Correct Answer – D
This can be understood on the basis of `E^(@)` values for `M^(2+)//M`
`E^(@)//V ” ” Cr ” ” Mn ” ” Fe” ” Co`
`M^(2+)//M ” ” -0.90 ” “-1.18 ” ” -0.44 ” ” -0.28`
`E^(@)` value for Mn is more negative than expected from general trend due to extra stability of half-filled electronic configuration of `Mn^(2+)` ion.
Thus, the correct order should be
`Mn gt Cr gt Fe gt Co`
An examination of `E^(@)` values for redox couple `M^(3+)//M^(2+)` shows that `Cr^(2+)` is strong reducing agent `(E^(@))_(M^(3+)//M^(2+))=0.41 V` and liberates `H_(2)` from dilute acids.
`2Cr^(2+)(aq) + 2H^(+) (aq) to 2Cr^(3+)(aq) + H_(2) uarr (g)`
`therefore` The correct order is `Mn gt Fe gt Cr gt Co`.