$ M n^{2+} $ compounds are more stable than $ F e^{2+} $ compounds towards oxidation to their $ +3 $ state, because

Question

$ M n^{2+} $ compounds are more stable than $ F e^{2+} $ compounds towards oxidation to their $ +3 $ state, because, $ M n^{2+} $ is more stable with high $ 3 r d $ lonisation energy., $ M n^{2+} $ is bigger in size., $ M n^{2+} $ has completely filled d-orbitals., $ M n^{2+} $ does not exist.

MARKS App · Accepted Answer

Electronic configuration of $ M n^{2+} $ is $ 4 s^{0} 3 d^{5} $ : \begin{array}{|l|l|l|l|l|} \hline $ 1 $ & $ 1 $ & $ 1 $ & $ 1 $ & $ 1 $ \ \hline \end{array} Electronic configuration of $ F e^{2+} $ is $ 4 s^{0} 3 d^{6}: $ \begin{array}{|ll|l|l|l|l|} \hline $ 1 $ & $ \downarrow $ & $ 1 $ & $ 1 $ & $ 1 $ & $ 1 $ \ \hline \end{array} Since, in $ M n^{2+} $ the d-orbital is completely half-filled, therefore, its third ionization energy is high and hence, it stable towards oxidation to $ +3 $ oxidation states, whereas, in $ \mathrm{Fe}^{2+} $, the d-orbital is partially filled and only one electron is paired, therefore, its third ionization energy is lower than $ \mathrm{Mn}^{2+} $ ion and hence, it is less stable towards $ +3 $ oxidation states.

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