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) \({\rm{XeO}}{{\rm{F}}_{\rm{4}}}\)

Short 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-by-step solution

 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)

 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.