Why is the oxidation state of noble gas zero?

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Why is the oxidation state of noble gas zero?

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Answer 1 · 2014-11-27T23:45:34

The oxidation state of a noble gas is not always zero.

The high electronegativity values of oxygen and fluorine led to research in the formation of possible compounds involving group 18 elements.

Here are some examples:

For the +2 state: $K r F_{2}$ $KrF_2$ , $X e F_{2}$ $XeF_2$ , $R n F_{2}$ $RnF_2$

For the +4 state: $X e F_{4}$ $XeF_4$ , $X e O F_{2}$ $XeOF_2$

For the +6 state $X e F_{6}$ $XeF_6$ , $X e O_{3}$ $XeO_3$ , $X e O F_{4}$ $XeOF_4$

For the +8 state $X e O_{4}$ $XeO_4$

You might think that these compounds violate the so - called "octet rule" which is true .

A rule is not a "law" in that it is not applicable in all cases. There are many more cases where the octet rule does not apply.

For this reason the name of the group 18 elements was changed from the "Inert Gases" to the "Noble Gases" to reflect the fact that they can exhibit non - zero oxidation states.

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Why is the oxidation state of noble gas zero?

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