Lesly Cormier
Nov 13, 2024
What is the magnitude of delta for [Ti(H2O)6 ]^{3 + } in in kJ/mol?
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(H_2O)_6]^{3+}) in kJ/mol?
1 Answers
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
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