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Question: A compound of \[Xe\] and \[F\] is found to have \(53.5%\) \[Xe\]. What is the oxidation number of \[...

A compound of XeXe and FF is found to have 53.553.5% XeXe. What is the oxidation number of XeXe in this compound?
(A)4\left( A \right)\,\,-4
(B)0\left( B \right)\,\,0
(C)+4\left( C \right)\,\,+4
(D)+6\left( D \right)\,\,+6

Explanation

Solution

To find out the oxidation number of any element say xenon here, from the given compound, first you need to find out the mass percentage of other elements in the given compound say fluorine here and then find number of moles of xenon, then form the molecular formula.

Complete step by step answer:
Let us assume that the given mass of the compounds is 100g100g
We are given with mass of xenon as 53.353.3%
So, from this we can find out the mass of fluorine:
Therefore, mass of fluorine=Total massmass of xenon\text{mass of fluorine}=\text{Total mass}-\text{mass of xenon}
=10053.5=46.5g=100-53.5=46.5g
Now, we know that the atomic mass of fluorine is 1919 and xenon is 131131
So, with the help of atomic mass and given mass we can calculate the number of moles of the given element as,
We know,
No. of moles=Given massMolar mass\text{No}\text{. of moles=}\dfrac{\text{Given mass}}{\text{Molar mass}}
Substituting the values in above equation, for calculating the moles of xenon we get:
No. of moles of Xe =53.5131=0.41 moles\text{No}\text{. of moles of Xe =}\dfrac{\text{53}\text{.5}}{131}=0.41\text{ moles}
Similarly, we find moles of fluorine as
No. of moles of F = 46.519=2.45 moles\text{No}\text{. of moles of F = }\dfrac{46.5}{19}=2.45\text{ moles}
Now, we need to convert moles in simple ratio so we will divide it by 0.410.41 which is the number of moles of xenon.
 Moles of Xe = 0.410.41=1\therefore \text{ Moles of Xe = }\dfrac{0.41}{0.41}=1
And, Moles of F =2.450.41=5.976\text{Moles of F =}\dfrac{2.45}{0.41}=5.97\approx 6
From here, we can say that the molecular formula for the given compound isXeF6Xe{{F}_{6}}.
Now we have to find the oxidation number of XeXe in this compound
So, let xx be the oxidation number of Xenon in XeF6Xe{{F}_{6}}
Fluorine has (1)\left( -1 \right) oxidation number and we know, the overall sum in any neutral element will be zero.
XeF6\Rightarrow Xe{{F}_{6}}
Substituting the values, we get:
x+(1)×6=0x+(-1)\times 6=0
x6=0\Rightarrow x-6=0
Calculating value of x as
x=+6\Rightarrow x=+6
Therefore, the oxidation state of XeXe is +6.

So, the correct answer is Option D.

Note: Oxidation number is defined as the electrons that atoms can share, lose or gain while forming chemical bonds with other atoms of a different element.
You can assume the mass of compounds as 100100 that is the total mass of the whole compound to find the mass of another element if the mass of one of the elements is given. And remember the atomic number or atomic mass of xenon and fluorine are 131131 and 1919 respectively.