Question

Maximum covalent character is associated with the compound
A. NaI
B. \[Mg{{I}_{2}}\]
C. \[AlC{{l}_{3}}\]
D. \[~Al{{I}_{3}}\]

Answer 1

Hint: This question will be solved as per Fajan’s rule. According to Fajan’s rule, higher the charge and smaller the size of cation, will form stronger covalent bonds with cations having higher charge and larger size. So, cation has tendency to polarise (has polarising power) and anion has great polarizability.

Complete Step by Step Solution:
The given cation of all the given molecules are\[N{{a}^{+}}\],\[M{{g}^{+2}}\] and\[A{{l}^{+3}}\]. From all the cations, \[A{{l}^{+3}}\] has higher charge means, from it 3 electrons has been removed and due to which its size will decrease. Not only this, its size is all very smaller than the other two cations as it belongs to group 13 (right side of period 2) and to 2ndperiod from which other two cautions also belong. Thus, aluminium has greater tendency to polarise an anion as it is most electronegative than other two cations (\[N{{a}^{+}}\]and\[M{{g}^{+2}}\]).

Now given anion of all the options are, iodine (I) and chlorine (Cl). Both chlorine and iodine belong to \[{{17}^{th}}\]group where iodine is at the end of this group. Due to this, iodine size is much greater than chlorine. And charge on chlorine and iodine is the same.

As Aluminium has higher charge and smaller size on the other hand iodine is much greater in size than chlorine, thus, both will form a bond with maximum covalent character.
Thus, the correct option is D.

Note: Another point of view is that electronegativity of iodine is greater than aluminium. An electron has been released from aluminium (3 electrons) which tends to increase its electronegativity. This makes some electronegativity of iodine equal to electronegativity of aluminium. As there is very small electronegativity difference, they both will form a bond of maximum covalent character.