7.17  Justify the placement of O, S, Se, Teand Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answers (1)

 

  • Electronic Configuration-
    O, S, Se, Te and Po , all have six valance electron each. The general electronic configuration of these elements is ns^2,\ np^4, where  n varies from 2 to 6.
  • Oxidation state-
    As all of these elements have six valence electrons, they should display an oxidation state of -2. The stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements show +2, +4 and +6 oxidation state due to availability of d-orbitals. It also exhibits the oxidation state of -1 (H_{2}O_{2}), zero (O_{2}), and +2 (OF_{2})
  •  Hydrides-
    They all form hydrides of formula H_{2}E, where E = O, S, Se, Te, Po. Oxygen and sulphur also form hydrides of type H_{2}E_{2} . These hydrides are volatile in nature.

 

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