Question

Consider the following test for a group-IV cation. \[ \text{M}^{2+} + \text{H}_2\text{S} \rightarrow \text{A (Black precipitate)} + \text{byproduct} \] \[ \text{A + aqua regia} \rightarrow \text{B + NOCl + S + H}_2\text{O} \] \[ \text{B + KNO}_2 + \text{CH}_3\text{COOH} \rightarrow \text{C + byproduct} \] The spin-only magnetic moment value of the metal complex \( \text{C} \) is \( \dots \dots \dots \) BM. (Nearest integer)

Answer 1

Correct Answer: 0

From the reactions, the metal complex formed is likely in the 4$^+$ oxidation state, which leads to no unpaired electrons. For such complexes, the spin-only magnetic moment is zero because all electrons are paired. Hence, the magnetic moment is:
\( \mu = 0 { BM}\)
Thus, the correct answer is (0).