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

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

Gadolinium, $\text{Gd}$, has the electron configuration

$\left[X e\right] 4 {f}^{7} 5 {d}^{1} 6 {s}^{2}$.

The $6 s$ and $5 d$ are higher in energy than the $4 f$, so those electrons are typically lost first in an ionization. As a result, ${\text{Gd}}^{3 +}$ has:

$\left[X e\right] 4 {f}^{7}$

On NIST, by searching "$\text{Gd IV}$", one can find the energy states of ${\text{Gd}}^{3 +}$.

As it turns out, the next energy state higher in energy is over ${\text{33000 cm}}^{- 1}$ 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"

$=$ $\textcolor{red}{\text{3030.3 nm}}$

And this wavelength is far higher than the visible region (it's in the infrared). It cannot be seen, so ${\text{Gd}}^{3 +}$ 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.