why is the ionization energy of Sb greater than that of Bi although that of In is less than Tl and S

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Discussion Response Post to: why is the ionization energy of Sb greater than that of Bi although that of In is less than Tl and S

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rtrikha (2958)

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This is due to Inert pair effect

· The inert-pair effect is the tendency to form ions two units lower in charge than expected from the group number. It is most pronounced for heavy elements in the p-block in groups 13-15 (In, Tl, Sn, Pb, Sb, and Bi). These elements have the tendency to form compounds in which the oxidation numbers differ by 2. Example: In⁺, and In³⁺, Tl⁺ and Tl³⁺, Sn²⁺ and Sn^4+, Pb²⁺ and Pb⁴⁺, Sb³⁺ and Sb⁵⁺, and Bi³⁺ and Bi⁵⁺.

· The inert pair effect is due to the different energies of the valence p- and s-electrons. In the later periods of the periodic table, valence s-electrons are very low in energy compared to the valence p-electrons. They may therefore remain attached to the atom, which makes them hard to remove. These s-electrons are called the inert pair or lazy pair.

· Indium is in the group 3A, which means that it should form In³⁺ cation. Instead indium forms In⁺. This can be explained by looking at the electron configuration of the parent element and its ions. In is: [Kr] 4d¹⁰5s²5p¹, In⁺ is: [Kr] 4d¹⁰5s²5p⁰, and In³⁺ is: [Kr] 4d¹⁰5s⁰5p⁰. It is easier for Indium to loose the 5p electron than the 5s pair of electrons (lazy pair).

Rakesh Trikha

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