Q30P - Chemistry: Molecular Nature Of Matter And Change

Question

Which member of each pair has more ionic character in its bonds: 

(a) BeF₂ or CaF₂; (b) PbF₂ or PbF₄; (c) GeF₄ or PF₃?

Step-by-Step Solution

Answer

The compound with more ionic character in its bonds is shown below:

(a) CaF₂

(b) PbF₂

(c) GeF₄

Step 1: Electronegativity difference

Electronegativity refers to the chemical process that determines the tendency of an atom involved in an ionic and covalent bond to attract the bonding electron toward itself.

If the electronegativity difference between the two bonded atoms is smaller then it results in a covalent bond whereas if the electronegativity difference between the two bonded atoms is larger it results in an ionic bond with a more ionic character.

The electronegativity difference is denoted by the symbol $∆X$ and it is given by:

$\mathbf{∆}\mathit{X}\mathbf{=}{\mathit{X}}_{\mathbf{2}}\mathbf{-}{\mathit{X}}_{\mathbf{1}}$

Where X₁ and X₂ represents electronegativity of first and second compound respectively.

Step 2: Fajan’s rule

Fajans rule is used in inorganic chemistry to predict the bond character whether it is covalent or ionic in the chemical bond, the conditions which favor the ionic character in a compound depend upon the size of cation and anion.

• Lower the polarizing power of cation results in more ionic character in a chemical bond 
• Larger the size of the cation and smaller the size anion contributes to the more ionic character in the chemical bond.

Step 3: Identify the compound with more ionic character in each of the following given pair

(a) CaF₂ will show more _­ionic character than CaF_{2 ,} because smaller size of Be⁺ as compare to Ce⁺ ion, therefore polarization occur more in BeF₂ as compare to CaF_2. Thus, it becomes less ionic than CaF₂ as we know when polarization increases ionic character decreases.

 

(b) PbF₂ has more ionic character than PbF₄, because increasing oxidation number on central atom decreases ionic character. PbF₄ has a +4 oxidation state whereas PbF₂ has a +2 oxidation state on its central atom.

 

(c) GeF₄ will show more ionic character than PF_3, because electronegativity difference of GeF₄ is more as compare to PF_3. Thus it has more ionic character in its chemical bonds.

$\begin{array}{rcl}∆XGe{F}_4& =& \left(4\times 3.98\right)-2.01\\ & =& 15.98-2.01\\ & =& 13.91\end{array}$           >        $\begin{array}{rcl}∆XP{F}_4& =& \left(3\times 3.98\right)-2.01\\ & =& 11.94-2.01\\ & =& 9.84\end{array}$