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Question

(d)$_{25}Mn = 3{d^5}4{s^2}$

After losing two electron electronic configuration will be like this ${(_{25}}M{n^{ + 2 }}3{d^5})$ and this is most sta

Vedclass · Accepted Answer

Manganese $(Z = 25)$

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(d)$_{25}Mn = 3{d^5}4{s^2}$

After losing two electron electronic configuration will be like this ${(_{25}}M{n^{ + 2 }}3{d^5})$ and this is most stable configuration due to half filled orbitals hence third ionization enthalpy will be highest in this case.

Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy

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