The elements in Group 16 of the periodic table, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). Generally, these elements are known to commonly exhibit a -2 oxidation state because they have six electrons in their outermost shell and can gain two electrons to complete their octet, thereby achieving a stable electronic configuration similar to the nearest noble gas.

Oxygen, being the most electronegative element in this group, predominantly shows a -2 oxidation state in most of its compounds, such as water ($\mathrm{H_2O}$) and oxides like $\mathrm{SO_2}$.

Selenium and Tellurium also typically exhibit the -2 oxidation state. For instance, in compounds like $\mathrm{H_2Se}$ (hydrogen selenide) and $\mathrm{H_2Te}$ (hydrogen telluride), these elements are in the -2 state.

Polonium (Po), on the other hand, is different from the lighter chalcogens. Being a metalloid with more metallic character, it does not favor the -2 oxidation state nearly as strongly as the other elements in its group. Polonium most commonly exhibits +2 and +4 oxidation states in its compounds, like $\mathrm{PoO_2}$ and $\mathrm{PoCl_2}$. Due to the relativistic effects and its position in the periodic table, the -2 oxidation state is unstable and rare in polonium compounds.

Given this, the correct answer is Option D: Po. Polonium does not commonly exhibit the -2 oxidation state like the other Group 16 elements.