JEE Main 2022 — Coordination Compounds Question with Solution

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.

${\mathrm{CoCl}}_3\cdot 4{\mathrm{NH}}_3\underset{\text{ excess }}{\stackrel{{\mathrm{AgNO}}_3}{⟶}}\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\cdot {\mathrm{Cl}}_2\right]+\mathrm{AgCl}$

$${{\mathrm{NiCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}\underset{\text{ excess }}{\stackrel{{\mathrm{AgNO}}_3}{⟶}}[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6]}^{2+}+2\mathrm{AgCl}$$

${\mathrm{PtCl}}_4\cdot 2\mathrm{HCl}⟶{\left[{\mathrm{PtCl}}_6\right]}^{4-}+\mathrm{No}\mathrm{AgCl}\mathrm{ppt}$

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.