Question
Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their $+3$ state?
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Answer
Electronic configuration of $Mn^{2+}$ is $[Ar]^{18 } \ 3d^5$
Electronic configuration of $Fe^{2+}$ is $[Ar]^{18 }b \ 3d^6$
It is known that half$-$filled and fully$-$filled orbitals are more stable. Therefore, $Mn$ in $(+2)$ state has a stable $d^5$ configuration. This is the reason $Mn^{2+}$ shows resistance to oxidation to $Mn^{3+}$ Also, $Fe^{2+}$ has $3d^6$ configuration and by losing one electron, its configuration changes to a more stable $3d^{5 }$ configuration. Therefore, $Fe^{2+}$ easily gets oxidized to $Fe^{+3}$ oxidation state.
Electronic configuration of $Fe^{2+}$ is $[Ar]^{18 }b \ 3d^6$
It is known that half$-$filled and fully$-$filled orbitals are more stable. Therefore, $Mn$ in $(+2)$ state has a stable $d^5$ configuration. This is the reason $Mn^{2+}$ shows resistance to oxidation to $Mn^{3+}$ Also, $Fe^{2+}$ has $3d^6$ configuration and by losing one electron, its configuration changes to a more stable $3d^{5 }$ configuration. Therefore, $Fe^{2+}$ easily gets oxidized to $Fe^{+3}$ oxidation state.
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