Crystal Field Stabilization Energy (cfse)

Question

Approximate CFSE (in terms of Δo) for a high-spin d5 octahedral complex is:

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1. **High Spin $d^5$ Configuration**: Since it is high spin, electrons do not pair up in the lower level. Distribution is $t_{2g}^3 e_g^2$.
2. **CFSE Calculation Formula**: $CFSE = [(-0.4 	imes n_{t_{2g}}) + (0.6 	imes n_{e_g})] \Delta_o$.
3. **Application**:
   - $n_{t_{2g}} = 3$ (each stabilized by $-0.4 \Delta_o$)
   - $n_{e_g} = 2$ (each destabilized by $+0.6 \Delta_o$)
4. **Math**: $CFSE = [(-0.4 	imes 3) + (0.6 	imes 2)] \Delta_o = (-1.2 + 1.2) \Delta_o = 0$.
5. **Significance**: Elements like $Mn^{2+}$ and $Fe^{3+}$ in a high-spin state have zero stabilization energy, making them more kinetically labile.

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Crystal Field Stabilization Energy (cfse)

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