Question

Boron can form only $B{{F}_{4}}^{-}$ . But Al can form $AlF_{6}^{3-}$ explain.

Answer 1

Hint: There is an observed trend in the chemical behavior of group 13 elements. All tri chlorides, bromides, and iodides of these group elements are covalent compounds and hydrolyzed in water. Due to electron-deficiency of monoatomic trihalides it acts as strong Lewis acids.

Complete step by step solution:
-Boron (B) and Aluminum (Al) are the elements that belong to group 13 and the maximum covalency of this group is +3. Due to the unavailability of d-orbitals, the maximum covalency of the first member of each group of s-block and p-block elements is +4. However, other members can expand their valence shell to accommodate more than 4pairs of electrons due to the availability of d-orbitals.
-B electronic configuration as follows: $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
-Whereas electronic configuration of Al as follows,
$[Ne]3{{s}^{2}}3{{p}^{1}}$
-This is observed that Al can expand its octet fulfilled by the availability of 3d-orbitals. So, the maxim valence of Al becomes +6, and Al forms $AlF_{6}^{3-}$
-In the case of boron, it does not have a 2d-orbital but only consists of one s and three p-orbitals and hence it cannot expand to its octet. Therefore, the maximum covalence for B is 4. So, B cannot accommodate six fluorine atoms around it and boron forms only $B{{F}_{4}}^{-}$, cannot form $B{{F}_{6}}^{3-}$

Note: Since Al and other elements in the boron family consist of the availability of d orbitals. Except for B, all the other metal halides are dimerized through hydrogen bridging. These metal species complete its octet by accepting electrons from halogen in these halogen bridged molecules.