Question

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

Answer 1

Correct Answer - The electron configuration of `Mn^(3+)` is `[Ar]^(18)3d^(4)` so n=4; The electron configuration of `Cr^(3+)` is `[Ar]6(18)3d^(3)` so n=3; The electron configuration of `N^(3+)` is [Ar]^(18)3d^(2) so n=2; The electron configuration of `Ti^(3+)` is `[Ar]^(18)3d^(1)`so n=1. `Cr^(3+)` most stable in aqeous solution because it has half filled `t_(2g)^(3)` energy level of `3d` orbitals in octahedral crystal field spliting and according to crystal field theory (CFT) it has highest value of `CFSE` i.e` -1.2Delta_(@)`.