Which one of the following is the oxidation state of oxygen in $$OF_2$$​​

Correct option is A

The correct answer is (A) +2

Explanation:
In the compound OF₂ (oxygen difluoride), the oxidation state of oxygen can be determined by considering the oxidation state of fluorine, which is typically -1.

Let the oxidation state of oxygen be x. The total oxidation state of the compound must be zero, as OF₂ is neutral.

The sum of oxidation numbers is:

$OxidationnumberofO+2\times (OxidationnumberofF)=0$

Substituting the known values:

​$$x+2(−1)=0$$

$$x−2=0$$

$$x=+2$$​​​

Thus, the oxidation state of oxygen in OF₂ is +2.

Information Booster:

• In OF₂, fluorine is more electronegative than oxygen, so it takes an oxidation state of -1.

• The oxidation state of oxygen in OF₂ is +2, which is unusual, as oxygen typically has an oxidation state of -2 in most compounds.

• OF₂ is a highly reactive compound, and oxygen in it exhibits an oxidation state that is higher than usual because fluorine is the most electronegative element.

• This high oxidation state of oxygen is one of the unique characteristics of oxygen difluoride.

• OF₂ is used in certain specialized applications, including as a fluorinating agent in chemical reactions.