Which one of the following is an oxide?

First, we recall the basic definition: an oxide is a compound in which the oxidation state of oxygen is $$-2$$. If the oxidation state of oxygen differs (for example $$-1$$ in peroxides or $$-\dfrac12$$ in superoxides), the substance is not classified as a simple oxide.

Now we test each option by calculating the oxidation state of oxygen in that compound, using the well-known rule that the sum of all oxidation states in a neutral molecule is zero.

Option A : $$SiO_2$$

Let the oxidation state of silicon be $$x$$. Each oxygen atom is assumed to have oxidation state $$-2$$ for the moment, so:

$$x + 2(-2) = 0$$

$$x - 4 = 0$$

$$x = +4$$

The oxidation state of oxygen remains $$-2$$, exactly matching the definition of an oxide. Hence $$SiO_2$$ is indeed an oxide.

Option B : $$KO_2$$

Potassium almost always has oxidation state $$+1$$. Let the oxidation state of each oxygen atom be $$y$$. We write:

$$+1 + 2y = 0$$

$$2y = -1$$

$$y = -\dfrac12$$

The oxidation state $$-\dfrac12$$ identifies this compound as a superoxide, not a normal oxide.

Option C : $$BaO_2$$

Barium belongs to group 2, so its usual oxidation state is $$+2$$. Setting the oxidation state of each oxygen atom as $$y$$:

$$+2 + 2y = 0$$

$$2y = -2$$

$$y = -1$$

Because the oxidation state of oxygen is $$-1$$, this is a peroxide, not an oxide.

Option D : $$CsO_2$$

Caesium, like all group 1 metals, has oxidation state $$+1$$. Taking again the oxidation state of each oxygen atom as $$y$$, we get:

$$+1 + 2y = 0$$

$$2y = -1$$

$$y = -\dfrac12$$

Here too, the value of $$-\dfrac12$$ for oxygen tells us that $$CsO_2$$ is a superoxide.

Comparing all four cases, only $$SiO_2$$ contains oxygen with oxidation state $$-2$$, so it alone fulfills the criterion for being an oxide.

Hence, the correct answer is Option A.