1.

A magnetic moment of 1.73 BM will be shown by one among the following :

Answer»

`[Cu(NH_(3))_(4)]^(2+)`
`[Ni(CN)_(4)]^(2-)`
`TiCl_(4)`
`[CoCl_(6)]^(4-)`

SOLUTION :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+)`.


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