The element that exhibits the only oxidation state in compounds is: hydrogen oxygen fluorine copper.

This element is fluorine, for it is the most electronegative of all elements, it has a small radius. Because of this, it is unprofitable and difficult for him to give an electron. He takes an electron from other compounds to himself.

Let’s go through all the elements:

1) H2. It can exhibit both +1 and -1 oxidation states. Even in the periodic table, it is in both the first group and the seventh. What is an advantageous energy position? This is when it is easy for an electron to donate electrons from the outer shell (if there are less than half of them) or to get them, if more than half. And hydrogen has only one. It is easy for him to do both. Therefore, it can both accept the oxidation state +1 (give up an electron) and -1 (accept). Examples: H2O => H (+1). CaH2 => H (-1).

2) Oxygen in almost all compounds takes the oxidation state -2. But it can have -1/2, and -1, +1, +2.

In superoxides it has st. approx -1/2: NaO2.

In peroxides -1: Na2O2.

In oxides -2: CaO

If I am not mistaken, in ozonides it can even show st.ok -1/3. And about article ok +1/2 I once read it in the literature, but I don’t remember an example.

And finally, the OF2 compound is a compound where two elements close in electronegativity are connected to each other. Since fluorine is more than EO, then it will pull electrons from oxygen.

3) From the above examples and reasoning, we can conclude that only fluorine (F) has a single oxidation state: F (-1).