Find the hybridization state of the central atom in [Co(NH3)6]³+. Also find its magnetic nature and geometry
Anushka , 7 Years ago
Grade 12
Inorganic Chemistry> Find the hybridization state of the centr...
1 Answers
Arun
Last Activity: 7 Years ago
The orbitals involved in hybridisation are determined by the ligand. Ammonia is a strong field ligand and it can pair up electrons of the inner d-orbital while fluoride being a weak field ligand is unable to do so. Co3+ has 6 electrons in 3d orbital out of which four are unpaired. Strong ligands like ammonia can pair up these 4 unpaired electrons and free up two 3d orbitals. These 3d orbitals are involved in hybridisation along with one 4s and three 4p orbitals forming an inner orbital complex.
Weak ligands cannot pair up these inner 3d electrons and utilize one 4s, three 4p and two 4d orbitals in hybridization forming an outer orbital complex.
So, hybridization is d2sp3 for ammonia and sp3d2 for fluoride.
Weak ligands cannot pair up these inner 3d electrons and utilize one 4s, three 4p and two 4d orbitals in hybridization forming an outer orbital complex.
So, hybridization is d2sp3 for ammonia and sp3d2 for fluoride.
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