Question

(a) Complete the following equations :

(i) $C{r}_2{O}_7^{2-}+2O{H}^{-}\to$

(ii) $Mn{O}_4^{-}+4H^{+}+3e^{-}\to$

(b) Account for the following :

(i) Zn is not considered as a transition element.

(ii) Transition metals form a large number of complexes.

(iii) The $E^0$ value for the $M{n}^{3+}/M{n}^{2+}$ couple is much more positive than that for $C{r}^{3+}/C{r}^{2+}$ couple.

Answer 1

(a) (i) $C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$

(ii) $Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$

(b) (i) Because Zn does not have partly filled d-subshell in its atomic state $\left(3d^{10}4s^2\right)$ or in its common oxidation state i.e., $Z{n}^{2+}\left(3d^{10}\right)$.

(ii) It is because of smaller size of their metal ions, their high ionic charges and availability of vacant d-orbitals.

(iii) It is because $M{n}^{2+}$ is more stable than $M{n}^{3+}$ as $M{n}^{2+}$ has stable configuration $3d^5$. So $I{E}_3$ of Mn is very high. On the other hand $C{r}^{2+}$ has configuration $3d^4$. It can easily lose electron to form $C{r}^{3+}$ due to stable $3d^3$ configuration $\left(t_{{2g}^3}\right)$. Thus, $E^o$ value of $M{n}^{3+}/M{n}^{2+}$ is more positive than that for $C{r}^{3+}/C{r}^{2+}$ couple.