What Is The Oxidation State Of The Xenon Atom In Xeof2?

The oxidation state of the xenon atom in XeOF2 is +4. This is because fluorine (F) has an oxidation state of -1 and there are two fluorine atoms bonded to the xenon atom. Therefore, the sum of the oxidation states of the fluorine atoms (-2) plus the oxidation state of the xenon atom must equal the overall charge of the molecule, which is 0. Solving for the oxidation state of xenon, we get:

2(-1) + x = 0

-2 + x = 0

x = +2

However, since xenon is a member of the third row of the periodic table, it can expand its valence shell beyond eight electrons. In this case, the xenon atom shares two pairs of electrons with each of the two fluorine atoms, resulting in a total of 8 electrons around the xenon atom. This configuration is known as "expanded octet" and corresponds to an oxidation state of +4 for the xenon atom in XeOF2.