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

Correct Answer: 7

Step 1: Determine the oxidation state of Mn in each complex. - Both complexes have charge -3 and ligands Br\(^-\) and CN\(^-\) are monodentate anions with charge -1 each. For \([\text{Mn}(\text{Br})_6]^{3-}\): \[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \] Similarly, for \([\text{Mn}(\text{CN})_6]^{3-}\): \[ x + 6(-1) = -3 \implies x = +3 \] So, Mn is +3 in both complexes. Step 2: Determine the electronic configuration of \(Mn^{3+}\). Atomic number of Mn = 25 Electronic configuration of Mn: \([Ar]\,3d^5\,4s^2\) For \(Mn^{3+}\), remove 3 electrons: \[ 3d^4 \quad \text{configuration} \] Step 3: Determine the nature of ligands and spin state. - Br\(^-\) is a weak field ligand → high spin complex - CN\(^-\) is a strong field ligand → low spin complex Step 4: Calculate spin only magnetic moment \(\mu_s\) using formula: \[ \mu_s = \sqrt{n(n+2)} \quad \text{B.M.} \] Where \(n\) = number of unpaired electrons. For \([\text{Mn}(\text{Br})_6]^{3-}\) (high spin \(d^4\)): Electron arrangement: \(t_{2g}^3 e_g^1\) → 4 unpaired electrons \[ \mu_s = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \approx 5 \, \text{B.M.} \] For \([\text{Mn}(\text{CN})_6]^{3-}\) (low spin \(d^4\)): Electron arrangement: \(t_{2g}^4 e_g^0\) → 2 unpaired electrons \[ \mu_s = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \approx 2.8 \, \text{B.M.} \] Step 5: Sum of magnetic moments: \[ 5 + 2.8 = 7.8 \approx 7 \]