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Question
Among the second period elements the actual ionization enthalpies are in the
order Li < B < Be < C < O < N < F < Ne.
Explain why O has lower ΔiH than N and F?
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Solution 1
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 ΔiH 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 ΔiH than fluorine.
Solution 2
The electronic configuration of
N7 = 1s2 2s2 2px1 2py1 2pz1
O8 = 1s2 2s2 2px2 2py1 2pz1
We can see that in case of nitrogen 2p-orbitals are exactly half filled. Therefore, it is difficult to remove an electron from N than from O. As a result ∆iH1 of N is higher than that of O.
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