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).

Question

A short bar magnet placed with its axis at $30^\circ$ with an external magnetic field of $800$ Gauss experiences a torque of $0.016\,\text{N m}$. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3}\,\text{J}$. The value of $\alpha$ is ___.

Answer 1

The magnetic moment ($\mu$) is calculated using the formula: \[ \mu = \sqrt{n(n+2)} { B.M.} \] where $ n $ represents the number of unpaired electrons.
- $ Mn_2O_3 $: Manganese is in the $ +3 $ oxidation state ($d^4$), resulting in 4 unpaired electrons.
- $ TiO $: Titanium is in the $ +2 $ oxidation state ($d^2$), resulting in 2 unpaired electrons.
- $ VO $: Vanadium is in the $ +2 $ oxidation state ($d^3$), resulting in 3 unpaired electrons.
Considering $ Mn_2O_3 $ exhibits the highest oxidation state and strongest oxidizing capacity, its magnetic moment is calculated as: \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \] Rounded to the nearest integer, the magnetic moment is 4 B.M.

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).

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