Question

Explain why iron, cobalt, nickel do not show the expected highest oxidation states of $+ 8,{\text{ }} + 9,{\text{ }}and{\text{ }} + 10$respectively?

Answer 1

Iron, cobalt and nickel are the elements which belong to the d-block series of periodic table and they have electronic configuration involving d-electrons and show the number of oxidation states. These elements and their electronic configuration (outermost) could take to the answer directly. Also, the half-filled and fully filled orbitals are more stable compared to asymmetrical filled orbitals.

Complete step by step answer:
-In the given question, the elements provided are D-block elements which are also called as transition elements because they lie between s and p- block elements. It is a property of transition elements that they have variable oxidation states and they can show different oxidation states because of their d-electronic configuration.
-Also, according to the question, iron, cobalt and nickel do not show a number of oxidation states but they show very limited oxidation states. We will look upon the electronic configuration of these elements which would lead to the answer directly.
-So, the electronic configuration of iron i.e. $Fe$is $\left[ {Ar} \right]3{d^6}4{s^2}$ and in this we can see that it has $8$ electrons in its outermost shell and therefore by losing these two electrons it will show oxidation state of $+ 2$ and also there are $6$ electrons left in d-orbital and it can lose 1 more electron and the configuration will become $\left[ {Ar} \right]3{d^5}$
-So, these two are stable oxidation states and therefore will not show oxidation state of $+ 8$ . We know that half-filled configuration is very stable and d electrons can have $10$ electrons and here it has $5$ which is half of it. So, it will gain stability and therefore will not show any oxidation state further instead of only $+ 2{\text{ }}and{\text{ }} + 3$.
-Similarly, the electronic configuration of Cobalt i.e. $Co$= $\left[ {Ar} \right]3{d^7}4{s^2}$ and in this there are $9$ outermost electrons which could be removed but cobalt only forms $+ 2$ and $+ 3$ oxidation state and does not loses electrons further and therefore, it will not show any oxidation state rather than $+ 2{\text{ }}and{\text{ }} + 3$ because of stability of these oxidation states.
-Electronic configuration of Nickel i.e. $Ni{\text{ }} = {\text{ }}\left[ {Ar} \right]3{d^8}4{s^2}$ and we see that there are $10$ available electrons in the outermost shell and it will only forms the $+ 2$ oxidation state generally and does not form any other oxidation state because there are already 8 electrons in d orbital and to remove these, very high energy will be required and will not be that much stable. So, it will also not show the oxidation state of $+ 10$ .

Note:
The configuration in which half filled or fully filled are highly stable because of symmetrical distribution of the electrons in orbitals and symmetry leads to greater stability. Also, there is exchange energy which is energy released when electrons in different orbitals with different spin exchange their position within the orbital. These were few elements of d-block series and there are other properties also which are quite different in d-block like size, ionization energy, reduction potential.