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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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`ns_(2)` `np_(4)`, where n varies from 2 to 6.
(ii) Oxidation state:
As these elements have six valence electrons `(ns_(2) np_(4))`, 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` `(H_(2)O_(2))`, zero `(O_(2))`, and `+2 (OF_(2))`. 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:
hese elements 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 quite volatile in nature.

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