Question

Hybridisation of Al in [Al (H2O)6)]^+3???​

Answer 1

Answer:

              Orbitals which are not occupied only those can undergo Hybridisation. Ligads will only donate the electrons for the purpose of shielding the nuclei and will require the unoccupied orbitals for the same.

                In it's ground state, the Aluminium is 10+2+1 = 13, which is how many electrons Al in its ground state has.  Ground state Al has of course 1 unpaired electron due to the fact that the 3s orbital is full and there is one electron in the p orbital, and obviously the core electrons are all paired.

                   [Al]$3s^2$$sp^1$

                 However, if we remove 12 valence electrons  it accepts additional electrons from hexahydrate, and because of this it gets an electronic structure which is within transition elements, which allows $sp^3d^2$  hybrid orbitals to form and result is octahedral complexes.

                    Since water is neutral, the addition of these electrons does not change the charge.

Answer 2

Hybridization of Aluminium is Sp^3 d^2.

Explanation:

• The given compound is:
• [Al (H2O)6)]^+3
• In this compound the valency of aluminium "Al" is +3.
• At normal conditions Aluminium can not exit with valency more than 3.
• It has poly-valent configuration under normal conditions.
• Thus we can form [He]1s2, 2s2, 2p3, 3s1, 3p3, 4s1, 3d1 = 6 electrons are required for filling bold styled orbitals.
• These orbitals will be filled by neutral ligands with coordination no. of 6 just like H2O ligands.