2HgO → 2Hg + O₂, in the reactant Hg act as

$2\stackrel{+2}{\mathrm{Hg}}\stackrel{-2}{\mathrm{O}} \to 2\stackrel{0}{\mathrm{Hg}} + {\stackrel{0}{\mathrm{O}}}_2$

In the given reaction, Hg reduced itself and oxidizes O from -2 to zero. Hence, option-A is correct.