Question

The compound having more covalent character is:
(A) $SnC{{l}_{2}}$
(B) $SnC{{l}_{4}}$
(C) $PbC{{l}_{2}}$
(D) $PbC{{l}_{4}}$

Answer 1

Recollect about the trends in periodic table. As we move from top to bottom in a periodic table, atomic size increases. Take a look at the metal atoms and identify in which group/s they belong. Think about Fajan’s rule. We know that chlorine is electronegative, so think about each compound and then come to a conclusion.

Complete step by step answer:
- Sn is Tin and Pb is Lead. Both tin and lead belong to the carbon family. Tin has atomic number 50 and lead has atomic number 82. Tin belongs to the fifth period and lead belongs to the sixth period of the periodic table.
- Both tin and lead form stable compounds with chlorine. To determine which of the given compounds has more covalent character, let’s take a look at Fajan’s rule.
- According to Fajan's rule, greater the charge on the cation, more is its potential to polarize the bonding electrons towards itself giving the compound a higher covalent character.
- From Fajan’s rule we can conclude that, $SnC{{l}_{4}}$ is more covalent than $SnC{{l}_{2}}$ and similarly, $PbC{{l}_{4}}$ is more covalent than $PbC{{l}_{2}}$ because in $SnC{{l}_{2}}$ and $PbC{{l}_{2}}$, cations have +2 charge whereas in $SnC{{l}_{4}}$ and $PbC{{l}_{4}}$, cations have +4 charge.
- We know that as we move from top to bottom in the periodic table, atomic size increases and hence, covalent character also decreases. Therefore, $SnC{{l}_{4}}$ will be more covalent than $PbC{{l}_{4}}$.
- Therefore, the compound having more covalent character is $SnC{{l}_{4}}$.
So, the correct answer is “Option B”.

Note: Remember according to Fajan’s rule, higher the charge and smaller the size of cation, more is the covalent character of the compounds due to higher ability of cations to polarize the bonding electrons towards itself. In case of anions, smaller the charge and larger the size of anion, more is the covalent character of the compound.