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

The magnetic moment (\(\mu\)) is given by: \[ \mu = \sqrt{n(n+2)} { B.M.} \] 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.