Question

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

• A. 1.73 BM
• B. 5.90 BM
• C. 4.90 BM
• D. 3.87 BM

Answer 1

The Correct Option is A

To determine the spin-only magnetic moment of Hexacyanomanganate(II) ion, we first need to understand the electronic configuration of Manganese in this complex. The chemical formula of the ion is [Mn(CN)₆]^4-. Manganese in this complex is in the +2 oxidation state, which means it has lost two electrons from its 3d⁵4s² configuration, resulting in a 3d⁵ configuration.

Now, the magnetic moment (μ) can be calculated using the spin-only formula:

\(μ = \sqrt {n(n+2)}\) BM

where n is the number of unpaired electrons. For a 3d⁵ configuration, there are 5 unpaired electrons.

Substituting n = 1 (due to low-spin configuration in strong field ligand cyanide environment) into the formula, we get:

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

Therefore, the spin-only magnetic moment of Hexacyanomanganate(II) ion is \(1.73\) BM.

Answer 2

To determine the spin-only magnetic moment of the Hexacyanomanganate(II) ion, [Mn(CN)₆]^4-, we need to find the number of unpaired electrons in the Mn²⁺ ion.

1. Electronic configuration of Mn:

Mn (Z = 25): [Ar] 3d⁵ 4s²

2. Electronic configuration of Mn²⁺:

Mn²⁺: [Ar] 3d⁵

3. Effect of CN^- ligand:

CN^- is a strong field ligand. In the presence of strong field ligands, the d-electrons pair up. However, since there are 5 d-electrons, there will be one unpaired electron in the t_2g orbital. The configuration will be t_2g⁵eg⁰

4. Number of unpaired electrons (n):

\(n = 1\)

5. Spin-only magnetic moment formula:

\(μ_s = \sqrt {[n(n+2)]}\) BM

6. Calculation:

\(μ_s = \sqrt {[1(1+2)]}\) BM

\(μ_s = \sqrt {[1(3)]}\) BM

\(μ_s = \sqrt3\) BM

\(μ_s ≈ 1.73\) BM

Therefore, the spin-only magnetic moment of [Mn(CN)₆]^4- is approximately 1.73 BM.

The correct answer is (B) : 1.73 BM