Calculate the number of unpaired electrons in the following gaseous ions `Mn^(3+), Cr^(3+), V^(3+)` and `Ti^(3+)`. Which one of these is the most stable in aqueous solution ?

Explanation

The electronic configurations of the given ions are as follow :
`Mn^(3+ (Z = 25) = [Ar] 3d^(4)`
`Cr^(3+) (Z = 24) = [Ar] 3d^(3)`
`V^(3) (Z = 23) = [Ar] 3d^(2)`
`Ti^(3+) (z = 22) = [Ar] 3d^(1)`
`:.` Number of unpaired electrons in `Mn^(+3) , Cr^(3+) V^(3+), Ti^(3+) ` are 4, 3, 2 and 1 respectively.
`Cr^(3+)` is the most stable among these in aqueous solution because it has half - filled `t_(2g) ` level `(- t_(2g)^(3))`