Question

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of $ + 2$ oxidation state will be there in which of the following order?
A.$Mn > Fe > Cr > Co$
B.$Fe > Mn > Co > Cr$
C.$Co > Mn > Fe > Cr$
D.$Cr > Mn > Co > Fe$

Answer 1

Hint:First of all, we need to write the electronic configuration of the given elements. The stability of an oxidation state depends on the electronic configuration of the element before and after attaining the given oxidation state. If the configuration becomes more stable after attaining the given oxidation state, then the oxidation state is stable.

Complete step by step answer:
The electronic configurations of Cr is:
${}^{24}Cr = [Ar]3{d^5}4{s^1}$
In the first excited state the $4s$ electron is excited and $3{d^5}$ configuration is attained which is a highly stable state due to the presence of half-filled d-orbitals.
In the second excited state, i.e., in $ + 2$ oxidation state, one electron from the d-orbital is lost and $3{d^4}$ configuration is achieved which is less stable than half${}^{26}Fe = [Ar]3{d^6}4{s^2}$-filled orbital. Therefore, $ + 2$ oxidation state is not stable in Cr.
The electronic configuration of Mn is:
${}^{25}Mn = [Ar]3{d^5}4{s^2}$
In the second excited state, that is on losing 2 electrons from its $4s$ subshell, manganese acquires $3{d^5}$ valence shell configuration which is very stable due to the half-filled valence shell.
Therefore, in Mn, $ + 2$ oxidation state is very stable.
The electronic configuration of Fe is
${}^{26}Fe = [Ar]3{d^6}4{s^2}$
The removal of 2 electrons results into $3{d^6}$ configuration which is less stable than ${d^5}$ but more stable than ${d^4}$ configuration, because the addition of one more electron to the half-filled state makes the atom quite unstable but more energy is required to excite an electron from half-filled state. Hence, the latter brings more instability.
The electronic configuration of Co is:
${}^{27}Co = [Ar]3{d^7}4{s^2}$
In $ + 2$ state, the valence shell configuration is $3{d^7}$, which is not stable because the ${d^6}$ configuration is already unstable due to the loss of half-filled state and addition of one more electron makes the atom highly unstable. So, $ + 2$ state in Co is the least stable among these elements.
Therefore, the order of stability is,
$Mn > Fe > Cr > Co$

Hence option B is correct.

Note:
The stability of these elements in $ + 2$ can also be determined on the basis of standard reduction potential. The more negative value of standard reduction potential of $ + 2$ state of these elements indicate that the ions are more stable. Values of standard reduction potentials of $C{r^{ + 2}},M{n^{ + 2}},F{e^{ + 2}}$ and $C{o^{ + 2}}$ are $ - 0.9, - 1.18, - 0.44$ and $ - 0.28$ respectively. Therefore order of stability is, $Mn > Fe > Cr > Co$.