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

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).

Answer 2

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).