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.

The elements of group 16 are collectively called chalcogens.

[i]. Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns² np⁴, where n varies from 2 to 6.

[ii]. Oxidation state: As these elements have six valence electrons [ns²np⁴], they should display an oxidation state of -2. However, only oxygen predominantly shows the oxidation state of -2 owing to its high Electronegativity. It also exhibits the oxidation state of -1 [H₂O₂], zero [O₂], and +2 [OF₂]. However, 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 of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.

[iii]. Formation of hydrides: These elements form hydrides of formula H₂E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H₂E_2. These hydrides are quite volatile in nature.