A magnetic moment of 1.73 B.M. will be shown by one among the following:

Options

A. `TiCl_4`
B. `[CoCl_6]^(4-)`
C. `[Cu(NH_3)_4]^(2+)`
D. `[Ni(CN)_4]^(2-)`

Answer

C

Explanation

`mu_(MM)=1.73B.M.=sqrt(3)BM`
`impliesforn=1(n=`no. of unpaired electrons).
a. In `TiCl_4:Ti(Z=22)implies3d^24s^2,Ti^(4+)=3d^(0)`
It does not have any unpaired electron.
So `mu_(MM)=0` (dimagnetic)
b.In `[Co^(2+)Cl_6^(-6)]^(4-):Co(Z=27)implies3d^74s^2,Co^(2+)=3d^7`
`Cl^(ɵ)` is a weak firld ligand, so pairing does not occur, it has `n=3`.
So `mu_(MM)=sqrt(n(n+2))BM=sqrt(15)BM`
c. In `[Cu^(2+)(Noverset(0)H_3)_4]^(2+),Cu(Z=29)implies3d^(10)4s^(1)impliesCu^(2+)=3d^(9)` It has one unpaired electron, so its `mu_(MM)=1.73BM`