Question

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 ____.

Answer 1

Step 1: Determine the Oxidation State of Mn

Both complexes, [Mn(Br)6]³⁻ and [Mn(CN)6]³⁻, have an overall charge of -3. The ligands Br⁻ and CN⁻ are monodentate anions, each with a charge of -1. Let’s calculate the oxidation state of Mn.

For [Mn(Br)6]³⁻:

\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

For [Mn(CN)6]³⁻:

\[ x + 6(-1) = -3 \implies x = +3 \]

Thus, Mn has an oxidation state of +3 in both complexes, a key concept for JEE Advanced coordination chemistry.

Step 2: Electron Configuration of Mn³⁺

The atomic number of Mn is 25, with an electronic configuration of:

\[ [Ar]\,3d^5\,4s^2 \]

For Mn³⁺, remove 2 electrons from 4s and 1 from 3d:

\[ [Ar]\,3d^4 \]

This d4 configuration is critical for determining the spin state and magnetic properties, a common topic in JEE Advanced inorganic chemistry.

Step 3: Nature of Ligands and Spin State

The nature of ligands determines whether the complex is high spin or low spin, a key concept in coordination compounds for JEE Advanced:

• Br⁻: Weak field ligand → high spin complex
• CN⁻: Strong field ligand → low spin complex

This distinction affects the electron arrangement and magnetic moment.

Step 4: Calculate Spin-Only Magnetic Moment

The spin-only magnetic moment (\(\mu_s\)) is calculated using:

\[ \mu_s = \sqrt{n(n+2)} \quad \text{B.M.} \]

Where \( n \) is the number of unpaired electrons.

For [Mn(Br)6]³⁻ (high spin, d⁴):

Electron arrangement in octahedral field: \( t_{2g}^3 e_g^1 \)

This gives 4 unpaired electrons:

\[ \mu_s = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \approx 5 \, \text{B.M.} \]

For [Mn(CN)6]³⁻ (low spin, d⁴):

Electron arrangement: \( t_{2g}^4 e_g^0 \)

This gives 2 unpaired electrons:

\[ \mu_s = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \approx 2.8 \, \text{B.M.} \]

These calculations align with JEE Advanced expectations for magnetic moment problems.

Step 5: Sum of Magnetic Moments

Add the magnetic moments of both complexes:

\[ 5 + 2.8 = 7.8 \approx 7 \]

Final Answer: The sum of the spin-only magnetic moments is approximately 7 B.M.