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Question
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
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Solution
- Electronic Configuration: All the elements, oxygen (O), sulphur (S), selenium (Se), tellurium (Te), and polonium (Po), belong to Group 16, known as chalcogens. They have six valence electrons, with the general outer electronic configuration ns2np4. where n varies from 2 to 6. This similarity in valence shell configuration justifies their placement in the same group.
- Oxidation state: Due to the presence of six valence electrons, these elements commonly exhibit an oxidation state of −2.
- Oxygen shows −2 oxidation state predominantly, due to its small size and high electronegativity. It also shows −1 (as in H2O2), 0 (as in O2), and +2 (in OF2) states.
- The −2 oxidation state becomes less stable down the group due to decreasing electronegativity.
- Heavier elements like S, Se, Te, and Po also exhibit +2, +4, and +6 oxidation states due to the availability of vacant d-orbitals for bonding.
- Formation of hydrides: All Group 16 elements form binary hydrides of the general formula H2E, where E = O, S, Se, Te, or Po, These hydrides are volatile and covalent in nature.
- Oxygen and sulphur also form peroxides (e.g. H2O2, H2S2).
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Volatility decreases down the group, while thermal stability and boiling point increase due to increasing molecular weight and decreasing hydrogen bonding.
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