Hybridization of (Co(H2O)6)2:

In order to determine the hybridization of the complex ion (Co(H2O)6)2, we need to analyze the central metal atom, cobalt (Co), and its surrounding ligands, which are water molecules (H2O).

1. Identify the central metal atom:
The central metal atom in this complex ion is cobalt (Co).

2. Determine the electron configuration of the central metal atom:
Cobalt (Co) has an atomic number of 27, which means it has 27 electrons. The electron configuration of cobalt is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7.

3. Count the number of valence electrons:
Since cobalt is in the +2 oxidation state, it has lost two electrons from its 4s orbital. Therefore, the number of valence electrons for the cobalt ion is 7.

4. Determine the number of ligands and their bonding orbitals:
The complex ion (Co(H2O)6)2 consists of six water molecules (H2O) acting as ligands. Each water molecule has a lone pair of electrons on the oxygen atom, which can be used for bonding.

5. Apply the concept of hybridization:
In order to form bonds with the ligands, the central metal atom undergoes hybridization. The type of hybridization depends on the number of bonding orbitals required to accommodate the ligands.

6. Determine the hybridization of the central metal atom:
Since cobalt (Co) has 7 valence electrons and needs to form six bonds with the water ligands, it undergoes sp3d2 hybridization. This means that one 4s orbital, three 4p orbitals, and two 4d orbitals of cobalt hybridize to form six new hybrid orbitals. These hybrid orbitals are known as sp3d2 orbitals.

7. Assign the hybrid orbitals to the ligands:
The six hybrid orbitals of cobalt overlap with the lone pairs of electrons on the oxygen atoms of the water ligands. This results in the formation of six sigma bonds between cobalt and the water molecules.

In summary, the hybridization of the central metal atom (cobalt) in the complex ion (Co(H2O)6)2 is sp3d2. This hybridization allows cobalt to form six sigma bonds with the six water ligands.