Chemistry - Complete d-block Question with Solution | TestHub

This can be understood on the basis of $E^{\circ }$ values for $M^{2+}/M$. $\begin{array}{lrrcr}E\%V& \mathrm{Cr}& \mathrm{Mn}& \mathrm{Fe}& \text{ Co }\\ M^{2+}/M& -0.90& -1.18& -0.44& -0.28\end{array}$ $E^{\circ }$ 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 $\mathrm{agent}\left(E_{M^{3+}}^{\circ }/M^{2+}=0.41\mathrm{V}\right)$ and liberates ${\mathrm{H}}_2$ from dilute acids. $$2{\mathrm{Cr}}^{2+}(aq)+2{\mathrm{H}}^{+}(aq)\to 2{\mathrm{Cr}}^{3+}(aq)+{\mathrm{H}}_2\uparrow (g)$$ $\therefore$ The correct order is $\mathrm{Mn}>\mathrm{Fe}>\mathrm{Cr}>\mathrm{Co}$.