The octahedral geometry of the complex [Co(NH3)6]2+ has hybridization:

Explanation

The octahedral geometry of the complex [Co(NH3)6]2+ is associated with the hybridization of the central cobalt (Co) atom. In an octahedral complex, the central metal atom is surrounded by six ligands arranged in an octahedral fashion, resulting in a geometry with six bonding pairs and no lone pairs.To determine the hybridization of the central cobalt atom, we can use the concept of hybridization theory, which describes the mixing of atomic orbitals to form new hybrid orbitals. In an octahedral complex, the central atom typically undergoes d2sp3 hybridization.Here's how the hybridization of [Co(NH3)6]2+ can be explained:1.Start with the electron configuration of the central cobalt atom (Co): [Ar] 3d7 4s2.2.In the octahedral complex, the six ammonia (NH3) ligands donate their electron pairs to the cobalt atom, resulting in the formation of six bonds. Each ammonia ligand contributes a lone pair of electrons.3.The cobalt atom promotes one of its 4s electrons to the empty 3d orbital, resulting in the following electron configuration: [Ar] 3d6 4s1.4. The six bonding pairs and one unpaired electron in the 3d orbital undergo hybridization to form a set of seven hybrid orbitals. These seven orbitals consist of six equivalent sp3d2 hybrid orbitals and one unhybridized 4s orbital.5. The six sp3d2 hybrid orbitals of cobalt overlap with the nitrogen (N) orbitals from the ammonia ligands to form six sigma (σ) bonds.In summary, the hybridization of the central cobalt atom in [Co(NH3)6]2+ is d2sp3.In this complex, cobalt is bonded to six ligands, NH3. The structure of the complex is given below: other words, the central atom is bonded to six other atoms (and no lone pairs). This corresponds to a hybridisation of sp3d2, which can be confirmed from the table below: option C is the correct answer.