Question

How would you account for the following:
A) Of the `d^(4)` species `Cr^(2+)` is strongly reducing while manganese(III) is strongly oxidizing.
B) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidized.
C) The `D^(1)` configuration is very unstable in ions.

Answer 1

(i) `Cr^(2+)` is strongly reducing in nature. It has a `d^(4)` configuration. While acting as a reducing agent, it gets oxidized to `Cr^(3+)` (electronic configuration, `d^(3)`). This `d^(3)` configuration can be written as `t_(2g)^(3)`configuration, which is a more stable configuration. In the case of `Mn^(3+) (d^(4))`, it acts as an oxidizing agent and gets reduced to `Mn^(2+) (d^(5))`. This has an exactly half-filled d-orbital and is highly stable.
(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the `3^(rd)` ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
(iii) The ions in `d^(1)` configuration tend to lose one more electron to get into stable `d^(0)` configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.