Number of electrons in 3d-orbital of `V^(2+),Cr^(2+)`,`Mn^(2+)` and `Fe^(2+)` are 3,4,5, and 6 respectively. Which of te following ions will have lgargest value of magnetic moment`(mu)`?

Options

A. `V^(2)`
B. `Cr^(2+)`
C. `Mn^(2+)`
D. `Fe^(2+)`

Answer

C

Explanation

Magnetic moment `(mu)=sqrt(n(n+2))BM` where, 'n' is the number of unpaired electrons.
`._(23)V^(2+)=[Ar]3d^(3)" "(n=3)`
`._(24)Cr^(2+)=[Ar]3d^(4) " "(n=4)`
`._(25)Mn^(2+)=[Ar]3d^(5) " "(n=5)`
`._(26)Fe^(2+)=[Ar]3d^(6) " "(n=4)`
Hence, magnetic moment will be maximum for `Mn^(2+)` (equal to 5.92BM)