Determining Ground State of $[Cr(H_2O)_6]^{2+}$

To find the ground state term symbol for the octahedral complex $[Cr(H_2O)_6]^{2+}$, we need to follow these steps based on Crystal Field Theory (CFT).

1. Determine the Oxidation State and d-electron Count

• Chromium (Cr) has the electronic configuration $[Ar] 3d^5 4s^1$.
• In the complex $[Cr(H_2O)_6]^{2+}$, Chromium is in the +2 oxidation state.
• Therefore, $Cr^{2+}$ has lost two electrons, resulting in a $3d^4$ configuration.
• The complex is octahedral, indicated by the coordination number 6.

2. Assess Ligand Field Strength and Spin State

• The ligand $H_2O$ (water) is a weak field ligand.
• For a weak field (low $\Delta_o$) and $d^4$ configuration, the complex is high spin. This means electrons occupy orbitals according to Hund's rule to maximize spin multiplicity before pairing occurs.
• In an octahedral field, the d-orbitals split into lower energy $t_{2g}$ and higher energy $e_g$ sets.
• The high spin $d^4$ electron configuration is $t_{2g}^3 e_g^1$.

3. Calculate Spin Multiplicity

• The configuration $t_{2g}^3 e_g^1$ has 4 unpaired electrons (3 in $t_{2g}$ and 1 in $e_g$).
• Spin multiplicity is calculated as $2S+1$, where $S$ is the total spin quantum number. For $n$ unpaired electrons, $S = n/2$.
• $S = 4/2 = 2$.
• Spin Multiplicity = $2(2) + 1 = 5$. This indicates the ground state term will have a spin multiplicity of 5 (a quintet state).

4. Determine the Orbital Term Symbol

• For a $d^4$ configuration in an octahedral ($O_h$) field, the high-spin electron distribution is $t_{2g}^3 e_g^1$.
• Reference tables or CFT correlation diagrams show that the ground state term symbol for a high-spin $d^4$ octahedral complex is $^5E_g$.
• The $E_g$ symbol arises from the combination of the $t_{2g}^3$ configuration (which contributes $A_{2g}$ as its ground term part) and the $e_g^1$ configuration (which contributes $E_g$). The direct product $A_{2g} \times E_g$ yields $E_g$.

Conclusion

Based on the $d^4$ configuration, weak field ligand, high spin state, and octahedral geometry, the ground state term symbol for $[Cr(H_2O)_6]^{2+}$ is $^5E_g$.