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3.16 Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (ii) O has lower than N and F?
Answers (1)
In oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital and in nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals.
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 sp-electrons from nitrogen.
Hence, oxygen has lower than nitrogen.
In case of fluorine the electron is added to the same shell, the increase in nuclear attraction dominates over the increase in electronic repulsion.
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 a fluorine atom than that required to remove an electron from oxygen atom.
Hence, oxygen has lower than fluorine.
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