Explain why #Gd^(3+)# is colorless?

1 Answer
Dec 30, 2017

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 #4f# electrons that start the process of absorption #-># emission are not far enough away in energy, so the wavelength of light associated with it is longer than the visible region (#400 - "700 nm"#).


Gadolinium, #"Gd"#, has the electron configuration

#[Xe] 4f^7 5d^1 6s^2#.

The #6s# and #5d# are higher in energy than the #4f#, so those electrons are typically lost first in an ionization. As a result, #"Gd"^(3+)# has:

#[Xe] 4f^7#

On NIST, by searching "#"Gd IV"#", one can find the energy states of #"Gd"^(3+)#.

As it turns out, the next energy state higher in energy is over #"33000 cm"^(-1)# higher, so the wavelength of light needed for the transition is 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 #"Gd"^(3+)# is colorless. In order to involve violet light (the color of the highest wavelength), one would need an energy state

#[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.