# 7.17  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.

• Electronic Configuration-
$O$, $S$, $Se$, $Te$ and $Po$ , all have six valance electron each. The general electronic configuration of these elements is $ns^2,\ np^4$, where  $n$ varies from 2 to 6.
• Oxidation state-
As all of these elements have six valence electrons, they should display an oxidation state of -2. 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 show +2, +4 and +6 oxidation state due to availability of $d$-orbitals. It also exhibits the oxidation state of -1 ($H_{2}O_{2}$), zero ($O_{2}$), and +2 ($OF_{2}$)
•  Hydrides-
They all 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 volatile in nature.

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