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Step-by-Step Explanation
Step 1: Identify the Complex and the Central Metal Ion
The complex in question is [Co(H_{2}O)_{6}]^{2+} , where cobalt is in the +2 oxidation state. So the central metal ion is Co^{2+} .
Step 2: Determine the d-Electron Count
A neutral cobalt (Co) atom has the electronic configuration [Ar]\,3d^{7}\,4s^{2} . In the +2 oxidation state, it loses two electrons (typically from the 4s orbital first), resulting in a 3d^{7} configuration for Co^{2+} .
Step 3: Octahedral Crystal Field Splitting
In an octahedral field, the five d -orbitals split into two sets: the lower-energy t_{2g} (three orbitals) and the higher-energy e_{g} (two orbitals). The energy gap between them is denoted by \Delta_{o} .
Step 4: High Spin vs. Low Spin
Water ( H_{2}O ) is a weak-field ligand, so \Delta_{o} is not large enough to force electrons to pair up at lower energy levels. Hence, Co^{2+} in [Co(H_{2}O)_{6}]^{2+} adopts a high-spin configuration for its 3d^{7} electrons.
Step 5: Electron Arrangement in the Octahedral Field
For a high-spin d^{7} system in an octahedral field:
Each of the three t_{2g} orbitals gets one electron (3 electrons total).
The two e_{g} orbitals each get one electron (2 electrons total).
The remaining two electrons then occupy the t_{2g} orbitals, pairing where necessary.
Thus, the electron configuration is t_{2g}^{5}\,e_{g}^{2} .
Step 6: Count the Unpaired Electrons in t_{2g}
In the t_{2g}^{5} configuration:
Two t_{2g} orbitals contain a pair of electrons each (4 electrons).
One t_{2g} orbital contains a single electron (1 unpaired electron).
Therefore, there is exactly 1 unpaired electron in the t_{2g} set of orbitals.
Final Answer
There is 1 unpaired electron in the t_{2g} orbitals of the [Co(H_{2}O)_{6}]^{2+} complex.
