Question

The difference in the 'spin-only' magnetic moment values of \( \text{KMnO}_4 \) and the manganese product formed during titration of \( \text{KMnO}_4 \) against oxalic acid in acidic medium is ______ BM (nearest integer).

Answer 1

Correct Answer: 6

Spin-only magnetic moment of Mn in KMnO$_4$:
\[\mu = 0\]
Spin-only magnetic moment of the manganese product formed during titration of KMnO$_4$ against oxalic acid in acidic medium:
\[\mu = 6 \, \text{BM}\]
Difference in magnetic moments:
\[6 - 0 = 6 \, \text{BM}\]

Answer 2

The problem asks for the difference between the 'spin-only' magnetic moment of manganese in \( \text{KMnO}_4 \) and in the product formed when \( \text{KMnO}_4 \) is titrated against oxalic acid in an acidic medium.

Concept Used:

The 'spin-only' magnetic moment (\(\mu\)) is calculated using the formula:

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

where \(n\) is the number of unpaired electrons in the d-orbitals of the metal ion. Bohr Magneton (BM) is the unit of magnetic moment.

To solve the problem, we need to:

1. Determine the oxidation state and the number of unpaired electrons for manganese in \( \text{KMnO}_4 \).
2. Identify the manganese-containing product of the redox reaction between \( \text{KMnO}_4 \) and oxalic acid in an acidic medium.
3. Determine the oxidation state and the number of unpaired electrons for manganese in the product.
4. Calculate the magnetic moment for both species and find the difference.

Step-by-Step Solution:

Step 1: Calculate the magnetic moment of Mn in \( \text{KMnO}_4 \).

First, we find the oxidation state of Mn in \( \text{KMnO}_4 \). Let the oxidation state be \(x\).

\[ (+1) + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7 \]

The atomic number of Mn is 25, and its ground state electronic configuration is \( [\text{Ar}] \, 3d^5 4s^2 \).

The electronic configuration of \( \text{Mn}^{7+} \) is \( [\text{Ar}] \, 3d^0 \).

The number of unpaired electrons, \( n_1 \), in \( \text{Mn}^{7+} \) is 0.

Therefore, the spin-only magnetic moment (\(\mu_1\)) for Mn in \( \text{KMnO}_4 \) is:

\[ \mu_1 = \sqrt{0(0+2)} = 0 \, \text{BM} \]

Step 2: Identify the product and calculate its magnetic moment.

In the titration of \( \text{KMnO}_4 \) with oxalic acid (\( \text{H}_2\text{C}_2\text{O}_4 \)) in an acidic medium (e.g., \( \text{H}_2\text{SO}_4 \)), the permanganate ion (\( \text{MnO}_4^- \)) acts as a strong oxidizing agent and is itself reduced. The half-reaction is:

\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

The manganese product formed is the \( \text{Mn}^{2+} \) ion.

The electronic configuration of \( \text{Mn}^{2+} \) is \( [\text{Ar}] \, 3d^5 \).

For a \( d^5 \) configuration, the electrons fill the five d-orbitals singly (according to Hund's rule), so the number of unpaired electrons, \( n_2 \), is 5.

The spin-only magnetic moment (\(\mu_2\)) for \( \text{Mn}^{2+} \) is:

\[ \mu_2 = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} \, \text{BM} \]

Step 3: Calculate the difference in the magnetic moment values.

The difference, \( \Delta\mu \), is:

\[ \Delta\mu = |\mu_2 - \mu_1| = |\sqrt{35} - 0| = \sqrt{35} \, \text{BM} \]

Final Computation & Result:

We need to find the value of \( \sqrt{35} \) and round it to the nearest integer.

\[ \sqrt{35} \approx 5.916 \, \text{BM} \]

Rounding 5.916 to the nearest integer gives 6.

The difference in the 'spin-only' magnetic moment values is 6 BM.