The non-metal that does not exhibit positive oxidation state is:

We begin by recalling the idea of oxidation state. The oxidation state (or oxidation number) of an element in a compound is a hypothetical charge assigned on the basis of a simple set of electron-book-keeping rules. A “positive oxidation state” means the element is treated as if it has lost one or more electrons.

Whether an element can show a positive oxidation state largely depends on its electronegativity. The general rule is:

$$\text{More electronegative element} \; \Rightarrow \; \text{negative oxidation state}$$

and only if an element is less electronegative than the partner bonded to it can it be forced into a positive oxidation state.

Now let us look at each non-metal in the options one by one.

Chlorine (Cl)
Chlorine has high electronegativity, but oxygen and fluorine are still more electronegative than chlorine. Therefore, in compounds such as $$Cl_2O$$, $$ClO_2$$, $$HClO_4$$, chlorine takes positive oxidation numbers $$+1, +4, +7$$ respectively. So chlorine does exhibit positive oxidation states.

Iodine (I)
Iodine is less electronegative than chlorine and bromine, and far less than oxygen and fluorine. Hence, in compounds such as $$I_2O_5$$ or $$HIO_3$$, iodine shows positive oxidation numbers $$+5$$. Therefore iodine can also possess positive oxidation states.

Oxygen (O)
Oxygen is the second most electronegative element, yet fluorine is still more electronegative than oxygen. Consequently, when oxygen is bonded to fluorine, as in $$OF_2$$, we must assign:

$$\text{Oxidation number of F} = -1$$ (because fluorine is the most electronegative and is always $$-1$$)

Using the algebraic sum rule $$\sum (\text{oxidation numbers}) = 0$$ for a neutral molecule, we get

$$2(-1) + (\text{oxidation number of O}) = 0$$

$$\Rightarrow \text{oxidation number of O} = +2$$

Thus oxygen also exhibits a positive oxidation state in $$OF_2$$.

Fluorine (F)
Fluorine is the most electronegative element of all ($$ \chi = 3.98$$ on the Pauling scale). Because no other element outranks fluorine in electronegativity, there is no partner to which fluorine can lose electron density. Therefore, fluorine is always assigned the oxidation number $$-1$$ in all its compounds; it never shows $$+1, +3, +5,$$ or any other positive value.

So among the given non-metals, fluorine uniquely lacks any positive oxidation state.

Hence, the correct answer is Option C.