Question

The colour of \[_{62}S{m^{3+}}\] is yellow. The colour of \[_{66}D{y^{3+}}\] is:
A. Colourless
B. yellow
C. red
D. blue

Answer 1

Hint:Colour in transition-series metal compounds is due to electronic transitions of two types:
1. Charge transfer transitions: An electron may jump from a predominantly ligand orbital to a predominantly metal orbital
2. F-F transitions: An electron jumps from one f-orbital to another.

Complete step by step solution:
Given elements are ‘f’ – block elements
 elements are also known by the name of inner transition elements. In these elements, the last electron usually enters the penultimate i.e. \[\left( {n{\text{ }}-{\text{ }}2} \right){\text{ }}f\;\] of the orbital. The General electronic configuration of elements is \[\left( {n-2} \right){\text{ }}{f^{1-14}}\left( {n-1} \right){\text{ }}{d^{0-1}}n{s^2}.\]
1. Colour in transition-series due to charge transfer transitions: An electron may jump from a predominantly ligand orbital to a predominantly metal orbital , giving rise to a ligand-to-metal charge-transfer (LMCT) transition
2. \[f - f\] transitions: The colour of lanthanide ions is due to the presence of partly filled\[\;f\]orbitals. it absorbs certain wavelengths from the visible region of the spectrum. This results in transitions from one\[\;4f\] orbital to another \[\;4f\] orbital known as \[f - f\] transition. Hence, when these electrons on losing energy reach towards lower energy then excess energy is released in the form of visible light. This visible light is responsible for imparting color to lanthanide compounds.
In \[S{m^{3+}}\] electronic configuration is \[\left[ {Xe} \right]4{f^5}\;\]
\[S{m^{3+}}\] has 5 unpaired electrons and it shows yellow colour.
\[D{y^{3+}}\] electronic configuration is \[\left[ {Xe} \right]4{f^9}\;\]
\[D{y^{3+}}\] also has 5 unpaired electrons and it will also show yellow colour.

So correct option is (B) yellow

Note:
The colour of lanthanide compounds are almost derived from charge transfer interactions between the metal and the ligand. Lanthanide loose their electron from outermost s and d electrons that’s why its oxidation state is \[+3\]