A magnetic moment of 1.73 BM will be shown by one among the following
- TiCl4
- [CoCl6]4-
- [Cu(NH3)4]2+
- [Ni(CN)4]2-
Answer
(3) [Cu(NH3)4]2+
Solution
= n(n+2) BM; = 1.73 BM ⇒ n = 1 unpaired electron.
TiCl4: Ti4+, 3d0 ⇒ n = 0
[CoCl6]4-: Co2+, 3d7 (weak-field Cl-) ⇒ n = 3
[Cu(NH3)4]2+: Cu2+, 3d9 ⇒ n = 1
[Ni(CN)4]2-: Ni2+, 3d8, square planar (dsp2) ⇒ n = 0
∴ [Cu(NH3)4]2+ shows = 1.73 BM.
Answer: (3)