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?

1 Answer
Dec 10, 2016

I got #"213.62 kJ/mol"#.


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.

http://i.stack.imgur.com/

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")#