In which compound of xenon does xenon have an oxidation state of +6?

Xenon, a noble gas, can form several compounds despite its inert nature. The oxidation state of xenon in a compound depends on the other atoms it is bonded with. XeOF₄ is a well-known xenon compound in which xenon exhibits an oxidation state of +6.

In XeOF₄, xenon is bonded to one oxygen atom and four fluorine atoms. Fluorine is the most electronegative element and always has an oxidation state of −1, while oxygen in this compound is −2. By applying the rule that the sum of oxidation states in a neutral molecule is zero, the oxidation state of xenon is calculated as:

Xe + (−2 for O) + 4 × (−1 for F) = 0 → Xe = +6

Other xenon compounds have different oxidation states. For example, XeF₄ has xenon in the +4 oxidation state, XeOF₂ has xenon in the +4 oxidation state, and Na₄XeO₆ contains xenon in the +8 oxidation state as part of a xenate ion.

XeOF₄ has a square pyramidal geometry according to the VSEPR theory, with xenon at the center, one oxygen atom forming a double bond, and four fluorine atoms forming the base. This compound is important in studying the chemistry of noble gases and demonstrates that xenon can achieve high oxidation states due to the availability of empty d-orbitals.

Understanding xenon compounds and their oxidation states is crucial in inorganic chemistry, as it challenges the earlier notion that noble gases are completely inert. XeOF₄ is widely studied in laboratory chemistry to explore noble gas reactivity, bonding, and molecular geometry.

In summary, XeOF₄ is the compound where xenon has an oxidation state of +6, showing the ability of xenon to form stable, high oxidation state compounds with oxygen and fluorine