Answer
418.2k+ views
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 $F{e^{3 + }}/F{e^{2 + }}$
Additional information:
In more practical terms we will define ionization power as the minimum electricity that an electron in a gaseous atom or ion has to immerse up to pop out of the influence of the nucleus. It is similarly sometimes known as ionization potential and is usually an endothermic technique.
What we can infer similarly is that ionization strength offers us an idea of the reactivity of chemical compounds. It can also be used to determine the electricity of chemical bonds. It is measured both in devices of electron volts or kilo per joule.
Depending on the ionization of molecules which often ends in changes in molecular geometry, ionization power can be either adiabatic ionization strength or vertical ionization strength of the molecule.
Note:
Ionization is a process that involves the removal of electrons present in an orbit of the atom to outside the atom. As the electron in each orbit it has characteristic energy and the ionization energy is equal to the difference of energy between the energy of the electron in the initial orbit and the energy of the electron present outside the atom.
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 $F{e^{3 + }}/F{e^{2 + }}$
Additional information:
In more practical terms we will define ionization power as the minimum electricity that an electron in a gaseous atom or ion has to immerse up to pop out of the influence of the nucleus. It is similarly sometimes known as ionization potential and is usually an endothermic technique.
What we can infer similarly is that ionization strength offers us an idea of the reactivity of chemical compounds. It can also be used to determine the electricity of chemical bonds. It is measured both in devices of electron volts or kilo per joule.
Depending on the ionization of molecules which often ends in changes in molecular geometry, ionization power can be either adiabatic ionization strength or vertical ionization strength of the molecule.
Note:
Ionization is a process that involves the removal of electrons present in an orbit of the atom to outside the atom. As the electron in each orbit it has characteristic energy and the ionization energy is equal to the difference of energy between the energy of the electron in the initial orbit and the energy of the electron present outside the atom.
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