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

Options

A. `[Cr(NH_(3))_(6)]^(3+),[Co[H_(2)O)_(6)]^(3)`
B. `[Cr(NH_(3))_(6)]^(3+),[Fe(H_(2)O)_(6)]^(2+)`
C. `[Co(NH_(3))_(6)]^(2+),[Cr(NH_(3))_(6)]^(3+)`
D. `[Mn(H_(2)O)_(6)]^(2+),[Fe(H_(2)O)_(6)]^(2+)`

Answer

C

Explanation

`mu=sqrt(n(n+2))`
No. of unpaired electrons are `:`
`(i)[Cr(NH_(3))_(6)]^(3+)Cr^(3+)(3d^(4)),n=3" "(ii)[Co(NH_(3))_(6)]^(2+)Co^(2+)(3d^(7)),n=3`
`(iii)Fe(H_(2)O)_(6)]^(2+)Fe^(2+)(3d^(7)),n=4" "(iv)[Mn(H_(2)O)_(6)]^(2+)Mn^(2+)(3d^(5)),n=5`
`(v)[Co(H_(2)O)_(6)]^(3+)Co^(3+)(3d^(6)),n=0(d^(2)sp^(3))`
Thus, `Cr^(3+)` and `Co^(2+)` have same number of unpaired electrons. Hence in pair `(C )` , both species will have same magnetic moment.