Q. A magnetic moment of 1.73 BM will be shown by one among the following
A) [Cu(NH₃)₄]²⁺
B) [Ni(CN)₄]²⁻
C) TiCl₄
D) [CoCl₆]⁴⁻
Explanation
[Solution from source] 1.73 BM = √(n(n + 2)) with n = 1 unpaired electron. Cu²⁺ (3d⁹) in [Cu(NH₃)₄]²⁺ has one unpaired electron. [Ni(CN)₄]²⁻ is square planar and diamagnetic, TiCl₄ has Ti⁴⁺ (3d⁰), and [CoCl₆]⁴⁻ has high-spin Co²⁺ (3d⁷) with three unpaired electrons. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.