# What is the Lewis dot structure of "H"_2"O"_2?

It is simply $H - O - O - H$, a neutral molecule, in which each $O$ bears $2$ lone pairs of electrons.
There are 14 valence electrons to distribute, 4 of these are conceived to constitute the $H - O$ bond, 2 of these the $O - O$ bond, and the 8 remaining are the 4 formal lone pairs on the oxygen atoms.
The oxidation state of oxygen in this molecule is $- I$; in fact this is why it is called a $\text{peroxide}$. Why does $O$ bear this formal oxidation state? For contrast, the oxidation of $O$ in water is certainly $- I I$.