The spin-only magnetic moment value of an octahedral complex among $$CoCl_3 \cdot 4NH_3$$, $$NiCl_2 \cdot 6H_2O$$ and $$PtCl_4 \cdot 2HCl$$, which upon reaction with excess of $$AgNO_3$$ gives 2 moles of AgCl is ______ B.M. (Nearest Integer)

We need to identify which complex gives 2 moles of AgCl with excess $$AgNO_3$$, then find its spin-only magnetic moment.

Write the coordination formulas and count ionizable chloride ions.

$$CoCl_3 \cdot 4NH_3$$ = $$[Co(NH_3)_4Cl_2]Cl$$ → 1 ionizable $$Cl^-$$ → 1 mol AgCl

$$NiCl_2 \cdot 6H_2O$$ = $$[Ni(H_2O)_6]Cl_2$$ → 2 ionizable $$Cl^-$$ → 2 mol AgCl

$$PtCl_4 \cdot 2HCl$$ = $$H_2[PtCl_6]$$ → 0 ionizable $$Cl^-$$ → 0 mol AgCl

So the complex that gives 2 moles of AgCl is $$[Ni(H_2O)_6]Cl_2$$.

Find the electronic configuration of $$Ni^{2+}$$.

Ni: [Ar] 3d$$^8$$ 4s$$^2$$

$$Ni^{2+}$$: [Ar] 3d$$^8$$ (2 unpaired electrons in octahedral field with weak ligand $$H_2O$$)

Calculate the spin-only magnetic moment.

$$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ B.M.}$$

Rounding to the nearest integer: $$\mu \approx \textbf{3}$$ B.M.