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: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Here, $ n $ represents the number of unpaired electrons.

For $ Mn_2O_3 $: The manganese (Mn) has a $ +3 $ oxidation state, corresponding to a $ d^4 $ configuration with 4 unpaired electrons.
For $ TiO $: The titanium (Ti) has a $ +2 $ oxidation state, corresponding to a $ d^2 $ configuration with 2 unpaired electrons.
For $ VO $: The vanadium (V) has a $ +2 $ oxidation state, corresponding to a $ d^3 $ configuration with 3 unpaired electrons.

Given that $ Mn_2O_3 $ exhibits the highest oxidation state and the strongest oxidizing power, its magnetic moment is computed as follows: \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \] Rounded to the nearest whole number, 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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Correct Answer: 4

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