Question

Which of the following compounds of group-14 elements would you expect to be ionic in character?
(A) $CC{l_4}$
(B) $SiC{l_4}$
(C) $PbC{l_2}$
(D) $PbC{l_4}$

Answer 1

Hint: The carbon group that is group 14 has metal, nonmetals and metalloids. Ionic character is determined by the transfer of valence electrons which form a chemical bond. The ionic character of a compound is determined either by the difference in their electronegativity or by following the Fajan’s rule.

Complete step by step answer:
So, the formation of ions by the transfer of valence electrons between atoms resulting in a chemical bond is known as ionic bonding. It depends on how evenly the electrons are shared among the atoms. Ionic character is determined by the difference in electronegativity of atoms. Greater the difference, greater will be ionic character. Ionic bonds are formed between metal and nonmetals whereas covalent bonds are formed between nonmetals. Ionic bonds are stronger covalent bonds as the ionic bonds have attraction due to oppositely charged ions whereas the covalent bonds are bonded due to sharing of electrons.
Group 14 elements are also known as carbon groups. It has carbon which is non-metal; silicon and germanium which are metalloids and the rest of the elements are metals. Group 14 elements have an oxidation state of $ + 4$ and $ + 2$ due to inert pair effect. These elements have higher boiling and melting points which decrease down the group because atomic forces tend to become less strong due to increasing size of the elements.
The Fajan’s rule was introduced to state that if in a compound the cation is larger in size and it has small positive charge and thus, if the size of anion is smaller than the cation then the compound is more iconic. For example let’s take $NaCl$ in which $Na + $ has a small positive charge and is a larger cation than the small $Cl - $ anion hence; we know that $NaCl$ is an ionic compound.
Now, for the given question
We see that in option (C) $PbC{l_2}$ $Pb$ has an oxidation state of +2 whereas in other options the non-metals have a charge of $ + 4$. As we know that Fajan’s rule states that the cation with smaller positive charge is a more ionic compound thus, we conclude that $PbC{l_2}$ is more ionic than the other options.
So, the correct option is, ‘(C) $PbC{l_2}$’.

Note: In group 14 electronegativity decreases down the group due to filling of d and f orbitals and increases on moving across from left to right though it remains same from silicon to lead. Electronegativity basically refers to how strongly an atom attracts electrons in a bond and the difference between electronegativity is used to calculate the ionic character of a compound.