electron from oxygen atom. Hence, oxygen has decrease ΔiH than fluorine.
