Question

The lanthanide ion that would show colour is:

{\mathbf{A}} - S{m^{3 + }} \\\ {\mathbf{B}} - L{a^{3 + }} \\\ {\mathbf{C}} - L{u^{3 + }} \\\ {\mathbf{D}} - G{d^{3 + }} \\\ \end{gathered} $$

Answer 1

The reason for colour in any metal compounds/ion is presence of free unpaired electrons in metal atom/ion. Check the number of unpaired electrons in each of the given lanthanide ions and decide the colour.

Complete answer:
The Lanthanides consist of the elements in the f-block of period six in the periodic table. While these metals can be considered as transition metals, because they have properties that set them apart from the rest of the elements. The Lanthanides elements were first classified as ‘rare earth elements’ due to the fact that they were obtained as reasonably rare minerals. However, this can be misleading since the Lanthanide elements have a practically unlimited abundance. The term Lanthanides was taken, originating from the first element of the series i.e. Lanthanum.
Lanthanides are different from the main group elements in the fact that they have electrons in the f orbital. After Lanthanum, the energy of the 4f subshell falls below that of the 5d subshell. This means that the electron starts to fill the 4f subshell before the 5d subshell. Let’s check the unpaired electrons in all the options provided.

Lanthanide ion	Valence shell electronic configuration
$S{m^{ + 3}}$	$4{f^5}$
$L{a^{ + 3}}$$$$$	$5{d^0}$
$L{u^{ + 3}}$	$4{f^{14}}$
$G{d^{ + 3}}$	$4{f^7}$

So, the unpaired electron is present in only $S{m^{ + 3}}$i.e. ($4{f^5}$)
Therefore, the lanthanide ion that would show colour is $S{m^{ + 3}}$

Hence, the correct option is (A) Sm3+ .

Note:
Samarium with 3 positive charges with $S{m^{ + 3}}$is colored, and forms a colored compound. This is because of the 5 unpaired electrons present in it with the configuration=$4{f^5}$ , and in this state the color it imparts is yellow.