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  • The hybridization state of aluminum in the octahedral complex   suggests:<st

The hybridization state of aluminum in the octahedral complex Al(H_{2}O)^{6+}  suggests:

Option: 1

The presence of sigma bonds only

 


Option: 2

The presence of pi bonds only


Option: 3

The presence of both sigma and pi bonds

 


Option: 4

The absence of any type of bonding


Answers (1)

best_answer
  • In an octahedral complex such as Al(H_{2}O)^{6+}, the central aluminum ion is surrounded by six water ligands (H_2O). In this case, aluminum undergoes sp^3d^2 hybridization, where one 3s orbital, three 3p orbitals, and two 3d orbitals mix to form six hybrid orbitals. These hybrid orbitals overlap with the ligand orbitals to form sigma bonds.
  • Sigma (\sigma)bonds are formed by the overlap of atomic orbitals along the axis connecting the bonded atoms. In the octahedral complex, the aluminum ion forms sigma bonds with the water ligands.
  • Therefore, the hybridization state of aluminum in the octahedral complex Al(H_{2}O)^{6+}  suggests the presence of sigma bonds only.
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