Q10.69P - Chemistry: Molecular Nature Of Matter And Change

Question

“Inert” xenon actually forms several compounds, especially with the highly electronegative elements oxygen and fluorine. The simple fluorides  ${\mathbf{XeF}}_2$, ${\mathbf{XeF}}_4$, ${\mathbf{XeF}}_6$  and   are all formed by the direct reaction of the elements. As you might expect from the size of the xenon atom, the Xe-F bond is not a strong one. Calculate the Xe-F bond energy in  ${\mathbf{XeF}}_6$, given that the heat of formation is - 402 kJ/mol.

Step-by-Step Solution

Answer

The bond energy of the Xe-F bond is - 67 kJ/mol

Step 1: Definition

The amount of energy required to separate the atoms forming a molecular bond into free atoms or the bond strength of a chemical bond is known as bond energy.

 

Step 2: Structure of XeF 6

${\mathbf{XeF}}_6$ has ${\mathbf{sp}}^3{\mathbf{d}}^3$  hybridization. Xe has 8 electrons in its valence shell, of which 6 electrons form 6 bonds with 6F atoms while the remaining 2 electrons remain as lone pair. Due to the presence of lone pair its structure is distorted octahedral.  

 

Step 2: Bond energy of Xe-F bond

Since ${\mathbf{XeF}}_6$  has 6 Xe-F bonds. The total heat of formation is the sum of energy released during the formation of all these six bonds.

So, the energy of each Xe-F bond is

 

$\frac{-402}{6}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{=}\mathbf{-}\mathbf{67}\mathbf{kJ}\mathbf{/}\mathbf{mol}$