Solution:

CaO2

Let expect oxidation number of O is x.

\[Oxidation\text{ }number\text{ }of\text{ }Ca\text{ }=\text{ }+2\]

Then, at that point, we have:

\[\begin{array}{*{35}{l}}

1\left( +2 \right)\text{ }+\text{ }2\left( x \right)\text{ }=\text{ }0  \\

~  \\

=\text{ }2\text{ }+\text{ }2x\text{ }=\text{ }0  \\

~  \\

=\text{ }2x\text{ }=\text{ }-\text{ }2  \\

~  \\

=\text{ }x\text{ }=\text{ }-\text{ }1  \\

\end{array}\]

Henceforth, Oxidation number of O is – 1