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The crystal field stabilisation energy (CFSE) and the 'spin only' magnetic moment of [FeF6]3[FeF_6]^{3-} ion respectively, are:

A

0Δo0 \Delta_o and 5.95.9 BM

B

0Δo0 \Delta_o and 1.731.73 BM

C

2.0Δo2.0 \Delta_o and 5.95.9 BM

D

2.0Δo2.0 \Delta_o and 1.731.73 BM

Step-by-Step Solution

In the complex [FeF6]3[FeF_6]^{3-}, the central metal ion is Iron (Fe) in the +3 oxidation state. The electronic configuration of Fe3+Fe^{3+} is 3d53d^5.

  1. Since the fluoride ion (FF^-) is a weak field ligand, the crystal field splitting energy (Δo\Delta_o) is less than the pairing energy (PP). Therefore, the pairing of electrons does not occur, and it forms a high-spin complex.
  2. The distribution of 5 electrons in the dd-orbitals will be t2g3eg2t_{2g}^3 e_g^2.
  3. Calculation of CFSE: CFSE=(0.4×nt2g+0.6×neg)ΔoCFSE = (-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g}) \Delta_o CFSE=(0.4×3+0.6×2)Δo=(1.2+1.2)Δo=0ΔoCFSE = (-0.4 \times 3 + 0.6 \times 2) \Delta_o = (-1.2 + 1.2) \Delta_o = 0 \Delta_o.
  4. Calculation of Magnetic Moment: The number of unpaired electrons (nn) in t2g3eg2t_{2g}^3 e_g^2 is 5. The spin-only magnetic moment (μ\mu) is given by μ=n(n+2)\mu = \sqrt{n(n+2)} BM. μ=5(5+2)=355.92\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 BM.

Therefore, the CFSE is 0Δo0 \Delta_o and the spin-only magnetic moment is 5.9 BM.

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