Explain why Gd^(3+) is colorless?
1 Answer
Well, as two orbitals get closer together in energy, light emission associated with those orbitals approaches higher wavelengths.
As it turns out, the symmetry-compatible orbitals higher in energy than
Gadolinium,
[Xe] 4f^7 5d^1 6s^2 .
The
[Xe] 4f^7
On NIST, by searching "
As it turns out, the next energy state higher in energy is over
1/(33000 cancel("cm"^(-1))) xx cancel"1 m"/(100 cancel"cm") xx (10^9 "nm")/cancel"1 m"
= color(red)"3030.3 nm"
And this wavelength is far higher than the visible region (it's in the infrared). It cannot be seen, so
[700 cancel"nm" xx cancel"1 m"/(10^9 cancel"nm") xx ("100 cm")/cancel"1 m"]^(-1) ~~ "14286 cm"^(-1)
higher than the ground state, so that the wavelength needed is short enough to see.