The oxidation state of O in \[\;{O_2}{F_2}\] is
A) +2
B) +1
C) -2
D) -1

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Hint: Oxidation state represents the degree of oxidation (loss of electrons) of an atom in a chemical compound and Fluorine is the most electronegative element on the periodic table, which means that it is a very strong oxidizing agent and accepts other elements electrons.

Complete answer:
Now we determine the most electronegative atom.
Oxygen is the second most electronegative element. It has an electronegativity of 3.5 on the Pauling scale. That means if oxygen combines with an element which is more electronegative than it will surely possess a positive oxidation state.
Fluorine being the first most electronegative element than oxygen (electronegativity of 4.0 on Pauling scale) will in any case (except in fluorine gas), oxidation state of fluorine is - 1 in all its compounds. So, the oxidation state of oxygen will be positive. Being in the group 17 fluoride ion would gain an electron to a -1 charge, so each has an oxidation number of -1.
Let us calculate the oxidation state of oxygen-
Let be ‘x’ is an oxidation number of oxygen.
Oxidation number of \[{F_2}\] is \[ - 1\].
Net charge is zero because the compound is neutral.
2x - 2 = 0
x = $\dfrac{2}{2}$
x = 1
So the oxidation number of oxygen in \[{O_2}{F_2}\] is + 1.

The oxygen has an oxidation number of + 1 each.

Note: Fluorine in its gaseous form ($F_2$) has zero oxidation state. The oxidation number of other halogens (Cl, Br, I ) is also -1, except when they are combined with oxygen. Oxygen in most of its compounds exist in -2 oxidation state.