Q112 E - Chemistry (OpenStax)

Question

What is the molecular structure of each of the following molecules?

        (a) \(Xe{F_2}\)        (b) \({\rm{Xe}}{{\rm{F}}_{\rm{4}}}\)      (c) \({\rm{Xe}}{{\rm{O}}_{\rm{3}}}\)        (d) \({\rm{Xe}}{{\rm{O}}_{\rm{4}}}\)        (e) \({\rm{XeO}}{{\rm{F}}_{\rm{4}}}\)

Step-by-Step Solution

Answer

The   molecular structure of the given molecule are shown below –

 

                              Molecule                               structure

1.        \({\rm{Xe}}{{\rm{F}}_{\rm{2}}}\)                     :               Linear 
2.       \({\rm{Xe}}{{\rm{F}}_{\rm{4}}}\)                     :            Square planar
3.       \({\rm{Xe}}{{\rm{O}}_{\rm{3}}}\)                     :            Trigonal Pyramidal
4.       \({\rm{Xe}}{{\rm{O}}_{\rm{4}}}\)                     :               Tetrahedral
5.      \({\rm{XeO}}{{\rm{F}}_{\rm{4}}}\)                    :           Square Pyramidal

Step 1 : VSEPR Theory :

The VSEPR theory is used to predict the structure of the Molecules from the electron pairs that surround the metal atom of the molecule.

  

 According to VSEPR Theory ,  

Empirical form of Hybridisation, H =  \(\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{V + X - C + A}}} \right)\)

  Where V = no. of valence electrons of central metal atom

               X = no. of monovalent atoms around the central atom

               C = Charge of the cation

               A = Charge of the anion

               No. of lone pairs = (H – no. of shared pair)

Step 2 : Structure determination :

Now let us see all the options, (a)    \({\rm{Xe}}{{\rm{F}}_{\rm{2}}}\):     H =  \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{(8 + 2) = 5 = s}}{{\rm{p}}^{^{\rm{3}}}}{\rm{d}}\)

                 Lone pair =  ( 5 -   2 ) = 3    .  

       The structure of this molecule is:                      

 so, it has a linear structure.

     (b)     \({\rm{Xe}}{{\rm{F}}_{\rm{4}}}\)  :   H =  \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{(8 + 4) = 6 = s}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}\)  and    Lone pair = ( 6 – 4 ) =2      .  

   The structure of the molecule is:

So, it has a square planar structure.

   (c)    \({\rm{Xe}}{{\rm{0}}_{\rm{3}}}\)    :      H =  \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times 8 = 4 = s}}{{\rm{p}}^{\rm{3}}}\)  ; Lone pair = (4 – 3 ) = 1

   The structure of the molecule is:  

So, it has a triagonal pyramidal structure.

(d)   \({\rm{Xe}}{{\rm{0}}_{\rm{4}}}\)       :      H =   \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times 8 = 4 = s}}{{\rm{p}}^{\rm{3}}}\);   Lone pair = (4 – 4) = 0

     The structure of the molecule is: 

   So, it has a tetrahedral structure.      

     

(e) \({\rm{Xe0}}{{\rm{F}}_{\rm{4}}}\)   :     H =   \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{(8 + 4) = 6 = s}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}\)   ;   Lone pair = ( 6 – 5 )  = 1

       The structure  Of the molecule is :

 So, it has a square pyramidal structure.

 

 Finally, we  can conclude that options (a),(b),(c),(d) and (e) have Linear, Square planar, Triagonal Pyramidal, Tetrahedral and Square Pyramidal respectively.