Question

Calculate the sum of magnetic moments (in Bohr Magnetons, B.M.) of the complexes: \[ [\mathrm{MnCl}_6]^{3-} \quad \text{and} \quad [\mathrm{Mn(CN)}_6]^{3-} \]

• A. 10.92 B.M.
• B. 6.93 B.M.
• C. 7.94 B.M.
• D. 8.94 B.M.

Hint:

Use ligand field strength to decide between high-spin and low-spin. Then apply \(\mu = \sqrt{n(n+2)}\) where \(n =\) number of unpaired electrons.

Answer 1

The Correct Option is C

Step 1: Determine oxidation state of Mn in both complexes:
In both cases:
\[ \text{Let oxidation state of Mn = } x \Rightarrow x + 6(-1) = -3 \Rightarrow x = +3 \] So, we are dealing with \( \mathrm{Mn^{3+}} \) which has atomic number 25.
Electronic configuration:
\[ \mathrm{Mn^{3+}}: [\mathrm{Ar}]\,3d^4 \]  

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Complex 1: \( [\mathrm{MnCl}_6]^{3-} \)
- \( \mathrm{Cl^-} \) is a weak field ligand → High spin complex.
- For high-spin \( d^4 \): number of unpaired electrons = 4
\[ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.} \] 

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Complex 2: \( [\mathrm{Mn(CN)}_6]^{3-} \)
- \( \mathrm{CN^-} \) is a strong field ligand → Low spin complex.
- For low-spin \( d^4 \), electrons pair in lower \( t_{2g} \) orbitals → 2 unpaired electrons
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, \text{B.M.} \] 

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Step 2: Total magnetic moment:
\[ \mu_{\text{total}} = 4.90 + 2.83 = 7.73 \, \text{B.M.} \approx \boxed{7.94 \, \text{B.M.}} \]