Ligand Field Stabilisation Energy (LFSE)
| Calculate the LFSE (in terms of Δo) for the indicated d-electron configuration and geometry | ||
| 1. | d3 octahedral | [Solution] |
| 2. | d5 high-spin octahedral | [Solution] |
| 3. | d5 low-spin octahedral | [Solution] |
| 4. | d2 tetrahedral | [Solution] |
| 5. | d4 tetrahedral | [Solution] |
| 6. | d7 tetrahedral | [Solution] |
| 7. | d10 octahedral | [Solution] |
| 8. | d9 octahedral | [Solution] |
| 9. | d5 tetrahedral | [Solution] |
| 10. | d8 tetrahedral | [Solution] |
| 11. | d7 high-spin octahedral | [Solution] |
| 12. | d6 low-spin octahedral | [Solution] |
| 13. | d1 octahedral | [Solution] |
| 14. | [Solution] | |
| 15. | [Solution] | |
Solutions
| 1. |
d3 octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 3 − 3/5 × 3 − 0 = 3 − 9/5 − 0 = 6/5Δo |
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| 2. |
d5 high-spin octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 5 − 3/5 × 5 − 2 = 5 − 15/5 − 2 = 0Δo |
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| 3. |
d5 low-spin octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 5 − 3/5 × 5 − 0 = 5 − 15/5 − 0 = 10/5Δo |
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| 4. |
d2 tetrahedral
tetrahedral: total electrons − 2/5 × total electrons − antibonding electrons 2 − 2/5 × 2 − 0 = 2 − 4/5 − 0 = 6/5Δt 6/5Δt × (4/9Δo)/Δt = 8/15Δo |
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| 5. |
d4 tetrahedral
tetrahedral: total electrons − 2/5 × total electrons − antibonding electrons 4 − 2/5 × 4 − 2 = 2 − 8/5 − 2 = 2/5Δt 2/5Δt × (4/9Δo)/Δt = 8/45Δo |
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| 6. |
d7 tetrahedral
tetrahedral: total electrons − 2/5 × total electrons − antibonding electrons 7 − 2/5 × 7 − 3 = 7 − 14/5 − 3 = 6/5Δt 6/5Δt × (4/9Δo)/Δt = 8/15Δo |
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| 7. |
d10 octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 10 − 3/5 × 10 − 4 = 10 − 30/5 − 4 = 0Δo |
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| 8. |
d10 octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 9 − 3/5 × 9 − 3 = 10 − 27/5 − 3 = 3/5Δo |
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| 9. |
d5 tetrahedral
tetrahedral: total electrons − 2/5 × total electrons − antibonding electrons 5 − 2/5 × 5 − 3 = 5 − 10/5 − 3 = 0Δt 0Δt × (4/9Δo)/Δt = 0Δo |
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| 10. |
d8 tetrahedral
tetrahedral: total electrons − 2/5 × total electrons − antibonding electrons 8 − 2/5 × 8 − 4 = 8 − 16/5 − 4 = 4/5Δt 4/5Δt × (4/9Δo)/Δt = 16/45Δo |
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| 11. |
d7 high-spin octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 7 − 3/5 × 7 − 2 = 7 − 21/5 − 2 = 4/5Δo |
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| 12. |
d6 low-spin octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 6 − 3/5 × 6 − 0 = 6 − 18/5 − 0 = 12/5Δo |
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| 13. |
d1 octahedral
octahedral: total electrons − 3/5 × total electrons − antibonding electrons 1 − 3/5 × 1 − 0 = 1 − 3/5 − 0 = 2/5Δo |
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| 14. | [Back to Questions] | ||||||||||||||||||||||||
| 15. | [Back to Questions] | ||||||||||||||||||||||||