The magnetic moment (\(\mu\)) is given by: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] where \( n \) is the number of unpaired electrons.
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.
(b) If \( \vec{L} \) is the angular momentum of the electron, show that:
\[ \vec{\mu} = -\frac{e}{2m} \vec{L} \]
The sum of the spin-only magnetic moment values (in B.M.) of $[\text{Mn}(\text{Br})_6]^{3-}$ and $[\text{Mn}(\text{CN})_6]^{3-}$ is ____.
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: