solns21b-ps6-Feb11

solns21b-ps6-Feb11 - Ch 21b Problem Set#6 Solution Set...

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Ch 21b Problem Set #6 Solution Set 2011 Originally prepared by: Rogier Braakman and Geoff Blake 1. Octahedral Crystal Field Orbitals: [Cu(H 2 O) 6 ] 2+ d 9 [V(H 2 O) 6 ] 3+ d 2 [Mn(H 2 O) 6 ] 2+ d 5 - high spin [Co(NH 3 ) 6 ] 3+ d 6 - low spin You might come up with a few reasons why [Mn(H 2 O) 6 ] 2+ makes a colorless solution. Remember that g to g transitions are forbidden, but all these other complexes overcome it using vibronic interactions. So what’s different about [Mn(H 2 O) 6 ] 2+ ? Notice that any transition between the e g and t 2g orbitals (emission or absorption) requires the electron to change spin so that it can pair up with the electron already in the orbital. Since this cannot happen, any d-d transition in this complex is spin-forbidden, and the solution is mostly colorless. Complexes in which absorption is spin forbidden absorb slightly due to spin-orbit coupling. Spin-orbit interactions are greater for heavier atoms and so first row transition metals (Mn 2+ , Fe 3+ , etc.) have smaller molar extinction coefficients ( ε max <1 L 3 mol -1 cm -1 ) than those of second or third row transition metals. Spin orbit coupling involves the coupling of the magnetic field generated by the spin of electrons
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solns21b-ps6-Feb11 - Ch 21b Problem Set#6 Solution Set...

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