What is the oxidation state of xenon in xenon | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The chemical formula for xenon monooxytetrafluoride is $XeOF_4$. Let the oxidation state of xenon be $x$. The oxidation state of oxygen is $-2$ an

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The chemical formula for xenon monooxytetrafluoride is $XeOF_4$. Let the oxidation state of xenon be $x$. The oxidation state of oxygen is $-2$ and the oxidation state of fluorine is $-1$. Setting the sum of oxidation states equal to the charge of the molecule $(0)$: $x + (-2) + 4(-1) = 0$ $x - 2 - 4 = 0$ $x - 6 = 0$ $x = +6$. Therefore,the oxidation state of xenon in $XeOF_4$ is $+6$.

What is the oxidation state of xenon in xenon monooxytetrafluoride?

Similar Questions

The reason for the lowest boiling point of $He$ is...

$SbF_5$ reacts with $XeF_4$ to form an adduct. The shapes of cation and anion in the adduct are respectively:

Identify the reaction in which oxygen is not one of the products.

The structures of $XeF_2$ and $XeF_4$ respectively are

Noble gases do not react with other elements because

Vedclass Test Series

Exam Paper Generator

Online Exam Module