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Why ${\text{C}}{{\text{r}}^{{\text{2 + }}}}$is a strong reducing agent whereas ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$is not? (Cr=24, Mn=25)

Last updated date: 24th Jun 2024
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Hint: Reducing agents are compounds which in turn undergoes oxidation to reduce other compounds. The oxidised ion must have a stable electronic configuration to be a stronger reducing agent. Oxidation process involves the loss of electrons.

Complete step by step answer: The Cr atom has an atomic number of 24 thus the electronic configuration can be written as
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{\text{5}}}{\text{4}}{{\text{s}}^{\text{1}}}$
So, only after losing two electrons we get ${\text{C}}{{\text{r}}^{{\text{2+ }}}}$ thus the electronic configuration can be written as
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{\text{4}}}$
To be a good reducing agent,${\text{C}}{{\text{r}}^{{\text{2+ }}}}$must get oxidised that is must lose an electron so if it loses an electron the electronic configuration becomes
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{\text{3}}}$
This half-filled electronic configuration of Cr contributes to its stability and thus it acts as a better reducing agent since half filled and full filled electronic configuration contributes to its stability.
In case of Mn atom, it has an atomic number of 25 thus the electronic configuration can be written as
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{\text{5}}}{\text{4}}{{\text{s}}^{\text{2}}}$
So, only after losing two electrons we get${\text{M}}{{\text{n}}^{{\text{2+ }}}}$thus the electronic configuration can be written as
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^5}$
To be a good reducing agent,${\text{M}}{{\text{n}}^{{\text{2 + }}}}$must get oxidised so if it loses an electron the electronic configuration becomes
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^4}$
This electronic configuration of Mn is not stable and thus it does not act as a better reducing agent.

Note: An oxidising agent is a one which undergoes reduction on itself but oxidises other compounds in its process. Oxidation is the process of gain of electrons whereas reduction is where the electrons are lost. Reducing agent reduces other compounds but in turn undergoes oxidation.