A magnetic moment of 1.73 BM will be shown by one among the folllowing -

Options

A. `[CoCl_(6)]^(4-)`
B. `[Cu(NH_(3))_(4)]^(2+)`
C. `[Ni(CN)_(4)]^(2-)`
D. `TiCl_(4)`

Answer

B

Explanation

The number of unpaired electrons of `Co^(2+), Cu^(2+), Ni^(2+) and Ti^(4+)` are 3, 1, 0 and 0 respectively. Hence the magnetic moment of `[Cu(NH_(3))_(4)]^(2+)` is `mu=sqrt(n(n+2))=sqrt(1(1+2))=1.73BM`.