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

Options

A. `|Cu(NH_(3))_(4)|^(2+)` - Square planar
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 electrom.
`:.` Magnetic moment `mu = sqrt(n(n+2))" "BM`
`:.1.73 = sqrt(n(n+2)) or n = 1`
So the complex containing one unpaired electron is `|Cu(NH_(3))_(4)|^(2+)` .