The spin-only magnetic moment of Hexacyanomanganate(II) ion is.........BM

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

To calculate the spin-only magnetic moment of the Hexacyanomanganate(II) ion, [Mn(CN)₆]^4-, we first determine the electronic configuration of Manganese. In this complex, Manganese is in the +2 oxidation state, possessing a 3d⁵ configuration after losing two electrons from its initial 3d⁵4s² state.

The magnetic moment (μ) is computed using the spin-only formula:

$μ = \sqrt {n(n+2)}$ BM

where 'n' represents the number of unpaired electrons. A 3d⁵ configuration typically has 5 unpaired electrons.

However, due to the strong field nature of the cyanide ligand, a low-spin configuration is observed, resulting in n = 1 unpaired electron. Substituting this value into the formula yields:

$μ = \sqrt {1(1+2)} = \sqrt 3 ≈ 1.73$ BM

Consequently, the spin-only magnetic moment of the Hexacyanomanganate(II) ion is determined to be $1.73$ BM.

The spin-only magnetic moment of Hexacyanomanganate(II) ion is.........BM

The Correct Option is A

Solution and Explanation

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