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 related with number of unpaired electrons as
`mu = sqrt(n(n+2))` BM
`(1.73)^2=n(n+2)`
on sloving n=1
Thus, the complex/compound havmg one unpaired electron exhibit a magnetic moment of 1.73 BM.
`In [Cu(NH_(3))_(4)]^(2+)`
`Cu^(2+)=[Ar] 3d^9`

(Although in the presence of strong field ligand `NH_3` , the unpaired electron gets excited to higher energy level but it still remams unpaired).
(b) `In [Ni (CN)_(4)]^2`
`Ni^(2+) = [Ar] 3d^8`

But `CN^-` bemg strong field hgand pair up the unpaired electrons and hence m this complex, number of unpaired electrons= 0.
In `[TiCl_(4)]`
`Ti^(4+) = [ Ar]`
no unpaired electron
In `[COCl_(6)]^(4-)]`
`Co^(2+)= [ Ar ]3d^7`

It contains three unpaired electrons.
Thus `[Cu(NH_(3))_4]^(2+)` is the complex that exhibits a magnetic moment 173 BM