Question
Question: The colour of \[_{62}S{m^{3+}}\] is yellow. The colour of \[_{66}D{y^{3+}}\] is: A. Colourless B...
The colour of 62Sm3+ is yellow. The colour of 66Dy3+ is:
A. Colourless
B. yellow
C. red
D. blue
Solution
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. (n − 2) f of the orbital. The General electronic configuration of elements is (n−2) f1−14(n−1) d0−1ns2.
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 filledforbitals. it absorbs certain wavelengths from the visible region of the spectrum. This results in transitions from one4f 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 Sm3+ electronic configuration is [Xe]4f5
Sm3+ has 5 unpaired electrons and it shows yellow colour.
Dy3+ electronic configuration is [Xe]4f9
Dy3+ 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