Question

The spin-only magnetic moment (\(\mu\)) value (B.M.) of the compound with the strongest oxidising power among \(Mn_2O_3\), \(TiO\), and \(VO\) is ________________________ B.M. (Nearest integer).

Hint:

For transition metals, the higher the oxidation state, the stronger the oxidising power.

Answer 1

Correct Answer: 4

The magnetic moment (\(\mu\)) is given by: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons.

• \( Mn_2O_3 \): Mn oxidation state is \( +3 \) (\(d^4\)), unpaired electrons = 4. 
• \( TiO \): Ti oxidation state is \( +2 \) (\(d^2\)), unpaired electrons = 2.
• \( VO \): V oxidation state is \( +2 \) (\(d^3\)), unpaired electrons = 3.

Since \( Mn_2O_3 \) has the highest oxidation state and strongest oxidising power, we calculate: \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \] Rounding to the nearest integer, the answer is 4 B.M.

Answer 2

Step 1: Identify oxidation states of metals in each compound.
For \( Mn_2O_3 \): Let the oxidation state of Mn be \( x \).
\[ 2x + 3(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow x = +3. \] Hence, oxidation state of Mn = +3.

For \( TiO \): Let oxidation state of Ti be \( x \).
\[ x + (-2) = 0 \Rightarrow x = +2. \] Hence, oxidation state of Ti = +2.

For \( VO \): Let oxidation state of V be \( x \).
\[ x + (-2) = 0 \Rightarrow x = +2. \] Hence, oxidation state of V = +2.

Step 2: Determine their electronic configurations and oxidation power.
- \( Mn^{3+} \): Atomic number 25 → \( [Ar]\,3d^4 \)
- \( Ti^{2+} \): Atomic number 22 → \( [Ar]\,3d^2 \)
- \( V^{2+} \): Atomic number 23 → \( [Ar]\,3d^3 \)

The oxidising power increases with the ease of reduction (i.e., higher oxidation state). Hence, \( Mn_2O_3 \) has the strongest oxidising power because Mn is in the +3 oxidation state (higher than +2 in the others).

Step 3: Find the magnetic moment of \( Mn^{3+} \).
Electronic configuration: \( [Ar]\,3d^4 \) → number of unpaired electrons \( n = 4 \).
Spin-only magnetic moment: \[ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \, \text{B.M.} \] The nearest integer value is 4.

Final Answer:
\[ \boxed{4 \, \text{B.M.}} \]