Question

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

Answer 1

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"`).

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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.