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

(i) Electronic configuration: They all have six electrons in the outermost shell and have ns² np⁴ general electronic configuration.
(ii) Oxidation state: The outer configuration of all these elements is ns² np⁴. Therefore, they complete their octet either by gaining two electrons or by sharing two electrons. Two types of oxidation states are shown by these elements.
(a) Negative oxidation state: Except the compound OF₂ oxygen shows-2 oxidation state in all its compounds. Due to hgh electronegativity, it forms O²' ion in most of the metal oxides.
The electronegativities of S, Se, Te are low hence their compounds even with most electropositive elements are not more than 50% ionic. Hence S^2–', Se^2–' and Te^2–' are less probable. Being a metal Po does not form Po²⁺ ion at all.
(b) Positive oxidation state: Oxygen does not show positive oxidation state except OF₂(O = + 2). With the increase in atomic number of electro negativity is decreasd in this group, hence the tendency to show the positive oxidation states will increase. S, Se, Te, Po show + 4, +6 oxidation state in addition to + 2.
(iii) Hydride formation: All the elements O, S, Se, Te and Po form M₂M type hydrides (where M = O, S, Se, Te and Po)