Final answer:
The transition metal M that satisfies the LFSE of -3/5 Δo for the complex [M(H₂O)₆]²⁺ is manganese (Mn) in its 2+ oxidation state, which has a d⁵ configuration and results in the required LFSE for a high spin complex.
Step-by-step explanation:
The question asks to identify the first row transition metal M with a complex [M(H₂O)₆]²⁺ that has a ligand field stabilization energy (LFSE) of -3/5 Δo. LFSE depends on the electron configuration of the metal ion in the crystal field of the ligand. Since a LFSE of -3/5 Δo suggests that the complex is a high spin complex with five unpaired electrons, the likely candidates would be those with a d⁵ electron configuration in their 2+ oxidation state.
One such example of a first row transition metal that satisfies this requirement is manganese (Mn). In its 2+ oxidation state, manganese has a d⁵ configuration, which would give the required LFSE. Similarly, iron (Fe) in the 2+ oxidation state also has a d⁶ configuration, but because the LFSE of high spin d⁶ is not -3/5 Δo, Fe²⁺ would not be the correct answer. The actual LFSE calculation also depends on whether the complex is high or low spin, but based on the question providing LFSE as -3/5 Δo, we can infer a d⁵ configuration is implied, aligning with Mn²⁺ in its high spin state.