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

Options

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

Answer

A

Explanation

In `[Cu(NH_(3))_(4)]^(2+), Cu^(2+)=3d^(9)`, i.e., it contains one unpaired electrons. Magnetic moment.
`therefore 1.73=sqrt(n(n+2))` or n=1
So the complex containing one unpaired one unpaired electron is `[Cu(NH_(3))_(4)]^(2+)`.