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Explain why (i) Be has higher i H than B? (ii) O has lower i H than N and F?

Question

Explain why (i) Be has higher i H than B? (ii) O has lower i H than N and F?

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Solution

(i) Be (Beryllium) has a higher ionization enthalpy (Δi H) than B (Boron) because of its electronic configuration. Beryllium has a completely filled s-orbital (1s2 2s2) which gives it extra stability. Therefore, it requires more energy to remove an electron from the s-orbital of Beryllium as compared to Boron (whose outermost electron is in a p-orbital).

(ii) O (Oxygen) has lower ionization enthalpy (Δi H) than N (Nitrogen) and F (Fluorine) due to its electronic configuration. Nitrogen has a half-filled p-orbital (2p3) and Fluorine has a completely filled p-orbital (2p5), both of which provide extra stability due to symmetry. Oxygen, on the other hand, has an electronic configuration of 2p4, which is neither half-filled nor completely filled, making it less stable. Therefore, it requires less energy to remove an electron from Oxygen as compared to Nitrogen and Fluorine.

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