Question

Which is more covalent $SnC{l_2}$ (or) $SnC{l_4}$?

Answer 1

Hint: We have to know that Fajans' rule predicts whether a compound bond will be covalent or ionic. A couple of ionic bonds have halfway covalent qualities which were first talked about by Kazimierz Fajans in $1923$. Around then with the assistance of X-beam crystallography, he had the option to anticipate ionic or covalent holding with the characteristics like ionic radius and atomic radius.

Complete answer:
Based on Fajan’s rule, we can classify a few compounds as ionic compounds or covalent compounds. We can express the rule based on three factors, which are:
Ionic size: Smaller the size of cation, the bigger the size of the anion, more noteworthy is the covalent character of the ionic bond.
The charge of Cation: Greater the charge of cation, more prominent is the covalent character of the ionic bond.
Electronic configuration: For cations with same charge and size, the one, with $\left( {n - 1} \right){d^n}n{s^0}$which is found on the transition elements have more prominent covalent character than the cation with $n{s^2}n{p^6}$ electronic arrangement, which is generally found in alkali metal or alkaline earth metals.
We can say that based on Fajan’s rule, the central metal that contains more positive oxidation number exhibits more polarizing power. So, this in turn exhibits more covalent character. In $SnC{l_4}$, the oxidation number of tin is $ + 4$ but the oxidation number of tin in $SnC{l_2}$ is $ + 2$. So, $SnC{l_4}$ is more covalent.

Note:
We can say polarizing power is the degree to which a cation could polarize an anion. Charge density and polarizing power are related to each other. Charge density is the proportion of charge to volume. Polarizing power is proportional to density of charge. When the density of charge is more, the polarizing power for that cation is greater.