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# To what extent do the electronic configuration decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with an example.

Last updated date: 11th Jun 2024
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Hint: Oxidation state can be calculated by calculating the change in the ion form reactant to the product side. With the help of electronic configuration, one can easily determine the capacity of the element of losing or accept the pair of electrons and then the oxidation state of the element.

-Here, manganese has the oxidation state of +2, +3 and +4. Among them, the +2 oxidation state is the most stable because the d - orbital is half-filled i.e. $\text{M}{{\text{n}}^{2+}}\text{ = (Ar) 3}{{\text{d}}^{5}}$, $\text{M}{{\text{n}}^{3+}}\text{ = (Ar) 3}{{\text{d}}^{4}}$ and $\text{M}{{\text{n}}^{4+}}\text{ = (Ar) 3}{{\text{d}}^{3}}$ .
Zn = $\text{(Ar) 3}{{\text{d}}^{10}}\text{ 4}{{\text{s}}^{^{2}}}$ and $\text{Z}{{\text{n}}^{2+}}\text{ = (Ar) 3}{{\text{d}}^{10}}$