(c) The trans-isomer of $\ce{Ni(en)2(NO2)2}$ D is red $(λ_\mathrm{max} = \pu{500 nm}).$ It can be heated to give a new trans-isomer E, which is violet $(λ_\mathrm{max} = \pu{560 nm}).$ Assume that, within each of D and E, the two $\ce{NO2-}$ ligands exhibit identical coordination modes.

(i) Rationalize the colours of D and E, explaining the difference in d-orbital splitting. [6]

(ii) Sketch structures for D and E. [2]

I've been stuck on this for some time to the stage that I'm probably just melting brain cells. Any help is appreciated.

My initial thoughts were that it was due to one end of the en ligands dissociating to form a 4-coordinate species in sq planar (owing to $\ce{Ni^2+} = \mathrm{d^8});$ however this would mean that it goes low-spin and then the splitting of d-orbitals becomes even greater, hence would expect a smaller UV-vis peak wavelength as a higher $E$ jump between occupied/unoccupied d-orbitals. (Δ(Sp) > Δ(Oct))

I'm probably going about it the wrong way so yes, any help appreciated! Would an MO approach be more useful here rather than CFT?


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