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

Options

A. `[Cr(H_(2)O)_(6)]^(2+)[ CoCI_(4)]^(2-)`
B. `[Cr(H_(2)O)_(6)]^(2+)[ Fe(H_(2)O)_(6)]^(2-)`
C. `[Mn(H_(2)O)_(6)]^(2+)[Cr(H_(2)O)_(6)]^(2-)`
D. `[CoCI_(4)]^(2-),[Fe(H_(2)O)_(6)]^(2-)`

Answer

b

Explanation

Spin only magnetic moment` mu=sqrt(n(n +2))`
where n= number of unpaied electrons. Number of unpaied in `Cr^(2+)([Ar] 3d^(4))` is 4 in `Co^(2+) ([Ar]rd^(1))` is 3. in `Fe^(2+)([Ar]3d^(6)` is 4. in `Mn^(2+)([Ar]3d^(6))`is 5.
As the numbe of unpaired elctrons in `Cr^(2+) and Fe^(2+)` are same, hence, `[Cr(H_(2)O)_(6)]^(2+) and [Fe(H_(2)O)_(6)]^(2+)` will have same magnetic moment.