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).
Correct Answer: 6
Approach Solution - 1
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:
\[6 - 0 = 6 \, \text{BM}\]
Approach Solution -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:
- Determine the oxidation state and the number of unpaired electrons for manganese in \( \text{KMnO}_4 \).
- Identify the manganese-containing product of the redox reaction between \( \text{KMnO}_4 \) and oxalic acid in an acidic medium.
- Determine the oxidation state and the number of unpaired electrons for manganese in the product.
- 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.
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