The calculated spin-only magnetic moments of $ K_3[Fe(OH)_6] $ and $ K_4[Fe(OH)_6] $ respectively are:

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 spin-only magnetic moment is calculated using the formula \[\mu = \sqrt{n(n+2)} \, \text{B.M.}\], where $n$ represents the number of unpaired electrons. For $ Fe^{3+} $ in $ K_3[Fe(OH)_6] $ and $ Fe^{2+} $ in $ K_4[Fe(OH)_6] $, the number of unpaired electrons are determined from their electronic configurations. $ Fe^{3+} $ has 5 unpaired electrons, yielding a magnetic moment of 5.92 B.M., while $ Fe^{2+} $ has 4 unpaired electrons, resulting in a magnetic moment of 4.90 B.M.

The calculated spin-only magnetic moments of \( K_3[Fe(OH)_6] \) and \( K_4[Fe(OH)_6] \) respectively are:

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The Correct Option is B

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