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CFSE of complex [mn (h_{2}*o) 6 ]cl 2 is o Zero Ο -1.2 Delta_{0} Ο -1.6 Delta_{0} Ο -0.4 Delta_{0}
Accepted Answer
The CFSE (Crystal Field Stabilization Energy) of the complex [Mn(H2O)6]Cl2 is zero. Here's why:
1. Mn in this complex has a +2 oxidation state.
2. Electronic Configuration: Mn²⁺ has a d⁵ configuration ([Ar] 3d⁵).
3. Octahedral Complex: The complex is octahedral, with six water ligands (H2O) surrounding the Mn²⁺ ion.
4. High Spin: Mn²⁺ in an octahedral complex will have a high spin configuration due to the weak field ligands (H2O).
5. CFSE Calculation: In a high spin d⁵ configuration, the electrons fill all the d orbitals individually, with one electron in each of the t2g and eg orbitals. Since there's an equal number of electrons in the t2g and eg orbitals, the CFSE is calculated as:
CFSE = (Number of t2g electrons
-0.4Δo) + (Number of eg electrons
0.6Δo) = (3
-0.4Δo) + (2
0.6Δo) = 0
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