Question

Which of the following compounds of elements in group IV is expected to be most ionic?
A.${\text{PbC}}{{\text{l}}_2}$
B.${\text{PbC}}{{\text{l}}_4}$
C.${\text{CC}}{{\text{l}}_{\text{4}}}$
D. ${\text{SiC}}{{\text{l}}_{\text{4}}}$

Answer 1

**To check the ionic character in the compound we need to look into Fajan’s rule. Greater is the charge on cation greater will be its polarizing power and hence more covalent character will be there. Check the charge on each of the elements and the one with lowest charge will be more ionic.

Complete step by step answer:
First of all let us look into a term called polarizing power. It is the ability of cation to polarize an anion. The postulates of Fajan’s rule are:
Size of ion: larger the size of anion and smaller the size of cation, makes the compound more covalent.
Charge on cation: the greater the charge on cation, the more covalent character will be there in the bond and the smaller is the charge on cation, lower will the covalent character or say higher will be the ionic character.
Now since the anion is the same in all the options, so we need to consider only charge on cation. Let us calculate charge on each of the cation given in question:
Since one chlorine ion carries a charge of $- 1$ so two will carry $- 2$ and hence ${\text{Pb}}$ will carry $+ 2$ charge in ${\text{PbC}}{{\text{l}}_2}$likewise in option 2 ${\text{PbC}}{{\text{l}}_4}$ will carry a charge of $+ 4$ . In ${\text{CC}}{{\text{l}}_{\text{4}}}$, ${\text{C}}$will carry $+ 4$ charge and in ${\text{SiC}}{{\text{l}}_{\text{4}}}$ too ${\text{Si}}$ will carry ${\text{ + 4}}$ charge. As all the cations carry a charge of ${\text{ + 4}}$ except ${\text{PbC}}{{\text{l}}_2}$in which ${\text{ + 2}}$ charge is there on ${\text{Pb}}$, which is lowest among all. Hence, it will have the most ionic character.

Thus, the correct option is option A.

Note:
Group 14 consists of 5 elements. They are Carbon, Silicon, Germanium, Tin and Lead. It is known as carbon family. Generally group 14 elements form tetra halides due to the presence of 4 electrons in their valence shell but the lower elements in this group form dihalide. It is because of inert pair effect that the stability of lower oxidation state of element increases while moving down the group and lead is the second last element of the group so can form halide as well due to inert pair effect.