Among the following complexes the one that shows zero crystal field stabilisation energy is

Question

BlackTom AI · Accepted Answer

In an octahedral crystal field t2g orbitals are stabilized by −0.4Δo per electron and eg orbitals are destabilized by +0.6Δo per electron; H2O is a weak-field ligand so all complexes here are high-spin. This is the key concept for computing crystal field stabilization energy (CFSE).

Step-by-step reasoning and calculations:
1. Determine d-electron count for each metal ion:
   - Mn3+: Mn (atomic) = 3d5 4s2 → Mn3+ = d4
   - Co2+: Co → 3d7 4s2 → Co2+ = d7
   - Fe3+: Fe → 3d6 4s2 → Fe3+ = d5
   - Co3+: Co3+ = d6

2. Assign high-spin electron distributions in octahedral field:
   - d4 (high-spin): t2g^3 eg^1
   - d7 (high-spin): t2g^5 eg^2
   - d5 (high-spin): t2g^3 eg^2
   - d6 (high-spin): t2g^4 eg^2

3. Compute CFSE = (n_t2g × −0.4Δo) + (n_eg × +0.6Δo) for each:
   - [Mn(H2O)6]3+ (d4): (3 × −0.4 + 1 × +0.6)Δo = (−1.2 + 0.6)Δo = −0.6Δo → nonzero
   - [Co(H2O)6]2+ (d7): (5 × −0.4 + 2 × +0.6)Δo = (−2.0 + 1.2)Δo = −0.8Δo → nonzero
   - [Fe(H2O)6]3+ (d5): (3 × −0.4 + 2 × +0.6)Δo = (−1.2 + 1.2)Δo = 0Δo → zero CFSE
   - [Co(H2O)6]3+ (d6): (4 × −0.4 + 2 × +0.6)Δo = (−1.6 + 1.2)Δo = −0.4Δo → nonzero

Why the correct answer is correct:
- For high-spin d5 (Fe3+), stabilization of three electrons in t2g exactly cancels the destabilization of two electrons in eg, giving net CFSE = 0. That is a special symmetry case for high-spin d5 in an octahedral field.

Why other options are incorrect (briefly):
- Mn3+ (d4), Co2+ (d7), and Co3+ (d6) all give nonzero net CFSE values as shown above, so they do not have zero CFSE.

Conclusion:
The complex with zero crystal field stabilization energy is [Fe(H2O)6]3+, matching the provided correct answer.

Among the following complexes the one that shows zero crystal field stabilisation energy is单项选择题

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