Question 15

NUMERICALMEDIUM

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br)6]3−[\text{Mn(Br)}_6]^{3-} and [Mn(CN)6]3−[\text{Mn(CN)}_6]^{3-} is _____.

Correct Answer: 7.73

Detailed Solution

  1. Identify the central metal ion: Mn3+Mn^{3+} has electronic configuration [Ar]3d4[Ar] 3d^4.

  2. For [Mn(Br)6]3−[\text{Mn(Br)}_6]^{3-}: Br−Br^- is a weak field ligand (WFL). The configuration is t2g3eg1t_{2g}^3 e_g^1. Number of unpaired electrons (nn) =4= 4. μ1=n(n+2)=4(4+2)=24≈4.899 B.M.\mu_1 = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899\text{ B.M.}

  3. For [Mn(CN)6]3−[\text{Mn(CN)}_6]^{3-}: CN−CN^- is a strong field ligand (SFL). The configuration is t2g4eg0t_{2g}^4 e_g^0. Number of unpaired electrons (nn) =2= 2. μ2=2(2+2)=8≈2.828 B.M.\mu_2 = \sqrt{2(2+2)} = \sqrt{8} \approx 2.828\text{ B.M.}

  4. Total sum =4.899+2.828=7.727 B.M.= 4.899 + 2.828 = 7.727\text{ B.M.}

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