Step 1: Determine the Oxidation State and Electron Configuration
For each complex, we calculate the oxidation state of the metal and its electron configuration:
- (A) \( [Mn(H_2O)_6]^{2+} \)
- Mn oxidation state: \( +2 \)
- Mn atomic number: \( 25 \) → Electron configuration: \( [Ar] 3d^5 \)
- \( t_{2g}^3 e_g^2 \) is not matching.
- (B) \( [Fe(H_2O)_6]^{2+} \)
- Fe oxidation state: \( +2 \)
- Fe atomic number: \( 26 \) → Electron configuration: \( [Ar] 3d^6 \)
- In an octahedral field, this configuration corresponds to \( t_{2g}^3 e_g^2 \) for high-spin Fe\(^{2+} \) → Correct choice.
- (C) \( [Co(NH_3)_6]^{3+} \)
- Co oxidation state: \( +3 \)
- Co atomic number: \( 27 \) → Electron configuration: \( [Ar] 3d^6 \)
- Ligand \( NH_3 \) is a strong field ligand, leading to a low-spin configuration \( t_{2g}^6 e_g^0 \) → Not matching.
- (D) \( [Ni(H_2O)_6]^{2+} \)
- Ni oxidation state: \( +2 \)
- Ni atomic number: \( 28 \) → Electron configuration: \( [Ar] 3d^8 \)
- The \( 3d^8 \) configuration leads to \( t_{2g}^6 e_g^2 \) → Not matching.
Conclusion
Thus, the correct answer is:
\[
[Fe(H_2O)_6]^{2+}
\]