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)

<p>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.</p> <p>Considering the complexes :</p> <ol> <li>CoCl₃.4NH₃ : This complex has 3 chloride ions, so it&#39;s not the one we are looking for.</li> <br/><li>NiCl₂.6H₂O : This complex has 2 chloride ions, so it&#39;s a potential candidate.</li> <br/><li>PtCl₄.2HCl : This complex has 4 chloride ions, so it&#39;s not the one we are looking for.</li> </ol> <p>Therefore, the complex we are interested in is NiCl₂.6H₂O.</p> $\mathrm{CoCl}_{3} \cdot 4 \mathrm{NH}_{3} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \cdot \mathrm{Cl}_{2}\right]+\mathrm{AgCl}$ <br/><br/> $$ \left.\left.\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \underset{\text { excess }}{\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow}}\right[ \mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+2 \mathrm{AgCl} $$ <br/><br/> $\mathrm{PtCl}_{4} \cdot 2 \mathrm{HCl} \longrightarrow\left[\mathrm{PtCl}_{6}\right]^{4-}+\mathrm{No} ~\mathrm{AgCl} ~\mathrm{ppt}$ <p>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⁺.</p> <p>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.</p> <p>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.</p> <p>The spin-only magnetic moment (μ) can be calculated using the formula :</p> <p>$\mu = \sqrt{n(n+2)} \, \text{B.M.}$</p> <p>where n is the number of unpaired electrons. In this case, n = 2, so</p> <p>$\mu = \sqrt{2 \times (2+2)} = \sqrt{8} \, \text{B.M.}$</p> <p>Rounding to the nearest integer, the spin-only magnetic moment is approximately 3 B.M.</p>