The spin-only magnetic moment value of an octahedral complex among CoCl₃.4NH₃, NiCl₂.6H₂O and PtCl₄.2HCl, which upon reaction with excess of AgNO₃ gives 2 moles of AgCl is ___________ B.M. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

First let AgNO₃ tell you which complex it is, then count unpaired electrons on that metal.

1. What AgCl counts. Ag⁺ precipitates only the chloride ions outside the coordination sphere. A chloride bonded to the metal stays put. So each compound is written as an octahedral (6-coordinate) complex, and you count the counter-ion chlorides.

   Compound	Octahedral complex	Free Cl⁻ per formula	mol AgCl
   CoCl₃.4NH₃	[Co(NH₃)₄Cl₂]Cl	1	1
   NiCl₂.6H₂O	[Ni(H₂O)₆]Cl₂	2	2
   PtCl₄.2HCl	H₂[PtCl₆]	0	0

2. Unpaired electrons. In [Ni(H₂O)₆]²⁺, nickel is Ni²⁺, a d⁸ ion. In an octahedral field this fills as t_(2g)⁶ e_g², and the two e_g electrons stay unpaired whether the ligand is weak or strong field. So n = 2.

   μ = √n(n+2) = √(2 × 4) = √8 ≈ 2.83 B.M.

3. So. The spin-only magnetic moment is 2.83 B.M., which rounds to 3.

The common slip: 0. Picking CoCl₃.4NH₃, where low-spin d⁶ Co³⁺ is diamagnetic, gives 0. But [Co(NH₃)₄Cl₂]Cl has only one free chloride, so it gives just 1 mol of AgCl.

The idea to keep. Moles of AgCl count only the chlorides outside the brackets. And an octahedral d⁸ ion always has 2 unpaired electrons, giving √8 ≈ 2.83 B.M.

Answer from the NTA final answer key. Solution by PrepWiser.