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# Give reasons. ${E^ \circ }$ value for $M{n^{3 + }}/M{n^{2 + }}$ couple is much more positive than that for $F{e^{3 + }}/F{e^{2 + }}$

Last updated date: 18th Jun 2024
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Hint: Ionization energy is easy watchwords and can be defined as a degree of the problem in moving away with an electron from an atom or ion or we can say as the tendency of an atom or ion to surrender an electron to another. The lack of electrons generally occurs in the ground state of the chemical species.

Complete step by step solution:
The relatively high value for $Mn$ shows that $M{n^{2 + }},d^5$ which is particularly stable and much larger than the third ionization energy of $Mn$ where the required change is from$d^5\;to\;d^4$. Because $M{n^{2 + }}$ is more stable than $M{n^{3 + }}$ due to their half-filled $d^5$ configuration whereas $F{e^{2 + }}$ becomes unstable after losing an electron from their half-filled orbital.
Hence, ${E^ \circ }$ value for $M{n^{3 + }}/M{n^{2 + }}$ couple is much more positive than that for $F{e^{3 + }}/F{e^{2 + }}$