Question

The oxidation number of Xe in \[\text{XeO}{{\text{F}}_{2}}\] is:
(A) 0
(B) 2
(C) 4
(D) 3

Answer 1

Hint: To calculate the oxidation number of any element we have to assume it as x and then by adding the oxidation number of other bonded elements multiplied by their respective number of atoms, we will get the oxidation number.

Complete step by step solution:
-In the given question, we have to find the oxidation number of xenon in xenon oxytetrafluoride.
-So, firstly we should know about the oxidation number which is also known as oxidation state.
-Oxidation number is defined as the net charge that is present on the atom after forming the ionic bond with the heteroatom.
-So, according to the rules of the oxidation state, the oxidation state of a charged ion is equal to the charge carried by it.
-For, example oxygen has the oxidation of -2 because it tends to accept the two electrons from the bonding atom.
-Similarly, the oxidation state of the halogen family is -1.
-Also, if there is any single element which is not bonded with the other atom, the oxidation state of the element will be 0.
-Now, let's assume that the oxidation state of xenon is x, then
$\text{XeO}{{\text{F}}_{2}}\text{ = x + (-2) + 2 }\cdot \text{ (-1) = 0}$
$\text{x = +4}$
-So, the oxidation number of the xenon is +4 in xenon oxytetrafluoride

Therefore, option (C) is the correct answer.

Note: Xenon oxytetrafluoride is considered as the reactive oxide of xenon, due to which it is unstable. 54 is the atomic number of xenon and also it is considered as the noble gas due to its stable, full-filled outermost shell.