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?
(At. no. $Cr=24,Mn=25,Fe=26,Co=27$ )

• A. Fe > Mn > Co > Cr
• B. Co > Mn > Fe > C
• C. Cr > Mn > Co > Fe
• D. Mn > Fe > Cr > Co

Answer 1

Answer: (D)

This can be understood on the basis of $E^{\circ }$ values for $M^{2+}/M$.

$E^{\circ }$ value for $Mn$ is more negative than expected from general trend due to extra stability of half-filled $M{n}^{2+}$ ion.
Thus, the correct order should be
$Mn>Cr>Fe>\text{ Co }$
An examination of $E^{\circ }$ values for redox couple $M^{3+}/M^{2+}$ shows that $C{r}^{2+}$ is strong reducing agent $\left(E_{M^{3+}/M^{2+}}^{\circ }=0.41V\right)$ and liberates $H_2$ from dilute acids.
$2C{r}^{2+}(aq)+2H^{+}(aq)⟶2C{r}^{3+}(aq)+H_2\uparrow (g)$
$\therefore$ The correct order is $Mn>Fe>Cr>Co$.