Statement I : Oxygen is always present in –2 oxidation state.

Statement II : Stability of oxidation state of group16 for +4 and +6 decreases down the group.

The correct option is: Statement II is correct and Statement I is false.

Analysis of Statement I:
Statement I states that oxygen is always present in the -2 oxidation state. This statement is false. Although the -2 oxidation state is the most common for oxygen due to its high electronegativity, it exhibits several other oxidation states in different chemical environments:
- In peroxides, such as hydrogen peroxide:
$H_2{O}_2$
oxygen exhibits an oxidation state of -1.
- In superoxides, such as potassium superoxide:
$K{O}_2$
oxygen exhibits an oxidation state of -1/2.
- In oxygen difluoride:
$O{F}_2$
oxygen exhibits an oxidation state of +2 (since fluorine is more electronegative than oxygen).
- In dioxygen difluoride:
$O_2{F}_2$
oxygen exhibits an oxidation state of +1.
- In elemental oxygen gas:
$O_2$
the oxidation state of oxygen is 0.

Analysis of Statement II:
Statement II states that the stability of the oxidation states of group 16 for +4 and +6 decreases down the group. This statement is correct. As we move down Group 16 (oxygen, sulfur, selenium, tellurium, and polonium), the stability of the highest +6 oxidation state decreases significantly. This trend is primarily due to the inert pair effect, which is the reluctance of the outermost s-electrons (ns²) to participate in bond formation due to the poor shielding of the nuclear charge by the inner d and f electrons. As a result, the +6 oxidation state becomes less stable down the group, and the overall stability of the higher positive oxidation states (+4 and +6) decreases in comparison to the lower, more stable states.