From the given information, we are looking for a complex which, upon reaction with excess of AgNO₃, gives 2 moles of AgCl. This implies that the complex has 2 chloride ions involved.

Considering the complexes :

1. CoCl₃.4NH₃ : This complex has 3 chloride ions, so it's not the one we are looking for.


2. NiCl₂.6H₂O : This complex has 2 chloride ions, so it's a potential candidate.


3. PtCl₄.2HCl : This complex has 4 chloride ions, so it's not the one we are looking for.

Therefore, the complex we are interested in is NiCl₂.6H₂O.

$\mathrm{CoCl}_{3} \cdot 4 \mathrm{NH}_{3} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \cdot \mathrm{Cl}_{2}\right]+\mathrm{AgCl}$

$$ \left.\left.\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\right[ \mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+2 \mathrm{AgCl} $$

$\mathrm{PtCl}_{4} \cdot 2 \mathrm{HCl} \longrightarrow\left[\mathrm{PtCl}_{6}\right]^{4-}+\mathrm{No} ~\mathrm{AgCl} ~\mathrm{ppt}$

The next step is to find the oxidation state of the nickel ion. The nickel ion must have a charge of +2 to balance the -2 charge from the two chloride ions, thus it is Ni²⁺.

In the case of Ni²⁺, the electron configuration is [Ar]3d⁸. For an octahedral complex, the d-orbitals split into two sets under the influence of ligands: the $e_g$ set which includes d(x²-y²) and d(z²) orbitals, and the $t_{2g}$ set which includes the d(xy), d(xz), and d(yz) orbitals. Electrons will occupy the lower energy $t_{2g}$ orbitals first.

The 3d⁸ electron configuration implies there are 8 electrons in the 3d orbitals. The first six electrons pair up in the three $t_{2g}$ orbitals, and the next two electrons will go into the two $e_g$ orbitals, with each one having one unpaired electron.

The spin-only magnetic moment (μ) can be calculated using the formula :

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

where n is the number of unpaired electrons. In this case, n = 2, so

$$ \mu = \sqrt{2 \times (2+2)} = \sqrt{8} \, \text{B.M.} $$

Rounding to the nearest integer, the spin-only magnetic moment is approximately 3 B.M.