Question

The covalent alkaline earth metal halide (X=Cl, Br, I) is :
(A) $Ca{{X}_{2}}$
(B) $Sr{{X}_{2}}$
(C) $Be{{X}_{2}}$
(D) $Mg{{X}_{2}}$

Answer 1

Hint: Polarising power is the extent to which a cation can polarise an anion. It is proportional to charge density. Charge density is the ratio of charge to volume. $Polarising\text{ }power\text{ }\alpha \text{ }Charge\text{ }density$ . More the charge density, greater is the polarising power for that cation.

Complete step by step answer: This question can be explained using Fajan’s rule. The postulates of Fajan’s rule are:
The rule can be stated on the basis of 3 factors, which are:
1.Size of the ion: Smaller the size of cation, the larger the size of the anion, greater is the covalent character of the ionic bond.
2.The charge of Cation: Greater the charge of cation, greater is the covalent character of the ionic bond.
3.Electronic configuration: For cations with same charge and size, the one, with $\left( n-1 \right){{d}^{n}}~n{{s}^{o}}~$ which is found in transition elements have greater covalent character than the cation with $n{{s}^{2}}~n{{p}^{6}}~$ electronic configuration, which is commonly found in alkali or alkaline earth metals. Now with the help of Fajan’s rule we will determine the covalent alkaline earth metal halide. In the given option, the halide atom is the same, therefore, covalent character will be decided on the basis of the metal atom. In metal atoms, charge and electronic configuration is the same for all metals, only change in size of the metal. According to fajan’s rule, smaller the size of cation, more is the covalent character. Within the given options, beryllium has the smallest size, therefore it will be the covalent alkaline earth metal halide.

Hence, the correct answer is the C option.

Note: If the cation is smaller, then we can say that the volume of the ion is less. If the volume is less, we can conclude that the charge density of the ion would be high. Since the charge density is high, the polarising power of the ion would be high. This makes the compound to be more covalent.