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.
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 group 16 elements are collectively known as chalcogens.
(i) The elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns2np4, where n can vary from 2 to 6 respectively.

(ii) Oxidation state:
As these elements have six valence electrons, ns2np4, they should display an oxidation state of −2 (in order to complete their octet). However, only oxygen predominantly shows the oxidation state of −2 because of its higher 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 such that there is an expansion electronic configuration.
(iii) Formation of hydrides:
The group-16 elements form hydrides with a formula H2M, where M = O, S, Se, Te, Po. Oxygen and sulphur also form variable hydrides of the form H2M2​. These hydrides are quite volatile in nature and rapidly vaporise.