Question

Consider the ions $F{{e}^{2+}}$, $F{{e}^{3+}}$, $C{{r}^{2+}}$,$C{{r}^{3+}}$, $M{{n}^{2+}}$ and $M{{n}^{3+}}$. Strongest reducing ion is ?
(a) $F{{e}^{2+}}$
(b) $M{{n}^{2+}}$
(c) $C{{r}^{2+}}$
(d) $M{{n}^{3+}}$

Answer 1

Hint: Strongest reducing ion is that which has a strong tendency to lose the electrons. Write the electronic configuration of all these given ions and then, you can easily identify the ion which will easily lose the electron and acts as a reducing ion.

Complete answer:
First of all, let’s discuss what is a reducing agent. By the reducing agent , we mean the tendency of the substance to lose electrons and undergo oxidation.
The substance that acts as a reducing agent itself gets oxidized i.e. loses electrons but reduces the others.
Now considering the statement;
We will discuss the given ions one by one as;
(a) $F{{e}^{2+}}$
Atomic number of $Fe$ is 26.
The ground state electronic configuration of $Fe$ is;${{[Ar]}_{18}}3{{d}^{6}}4{{s}^{2}}$
After the loss of two electrons,
The ground state electronic configuration of $F{{e}^{2+}}$ is; ${{[Ar]}_{18}}3{{d}^{6}}$
It has a strong tendency to lose one electron and attain the stable half-filled electronic configuration.
 (b) $F{{e}^{3+}}$
Atomic number of $Fe$ is 26.
The ground state electronic configuration of $Fe$ is;${{[Ar]}_{18}}3{{d}^{6}}4{{s}^{2}}$
After the loss of three electrons,
The ground state electronic configuration of $F{{e}^{+3}}$ is; ${{[Ar]}_{18}}3{{d}^{5}}$
Since, $F{{e}^{3+}}$has the stable half-filled electronic configuration, therefore, it has less tendency to lose the electron.
(c) $C{{r}^{2+}}$
Atomic number of $Cr$ is 24.
The ground state electronic configuration of $Cr$ is; ${{[Ar]}_{18}}3{{d}^{5}}4{{s}^{1}}$
After the loss of two electrons,
The ground state electronic configuration of $C{{r}^{2+}}$ is; ${{[Ar]}_{18}}3{{d}^{4}}$
Since, $C{{r}^{2+}}$has four unpaired electrons in its valence shell, so it has the tendency to lose the electron.
 (d) $C{{r}^{3+}}$
Atomic number of $Cr$ is 24.
The ground state electronic configuration of $Cr$ is;${{[Ar]}_{18}}3{{d}^{5}}4{{s}^{1}}$
After the loss of three electrons,
The ground state electronic configuration of $C{{r}^{3+}}$is; ${{[Ar]}_{18}}3{{d}^{3}}$
Since, $C{{r}^{3+}}$has three unpaired electrons in its valence shell, so it has the tendency to lose the electron.
(e) $M{{n}^{2+}}$
Atomic number of $Mn$ is 25.
The ground state electronic configuration of is $Mn$;${{[Ar]}_{18}}3{{d}^{5}}4{{s}^{2}}$
After the loss of two electrons,
The ground state electronic configuration of is $M{{n}^{2+}}$;${{[Ar]}_{18}}3{{d}^{5}}$
Since, $M{{n}^{2+}}$has the stable half-filled electronic configuration, therefore, it has less tendency to lose the electron.
(e) $M{{n}^{3+}}$
Atomic number of $Mn$ is 25.
The ground state electronic configuration of is $Mn$;${{[Ar]}_{18}}3{{d}^{5}}4{{s}^{2}}$
After the loss of two electrons,
The ground state electronic configuration of is $M{{n}^{3+}}$;${{[Ar]}_{18}}3{{d}^{4}}$
Since, $M{{n}^{3+}}$ has four unpaired electrons in its valence shell, so it has the tendency to lose the electron.
So, thus, from the above , it is clear that $F{{e}^{2+}}$ has a strong tendency to lose one electron and acts as the strongest reducing ion.

Hence, option (a) is correct.

Note: Don’t get confused in the reducing and the oxidizing agents. The substance which loses the electron acts as a reducing agent whereas on the contrary the substance which gains the electrons acts as an oxidizing agent.