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

Options

A. `[CoCl_(6)]^(4-)`
B. `[Cu(NH_(3))_(4)]^(2+)`
C. `[Ni(CN)_(4)]^(2-)`
D. `TiCl_(4)`

Answer

B

Explanation

Magnetic moment 1.73 BM
`mu=sqrt(n(n+2))B.M`
n=no. of unpaired `e^(-)`
`mu=1.73rArr1.73=sqrt(n(n+2))B.MrArrn=1`
`**[CoCl_(6)]^(4-)rarrCo^(+2),d^(7)`
`Cl^(-)` (weak field ligand) `t_(2)g^(5)eg^(2)`, unpaired `e^(-)=2`
* `[Cu(NH_(3))_(4)]^(2+)Cu^(+2)-d^(9)`
`NH_(3)` Strong field ligand, `dsp^(2)` hybridization
*one `e^(-)` of 3d jumps into 4p subshell, unpaired `e^(-)=1`
`**[Ni(CN)_(4)]^(2-)rarr Ni^(+2)-d^(8)`, unpaired `e^(-)=0`
`CN^(-)rarr` strong field ligand, `dsp^(2)` hybridisation
`**TiCl_(4)rarr Ti^(+4), d^(@)` hybridized, unpaired `e^(-)`= zero.