Question

Solutions of the `[V(OH_2)_6]^(3+)` ion are green and absorb light of wavelength 560 nm. What is the ligand field splitting in the complex in kJ/mol?

Answer 1

I got `"213.62 kJ/mol"`.

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Well, the HOMO-LUMO gap in transition metal complexes tends to correspond to the `t_(2g)-> e_g` excitation, i.e. the excitation of the `pi`-based `d`-orbital electrons to the `sigma`-based `d` orbitals.

Vanadium as a `+3` oxidation state has two `d` electrons in the octahedral splitting diagram one of which can be excited to the `e_g` orbitals after absorbing a certain amount of energy.

But that IS the octahedral ligand-field splitting energy, `Delta_o`, in the octahedral hexaaquavanadium(III) complex.

So, all you need to do is convert the wavelength to an energy. Recall:

`DeltaE = hnu = (hc)/lambda`

Therefore, we analogously have:

`Delta_o = (hc)/(lambda)`

`= ((6.626xx10^(-34) "J"cdotcancel"s")(2.998xx10^8 cancel"m""/"cancel"s"))/(560 cancel"nm" xx (cancel"1 m")/(10^9 cancel"nm"))`

`= 3.5473 xx 10^(-19) "J"`

Therefore, we need to convert to `"kJ/mol"`, that is, per mol of the complex:

`3.5473 xx 10^(-19) cancel"J" xx "1 kJ"/(1000 cancel"J") xx (6.0221413xx10^(23))/("1 mol complex")`

`=> color(blue)(Delta_o) =` `color(blue)("213.62 kJ/mol")`