JEE Chemistry 2024 Question with Solution

Given: A group-IV cation $M^{2+}$ gives a black precipitate $A$ with $H_2S$. Then:

$$M^{2+} + H_2S \rightarrow A + \text{byproduct}$$ $$A + \text{aqua regia} \rightarrow B + \text{NOCl} + S + H_2O$$ $$B + \text{KNO}_2 + \text{CH}_3\text{COOH} \rightarrow C + \text{byproduct}$$

Find: The spin-only magnetic moment of complex $C$ in BM.

Identify the cation: The black precipitate formed with $H_2S$ is taken as $\text{PbS}$, so $M^{2+} = \text{Pb}^{2+}$.

Apply the reaction sequence: On treatment with aqua regia, $A$ forms $B$ with $\text{NOCl}$, sulfur, and water. The solution identifies $B$ as $\text{Pb(NO}_3\text{)}_2$.

Then $B$ reacts with $\text{KNO}_2$ and $\text{CH}_3\text{COOH}$ to form complex $C$.

Magnetic moment calculation: The spin-only magnetic moment is

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

where $n$ is the number of unpaired electrons.

For lead in the relevant state used in the solution, all electrons are paired, so

$$n = 0$$

Hence,

$$\mu = \sqrt{0(0+2)} = 0 \, \text{BM}$$

Conclude: Therefore, the spin-only magnetic moment value of the metal complex $C$ is $0$.

Given: The reaction sequence leads to a lead-based complex $C$ according to the provided solution.

Find: The spin-only magnetic moment.

Use the direct relation:

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

If the complex has no unpaired electrons, then immediately

$$n = 0 \Rightarrow \mu = 0 \, \text{BM}$$

Conclude: The required nearest integer value is $0$.