$M n^{2 +}$ is much more resistant than $F e^{2 +}$ towards oxidation. Give reason.

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Solution

Electronic configuration of $M n^{2 +}$ is $[A r] 183 d^{5}$ .
Electronic configuration of $F e^{2 +}$ is $[A r] 183 d^{6}$ .

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 $M n^{2 +}$ shows resistance to oxidation to $M n^{3 +}$ . Also, $F e^{2 +}$ has $3 d^{6}$ configuration and by losing one electron, its configuration changes to a more stable $3 d^{5}$ configuration. Therefore, $F e^{2 +}$ easily gets oxidized to $F e^{+ 3}$ oxidation state.

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