the d-d transition of [Ti(H}2O)6 ]^{3 + } produces an absorption maximum at a wavelength of 500 nm
What is the magnitude of delta for [Ti(H2O)6 ]^{3 + } in in kJ/mol?
1 Answer
-
Ti(III) is [Ar]3d^1 and the only d-d transition is t2g → eg which is equivalent to Δoct
E = hν = hc/λ
Planck constant h = 6.626×10^-34 J s
c = speed of light = 2.998 × 10^8 m s-1
λ = 500 nm = 500 × 10^-9 m = 5× 10^-7 m
E = [6.626×10^-34 × 2.998 × 10^8]/(5 × 10^-7) = 3.97× 10^-19 J per photon (I know from experience this is in the correct ball park)
This gives the value in J per photon
To go from J per photon to kJ mol-1 multiply by 6.022 × 10^23 mol^-1 to give J mol-1, and divide by 10^3 to go from J mol-1 to kJ mol-1.
E = (3.97× 10^-19)(6.022 × 10^23)/1000 = 239 kJ mol^-1
cheers
