Which of the following statement are correct about `Cr^(2+)` (Z = 24) and `Mn^(3+)` (Z = 25) ?
(i) `Cr^(2+)` is a reducing agent
(ii) `Mn^(3+)` is an oxidizing agent
(iii) Both `Cr^(2+)` and `Mn^(3+)` exhibit `d^(4)` configuration
(iv) When `Cr^(2+)` is used as a reducing agent, the chromium ion attains `d^(5)` electronic configuration

Options

A. (i), (ii), (iii), (iv)
B. (ii), (iii), (iv)
C. (i), (ii), (iii)
D. (ii), (iii)

Answer

C

Explanation

`Cr(3d^(5) 4s^(1)) rarr Cr^(2+) (3d^(4))`
`Mn(3d^(5) 4s^(2)) rarr Mn^(3+) (3d^(4))`
Thus both have `d^(4)` configuration. `Cr^(2+)` is a reducing agent (i.e. it reduces the other and gets oxiudized) becasuse it gets oxidized to more stable `Cr^(3+)` . During this transformation th econfiguration of chromium changes form `d^(4)` to `d^(3)`, the latter having a half-filled `t_(2g)` level (strong field as well as weak field ligands) is relatively more stable. `Mn^(3+)` is an oxidizing agent (i.e. oxidizes the other and gets redused) because it gets reduced to more stable `Mn^(2+)`. The change form `Mn^(3+)` to `Mn^(2+)` results in the exactly half-filled `(d^(6))` configuration which has extra stability. Moreover, `d^(5)` configuration leads to half-filled `t_(2g)` level and half-filled `e_(g)` level in case of weake field ligands such as `H_(2)O`.