Q114 E - Chemistry (OpenStax)

Question

What  is the oxidation state of the nobel gas in each of the following ? 

    (a)   \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_2}\)           (b)  \({\rm{Kr}}{{\rm{F}}_{\rm{2}}}\)        (c) \({\rm{XeF}}_{\rm{3}}^{\rm{ + }}\)               (d) \(\)\({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\)

Step-by-Step Solution

Answer

The oxidation state  \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{_{\rm{2}}}}\),\({\rm{Kr}}{{\rm{F}}_{\rm{2}}}\) , \({\rm{XeF}}_{\rm{3}}^{\rm{ + }}\) and \({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\)  are +6 , +2 ,+4 , and +8    respectively.

Step 1: Oxidation state:

The oxidation state or oxidation number is a hypothetical charge of an atom when all of its bonds to different atoms are fully ionic. 

It describes degree of oxidation of an atom in a chemical compound. Conceptually, the oxidation no. may be positive, negative or zero.   

Step 2: Finding the oxidation state of given noble gases:

 Let oxidation state of Xe be ‘x’

 0 has (-2) and F has (-1) oxidation state.

We know that the algebraic sum of oxidation no. of all the atoms in a molecule is charge on the molecule. 

(a) \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{_{\rm{2}}}}\)                      

 ∴  \(x + 2( - 2) + 2( - 1) = 0\) or, x = +6

(b) \({\rm{Kr}}{{\rm{F}}_{\rm{2}}}\) :  Let oxidation state of Kr be y,    \(y + 2( - 1) = 0\)   or, y = +2   

 (c) \({\rm{XeF}}_{\rm{3}}^{\rm{ + }}\)  :\(x + 3( - 1) =  + 1\)  or, x = +4

 (d) \({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\):  \(x + 6( - 2) =  - 4\)   or, x = +8  

                                                                   

So, the oxidation state of  Xe  in \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{_{\rm{2}}}}\) , \({\rm{XeF}}_{\rm{3}}^{\rm{ + }}\) , \({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\) are +6 , +4  and  +8 respectively where oxidation state of Kr in  \({\rm{Kr}}{{\rm{F}}_{\rm{2}}}\)  is +2 .