For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be…
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?
(At. no. $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ )
Fe $ > \mathrm{Mn} > \mathrm{Co} > \mathrm{Cr}$
Co $ > \mathrm{Mn} > $ Fe $ > \mathrm{Cr}$
Cr $>$ Mn $>$ Co $>$ Fe
Mn $>$ Fe $>$ Cr $>$ Co
Solution
This can be understood on the basis of $E^{\circ}$ values for $M^{2+} / M$.
$\begin{array}{lrrcr}E^{\circ} V & \mathrm{Cr} & \mathrm{Mn} & \mathrm{Fe} & \text { Co } \\ M^{2+} / M & -0.90 & -1.18 & -0.44 & -0.28\end{array}$
$E^{\circ}$ value for $\mathrm{Mn}$ is more negative than expected from general trend due to extra stability of half-filled $\mathrm{Mn}^{2+}$ ion.
Thus the correct order should be,
$\mathrm{Mn} > \mathrm{Cr} > \mathrm{Fe} > \mathrm{CO}$
an examination of $E^{\circ}$ values for redox couple $M^{3+} / M^{2+}$ shows that $\mathrm{Cr}^{2+}$ is strong reducing $\operatorname{agent}\left(E_{M^{3+}}^{\circ} / M^{2+}=0.41 \mathrm{~V}\right)$ and liberates $\mathrm{H}_2$ from dilute acids.
$2 \mathrm{Cr}^{2+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow 2 \mathrm{Cr}^{3+}(a q)+\mathrm{H}_2 \uparrow(g)$
$\therefore$ The correct order is $\mathrm{Mn} > \mathrm{Fe} > \mathrm{Cr} > \mathrm{Co}$.