Question

Which one of the following elements is unable to form MF₆^3-ion ?

• A. Ga
• B. B
• C. Al
• D. In

Answer 1

The Correct Option is B

To determine which element among the given options is unable to form the complex ion MF₆^3-, we need to consider the electronic structure and availability of orbitals for each element to reach the required hybridization state.

1. Electronic Configuration:
   • Ga (Gallium): [Ar] 3d¹⁰ 4s² 4p¹
   • Al (Aluminum): [Ne] 3s² 3p¹
   • In (Indium): [Kr] 4d¹⁰ 5s² 5p¹
   • B (Boron): [He] 2s² 2p¹
2. Hybridization Requirement:

   The ion MF₆^3- suggests that the element M must be able to expand its valence shell to accommodate six ligands (fluorine atoms), typically through sp^3d^2 hybridization.

3. Analysis of Elements:
   • Ga, Al, and In: All these elements are in group 13 along with Boron but have empty d orbitals, which can be utilized for the sp^3d^2 hybridization necessary to form MF₆^3-.
   • B (Boron): Boron lacks d orbitals in its valence shell, having only the 2s and 2p orbitals. Therefore, it cannot achieve the required sp^3d^2 hybridization.
4. Conclusion:

   Based on the absence of available d orbitals in boron's valence shell, it is unable to form the MF₆^3- ion. Therefore, the correct answer is B (Boron).

Thus, Boron cannot form a hexafluorocomplex ion due to the lack of d orbitals for hybridization, contrasting with Gallium, Aluminum, and Indium, which can accommodate such structures.