# 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 **not far enough away in energy**, so the wavelength of light associated with it is **longer** than the visible region (

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 **somewhat smaller than**:

#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.