Question

Which compound is highest covalent?
A) $LiCl$
B) $LiF$
C) $LiBr$
D) $LiI$

Answer 1

Hint: We can compare the covalent or ionic nature of the compounds with the help of Fajan’s rule according to which if the smaller the size of cation and larger the size of anion, greater will be the covalent character of the compound.

Complete step by step solution:There are two basic postulates of Fajan’s rule which are stated as follows:
1. Size of the cation: Smaller the size of the cation in the molecule, results in the greater covalent character of the bond. This can be explained on the basis of polarizing power of cation. If the size of cation is smaller, the charge density on the nucleus will be high due to which polarizing power of the cation would be high resulting in greater covalent character.
2. Size of the anion: Larger the size of the anion in the molecule, greater is the covalent character of the bond. This can be explained on the basis of polarizability of the anion as larger the size of anion results in less effective nuclear charge which makes the valence electrons easily polarized by the cation which makes the bond more covalent.
Now among the given molecules, the size of the cation is the same in each molecule and we have to check the size of anion. The compound with the largest anion will be having the highest covalent character. As the mentioned anions in the compounds are from halogen family i.e., belongs to group 17 and we know that the size of the anion varies as follows:
${{F}^{-}}<{{Cl}^{-}}<{{Br}^{-}}<{{I}^{-}}$

Therefore, the compound having the largest anion is $LiI$ and thus will have the maximum covalent character.

Option ‘D’ is correct

Note: It is important to note that if the size of the cations in the two molecules is same then the cations with pseudo inert gas configuration $\left( n{{s}^{2}}{{p}^{6}}{{d}^{10}} \right)$ or having inert pair configuration $\left( (n-1){{d}^{10}}n{{s}^{2}} \right)$, have comparatively high polarizing power as compared to the cations having noble gas configuration $\left( n{{s}^{2}}n{{p}^{6}} \right)$.