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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.

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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.

Complete answer:
-The distribution of atoms in the atomic orbital is represented with the help of an electronic configuration.
-The orbitals which have half-filled and full - filled are considered as the stable electronic configuration because their reactivity is less in the environment
-Because their ability to accept electrons is less. Also, they will need high energy to release the outermost electron from the shell.
-For example, in manganese the electronic configuration of the element is .
-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 $\text{M}{{\text{n}}^{4+}}\text{ = (Ar) 3}{{\text{d}}^{3}}$ .
-So, +2 oxidation state has the most stable electronic configuration.
-Another example is of the zinc, zinc has the most stable electronic configuration at +2 oxidation state.
Zn = $\text{(Ar) 3}{{\text{d}}^{10}}\text{ 4}{{\text{s}}^{^{2}}}$ and $\text{Z}{{\text{n}}^{2+}}\text{ = (Ar) 3}{{\text{d}}^{10}}$
-So, we can see that +2 oxidation state has a full filled d - orbital so it has the most stable electronic configuration.

Note: Oxidation state is different from the valency because oxidation state tells the no. of the electron which is gained or lost by an atom to form a molecule whereas valency is the no. of electrons that are lost or gained. Oxidation can vary in an atom but the valency remains the same.

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