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.