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    1. Calculate the CFSE for both high spin and low spin octahedral complexes of Co(gly)63-. Which is preferred?

    10Dq = 21476 cm-1, P = 23,625 cm-1, d6 complex.

    H.S. CFSE = 4Dq - P = - 15034.6 cm-1 L.S. CFSE = 24Dq - 3P = -19332.6 cm-1
    H.S. preferred, as expected when comparing 10Dq with P

    2. Determine whether the Co+2 complex with phenanthroline will prefer to be octahedral or tetrahedral based on CFSE.

    d7 complex, 10DqOh = 12060 cm-1, P = 20800 cm-1: complex will be H.S.
    Oh CFSE = 8Dq - 2P = -31952 cm-1

    10DqTd = 5360 cm-1
    Td CFSE = 12Dq - 2P = - 35168 cm-1

    Oh preferred

    3. Draw the M.O. diagram for an octahedral complex with six sigma donor ligands. Draw the electrons for a d5 high spin complex. see class notes

    4. What are Jahn-Teller distortions? Where do they come from? Show 'z - in' and 'z-out' configurations. Which is preferred for a d9 configuration?

    Tetragonal distortions arising from a) different distances for the ligands, b) ligands of different field strength, and c) to remove degeneracy.

    d9 prefers z - out

    5. Draw the expected splitting for a Cu(ox)34- complex. Would the splitting pattern change if two of the Cu-O bond lengths were longer than the other four?

    Cu+2, d9 complex

    6. Using the Tanabe-Sugano diagrams, give the ground states for the free metal, high spin and low spin d5 complexes. What are the first excited state for each of these? List all possible transitions. Draw the microstates of the ground state for each of the three species.

    Ground State 1st Excited State
    Free metal ion 6S 4G
    H.S. 6A1g 4T1g (below 2Dq/B), 2T2g (above 2Dq/B)
    L.S. 2T2g 6A1g (below 3.6Dq/B), 4T1g (above 3.6Dq/B)

    H.S. - none
    L.S. - 7 possible transitions
    2T2g- 2A2g
    2T2g- 2T1g
    2T2g- 2Eg
    2T2g- 2T2g
    2T2g- 2T1g
    2T2g- 2A1g
    2T2g- 2A1g

    7. Determine the spin state for each of the following:

    a) Fe(OH2)63+ d5 H.S., 10Dq < P
    b) V(CN)63- d2 neither
    c) CuI42- d9 neither, even though Td always H.S, only one way to do diagram
    d) RuCl64- Ru+2 d6 2nd row transition metal, always L.S.

    8. Draw the expected splitting of the d orbitals for a trigonal bipyramidal geometry

    9. Draw the high spin and low spin configurations for a d4 ion in a tetrahedral field. Which is preferred and why?

    H.S. always perferred for Td geometry since 10DqTd is small compared to pairing energies.

    10. Determine the CFSE for the tetrahedral Co+2 complex with methylamine.

    10DqOh = 10530 cm-1, 10DqTd = 4680 cm-1, P = 20800 cm-1,
    CFSE = 12Dq - 2P = -35984 cm-1

    This page titled Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Draganjac via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.