The pair in which both species have the same magnetic moment (spin only value) is

Options

A. `[Cr(H_2O)_6]^(2+) , [CoCl_4]^(2-)`
B. `[Cr(H_2O)_6]^(2+) , [Fe(H_2O)_6]^(2+)`
C. `[Mn(H_2O)_6]^(2+) , [Cr(H_2O)_6]^(2+)`
D. `[CoCl_4]^(2-) , [Fe(H_2O)_6]^(2+)`

Answer

B

Explanation

`[Cr(H_2O)_6]^(2+)=Cr^(2+) , 3d^4` = four unpaired electrons
`[Fe(H_2O)_6]^(2+)= Fe^(2+) =3d^6` , Four unpaired electrons.
Hence, both the species have same magnetic moment.