Question

How would you account for the following?
(i) With the same d-orbital configuration $\left(d^4\right)C{r}^{2+}$ is reducing agent while $M{n}^{3+}$ is an oxidizing agent.
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

Answer 1

I) $C{r}^{2+}$ is strongly reducing in nature. It has a $d^4$ configuration. While acting as a reducing agent, it gets oxidized to $C{r}^{+3}$ (electronic configuration, $d^3$). This $d^3$ configuration can be written as ${t^3}_{2g}$ configuration, which is a more stable configuration.

In the case of $M{n}^{+3}(d^4$), it acts as an oxidizing agent and gets reduced to $M{n}^{2+}\left(d^5\right)$. This has an exactly half-filled d-orbital and has an extra-stability.

(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid seriesbecause they are bigger, the outermost shell is further away from the nucleus, that means the ionization energy is lower for the corresponding oxidation state compared to lanthanides.

III) Due to partial absorption of visible light, the electron from one orbital gets promoted to another orbital of the d-subshell. Due to presence of unpaired electron transition metals are coloured.