Question

What is the spin only magnetic moment of the metal ion in P and the oxidation number of Sulphur in the oxidised product Z?

• A. \(\mu = 0 \quad\) Oxidation number \(= +4\)
• B. \(\mu = 5.00 \, BM \quad\) Oxidation number \(= 0\)
• C. \(\mu = 5.92 \, BM \quad\) Oxidation number \(= +2\)
• D. \(\mu = 1.732 \, BM \quad\) Oxidation number \(= +6\)

Hint:

Use \(\mu = \sqrt{n(n+2)}\) to calculate magnetic moment. Strong oxidizing agents like \((S_2O_8)^{2-}\) usually oxidize sulphur to its highest oxidation state \(+6\).

Answer 1

The Correct Option is D

Step 1: Understanding the reaction sequence.
The given sequence involves pyrolysis with \(KOH/KNO_3\), followed by disproportionation, and then oxidation using \((S_2O_8)^{2-}\).
This type of sequence is characteristic of manganese chemistry.
Thus, compound \(P\) contains manganese ion.

Step 2: Identifying the species formed (P).
Under strong oxidizing alkaline conditions (\(KOH/KNO_3\)), manganese compounds form manganate ion:
\[ \mathrm{MnO_4^{2-}} \] So, \(P\) contains \(Mn^{6+}\).

Step 3: Disproportionation of manganate.
Manganate ion undergoes disproportionation in acidic medium:
\[ 3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O \] Here,
- \(MnO_4^-\) (purple)
- \(MnO_2\) (blackish brown solid)
Thus, the blackish brown compound \(X\) is \(MnO_2\).

Step 4: Determining unpaired electrons in \(Mn^{6+}\).
Electronic configuration of Mn:
\[ [Ar] 3d^5 4s^2 \] For \(Mn^{6+}\):
\[ 3d^1 \] So, number of unpaired electrons \(n = 1\).

Step 5: Calculating magnetic moment.
Spin only magnetic moment is given by:
\[ \mu = \sqrt{n(n+2)} \, BM \] \[ \mu = \sqrt{1(1+2)} = \sqrt{3} = 1.732 \, BM \]

Step 6: Oxidation of sulphur compound.
The species \(Y\) reacts with peroxodisulphate \((S_2O_8)^{2-}\), which is a strong oxidizing agent.
Sulphur is oxidized to its highest oxidation state.
The final product \(Z\) contains sulphur in oxidation state:
\[ +6 \quad (\text{as in } SO_4^{2-}) \]

Step 7: Final Answer.
Thus, the spin only magnetic moment is:
\[ \mu = 1.732 \, BM \] and oxidation number of sulphur in \(Z\) is:
\[ +6 \] \[ \boxed{\mu = 1.732 \, BM \quad \text{and oxidation number } = +6} \] Hence, the correct answer is option (D).