The magnetic moment of a transition metal compound has been...

The magnetic moment, ${\mu}_{s} = \sqrt{n ( n + 2 )} BM$ where $n$ is the number of unpaired electrons.

${Cr}^{+ 2}: \left[Ar\right], 3{d}^{4}, 4{s}^{°}n = 4, \mu = \sqrt{4 ( 4 + 2 )} = \sqrt{24} = 4 . 89BM{Mn}^{+ 2}: \left[Ar\right], 3{d}^{5}, 4{s}^{°}n = 5, \mu = \sqrt{5 ( 5 + 2 )} = \sqrt{35} = 5 . 91BM{V}^{+ 2}: \left[Ar\right], 3{d}^{3}, 4{s}^{°}n = 3, \mu = \sqrt{3 ( 3 + 2 )} = \sqrt{15} = 3 . 87BM{Ti}^{+ 2}: \left[Ar\right], 3{d}^{2}, 4{s}^{°}n = 2, \mu = \sqrt{2 ( 2 + 2 )} = \sqrt{8} = 2 . 82BM$