Question

With `3d^(4)` configuration, `Cr^(2+)` acts as a reducing agent but `Mn^(3+)` acts as an oxidising agent. Explain.

Answer 1

`Cr^(2+)(3d^(4)) to Cr^(3+)(3d^(3))+e^(-),Mn^(3+)(3d^(4))+e^(-) to Mn^(2+)(3d^(5))`
`Cr^(2+)` acts as a reducing agent due to greate stability of `Cr^(3+)` with exactly half-filled `t_(2g)` level. `Mn^(3+)` acts as an oxidising agent due to extra stability of `Mn^(2+)` ion with all half filled d-orbitals .