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.

(N/A) The elements of group $16$ are collectively called chalcogens.
$(i)$ Electronic configuration: 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 typically exhibit an oxidation state of $-2$. Oxygen,due to its high electronegativity,shows $-2$ as its most common state,but also exhibits $-1$ (in $H_{2}O_{2}$),$0$ (in $O_{2}$),and $+2$ (in $OF_{2}$). The stability of the $-2$ oxidation state decreases down the group due to decreasing electronegativity. Heavier elements show $+2, +4$,and $+6$ oxidation states due to the availability of $d$-orbitals.
$(iii)$ Formation of hydrides: These elements form hydrides of the general formula $H_{2}E$,where $E = O, S, Se, Te, Po$. Oxygen and sulfur also form hydrides of the type $H_{2}E_{2}$. These hydrides are generally volatile.