Q17 E - Chemistry (OpenStax)

Question

For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains \({\rm{77}}{\rm{.55\% }}\)Xe and \({\rm{22}}{\rm{.45\% \;F}}\)by mass.

(a) What is the formula of the compound?

(b) Write a Lewis structure for the compound.

(c) Predict the shape of the molecules of the compound.

(d) What hybridization is consistent with the shape you predicted?

Step-by-Step Solution

Answer

a. The molecular formula of the compound is \({\rm{Xe}}{{\rm{F}}_{\rm{2}}}\).

 

b. The Lewis structure is as follows:

 

c. The shape of the molecule is linear.

 

d. The hybridization is\({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;d}}\).

Step 1: Definition of Concept

Lewis structure: A Lewis Structure are quite generalised representation of the valence shell electrons in a molecule.

 

Hybridization: When pure orbitals (having comparable energy) are mixed together to produce an equivalent amount of impure orbitals (with equal energy and definite geometry).

Step 2: Find the formula of the compound

(a)

 

We will put the given data in tabular form in the question to find the molecular formula. 

 

Considering the given information:

 

\({\rm{\% }}\)of \({\rm{Xe = 77}}{\rm{.55}}\)

 

\({\rm{\% }}\) of \({\rm{F = 22}}{\rm{.45}}\)

 

Atomic mass of \({\rm{Xe = 131}}{\rm{.293amu}}\)

Atomic mass of \({\rm{F = 18}}{\rm{.99amu}}\)

 

Now,

 

Therefore, there is one Xe atom and two F atoms.

 

So, the formula is \({\rm{Xe}}{{\rm{F}}_{\rm{2}}}\).

 

Step 3: Draw a Lewis structure for the compound

Because fluorine has seven valence electrons and xenon has eight in its outermost orbit, the two fluorine atoms share one electron with xenon to form a noble gas. 

Therefore, the following is the Lewis dot structure:

Step 4: Predict the shape of the molecules of the compound

 

The Lewis dot structure is based on the Lewis dot structure.

Number of electrons in a bond pair \({\rm{ = 2}}\)

 

The number of electrons in a lone pair,

 

\(\begin{array}{l}{\rm{ = }}\frac{{{\rm{ Valenceelectronofcentralatom Xe - Number f bondpairelectron }}}}{{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{8 - 2}}}}{{\rm{2}}}\\{\rm{ = 3}}\end{array}\)

 

Total pair of electron \({\rm{ = 5}}\)

 

Structure = Trigonal bipyramidal

Therefore, the required we can see that the molecule's shape is linear.

 

 

Step 5: Find the hybridization of the molecule

The Lewis dot structure is based on the Lewis dot structure.

Number of electrons in a bond pair \({\rm{ = 2}}\)

 

The number of electrons in a lone pair,

 

\(\begin{array}{l}{\rm{ = }}\frac{{{\rm{ Valenceelectronofcentralatom Xe - Number f bondpairelectron }}}}{{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{8 - 2}}}}{{\rm{2}}}\\{\rm{ = 3}}\end{array}\)

 

Total pair of electron \({\rm{ = 5}}\)

 

Structure = Trigonal bipyramidal

Therefore, the required hybridization is\({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{\;d}}\).