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

Magnetic moment `mu ` is is related with number of unpaired electrons as
`mu = sqrt(n(n+2))BM `
`(1.73)^(2) = n(n+2)`
On solving , n=1
thus the complex / compound having one unpaired electron exhibit a magnetic moment of `1.73 ` BM
in `[Cu (NH_(3))_(4) ]^(2+)`
` Cu^(2+)= [Ar] 3d^(9)` =