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A magnetic moment of 1.73BM will be show...

A magnetic moment of `1.73BM` will be shown by one amongst the following

A

`[Ni(CN)_(4)]^(2-)`

B

`TiCl_(4)`

C

`[CoCl_(6)]^(4-)`

D

`[Cu(NH_(3))_(4)]^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
D
`[Cu(NH_(3))_(4)]^(2+)` hybridisation `dsp^(2)`
`Cu^(+2)-3d^(9)` has one unpaired `e^(-)`
So magnetic moment `mu=sqrt(n(n+2))=sqrt(3)=`
`sqrt(3)=1.73`
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