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.

Explanation

(i) Electronic configuration. All these elements have the same `ns^(2)np^(4)` (n = 2 to 6) valence shell electronic configuration and hence are justified to be placed in group 16 of the periodic table :
`{:(._(8)O = [He] 2s^(2)2p^(4) " ," ._(16)S = [Ne] 3s^(2)3p^(4)" ,"._(34)Se = [Ar] 3d^(10)4s^(2)4p^(4)),(._(52)Te=[Kr]4d^(10)5s^(2)5p^(4) " and"._(84)Po = [Xe] 4f^(14)5d^(10) 6s^(2)6p^(4)):}`
(ii) Oxidation states. They need two more electrons to form dinegative ions by acquiring the nearest inert gas configuration. Thus, the minimum oxidation state of these elements should be -2. Oxygen predominatly and sulphur to some extent being electronegative show an oxidation state of -2. Other elements of this group, being less electropositive than O and S, do not show negative oxidation states. Since these elements have six electrons in the valence shell, therefore, at the maximum they can show an oxidation state of +6. Other positive oxidation states shown by these elements are +2 and +4. However, oxygen due to the absence of d-orbitals does not show oxidation states of +4 and +6. Thus, on the basis of minimum ans maximum oxidation states, these elements are justified to be placed in the same group, i.e., gorup 16 of the periodic table.
(iii) Formation of hydrides. All the elements complete their respective octets by sharing two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula `EH_(2)`, i.e., `H_(2)O, H_(2)S, H_(2)Se, H_(2)Te and H_(2)PO`. Thus, on the basis of formation of hydrides of the general formula `EH_(2)`, these elements are justified to be placed in the same group, i.e., group 16 of the periodic table.