Question

Which of the following ions is colourless in solution?
A.${V^{3 + }}$
B.$C{r^{3 + }}$
C.$C{o^{2 + }}$
D.$S{c^{3 + }}$

Answer 1

These ions belong to $d -$ elements. So their coloring property can be explained by $d$ to $d$ transitions. If there is transition then only they will show coloured property otherwise they will remain colorless.

Complete step by step answer:
$d -$ block elements: The elements which are in $d -$ group i.e. group from $3$ to $12$, are known as $d -$ block elements. For example: scandium $(Sc)$, cobalt $(Co)$, chromium $(Cr)$, etc.
Ions: Ions are those atoms which are formed after losing or gaining the electrons. They are of two types: positive charge ion(i.e. cation) and negative charge ion(i.e. anion)
Cations: The ions formed after losing one or more electrons, are known as cations.
Anions: The ions formed after gaining of one or more electrons, are known as anions.
Generally cations and anions are formed because every atom wants to complete its octet. If after gaining or losing one, two or three electrons they can achieve the complete filled octet then only they will lose or gain.
In $d -$ block elements the ions will be coloured if they show $d - d$ transition.
$d - d$ transition: When electrons move from one $d -$ orbital to another $d -$ orbital, then this movement is known as $d - d$ transition.
The atomic number of $Sc$ is $21$. So its electronic configuration will be $Ar3{d^1}4{s^2}$. Here the symbol of $Ar$ represents the electronic configuration of noble gas Argon. After losing three electrons it will become $S{c^{3 + }}$ with electronic configuration as $Ar$. And the anion will not have $d -$ electrons. Since no $d -$ electrons are present so there will be no $d - d$ transitions. Hence it will be colorless.
The atomic number of $Co$ is $27$. So its electronic configuration will be $Ar3{d^7}4{s^2}$. Here the symbol of $Ar$ represents the electronic configuration of noble gas Argon. After losing two electrons it will become $C{o^{2 + }}$ with electronic configuration as $Ar3{d^7}4{s^0}$. And the anion will have five $d -$ electrons. Since $d -$ electrons are present so there will be $d - d$ transitions. Hence it will be coloured.
The atomic number of $Cr$ is $24$. So its electronic configuration will be $Ar3{d^4}4{s^2}$. Here the symbol of $Ar$ represents the electronic configuration of noble gas Argon. After losing three electrons it will become $C{r^{3 + }}$ with electronic configuration as $Ar3{d^3}4{s^0}$. And the anion will have one$d -$ electron. Since $d -$ electron are present so there will be $d - d$ transitions. Hence it will be coloured.
The atomic number of $V$ is $23$. So its electronic configuration will be $Ar3{d^3}4{s^2}$. Here the symbol of $Ar$ represents the electronic configuration of noble gas Argon. After losing three electrons it will become ${V^{3 + }}$ with electronic configuration as $Ar3{d^2}$. And the anion will have two $d -$ electrons. Since $d -$ electron are present so there will be $d - d$transitions. Hence it will be coloured.
Hence, the correct answer is option D.

Note:
The ions will be colored if they have unpaired $d -$ electrons. For example: if a paired number of $d -$electrons are present then there is no possibility for the movement of electrons. Hence they will not be colored.