$$M^{2+} + H_2S \to A \text{ (Black precipitate)} + \text{by product}$$
$$A + \text{aqua regia} \to B + NOCl + S + H_2O$$
$$B + KNO_2 + CH_3COOH \to C + \text{by product}$$
Consider the following test for a group-IV cation. The spin-only magnetic moment value of the metal complex C is ______ BM (Nearest integer)

• Identification of Cation ($$\text{M}^{2+}$$): Among group-IV cations ($$\text{Mn}^{2+}, \text{Zn}^{2+}, \text{Co}^{2+}, \text{Ni}^{2+}$$), only $$\text{Co}^{2+}$$ forms a black precipitate ($$\text{A} = \text{CoS}$$) that dissolves in aqua regia to yield $$\text{B} = \text{CoCl}_2$$, which then reacts with $$\text{KNO}_2$$ in acetic acid to give a brown precipitate.

• Complex Formation (C): The brown precipitate is potassium hexanitritocobaltate(III):
  $$\text{C} = \text{K}_3[\text{Co}(\text{NO}_2)_6]$$
• Magnetic Moment Calculation:
  • Oxidation state of Cobalt in the complex is $$+3$$.
  • Electronic configuration: $$\text{Co}^{3+} = [\text{Ar}] 3d^6$$.
  • Since $$\text{NO}_2^-$$ is a strong-field ligand, it causes complete pairing of electrons in the octahedral field ($$t_{2g}^6 e_g^0$$).
  • Number of unpaired electrons ($$n$$) = 0.
  • $$\text{Spin-only magnetic moment } (\mu) = \sqrt{n(n+2)} = 0\text{ BM}$$

Conclusion:

Because all six $$3d$$ electrons are fully paired up by the strong-field nitrito ligands, the complex is completely diamagnetic.

Answer: 0