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Step-by-Step Detailed Solution
Step 1: Write the d-Electronic Configuration of the Metal Ion
To identify the metal ion Mz+ based on the given CFSE and magnetic moment, start by listing the possible options. The correct answer is Co2+, so let us verify this systematically.
Co2+ has an electronic configuration:
[Ar] 3d74s0
Step 2: Determine the Splitting in an Octahedral Field
In an octahedral complex, the five d-orbitals of the metal split into two sets:
• Lower-energy set: $t_{2g}$ (three orbitals)
• Higher-energy set: $e_g$ (two orbitals).
For a weak-field ligand like H2O, the electron arrangement tends to maximize the number of unpaired electrons (high-spin complex).
Step 3: Distribute the Electrons in d-Orbitals for Co2+
Co2+ with 3d7 electrons in an octahedral field of a weak ligand (H2O) is arranged as:
$t_{2g}^5 \, e_{g}^2$
This indicates five electrons in the lower-energy $t_{2g}$ orbitals and two electrons in the higher-energy $e_g$ orbitals.
Step 4: Calculate the CFSE
For an octahedral complex, the CFSE (Crystal Field Stabilization Energy) in terms of $ \Delta_0 $ (octahedral crystal field splitting) is:
CFSE = [($-0.4 \Delta_0$) × (number of electrons in $t_{2g}$)] + [($+0.6 \Delta_0$) × (number of electrons in $e_g$)]
Here, there are 5 electrons in $t_{2g}$ and 2 electrons in $e_g$. Thus:
$ \text{CFSE} = [-0.4 \Delta_0 \times 5] + [0.6 \Delta_0 \times 2] \\ = -2.0 \Delta_0 + 1.2 \Delta_0 \\ = -0.8 \Delta_0 $
This matches the given CFSE of $-0.8 \Delta_0$.
Step 5: Calculate the Spin-Only Magnetic Moment
The spin-only magnetic moment $ \mu $ (in Bohr Magnetons, BM) is calculated using:
$ \mu = \sqrt{n(n + 2)} $
where $n$ is the number of unpaired electrons. In $t_{2g}^5\, e_g^2$, there are 3 unpaired electrons. Hence:
$ \mu = \sqrt{3 \times (3 + 2)} = \sqrt{15} = 3.87 \text{ BM} $
This matches the given magnetic moment of 3.87 BM.
Step 6: Conclusion
Both the calculated CFSE ($-0.8 \Delta_0$) and the spin-only magnetic moment (3.87 BM) are consistent with Co2+ in an octahedral complex with a weak-field ligand like H2O. Thus, the metal ion is correctly identified as Co2+.
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