Magnetic moment values for \( [Fe(H_2O)_6]^{3+} \) and \( [Fe(CN)_6]^{3-} \) are......
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.92 BM and 1.74 respectively
The magnetic moment of a complex of the first transition series is given by the spin-only formula, \( \mu = \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons. How many electrons stay unpaired in an octahedral complex depends on the ligand field strength: a weak field ligand gives a high spin complex, a strong field ligand forces pairing in the lower \( t_{2g} \) set and gives a low spin complex.
Both complexes contain Fe(III). Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): \( \mathrm{3d^{5}} \).
\( \ce{[Fe(H2O)6]^3+} \): \( \ce{H2O} \) is a weak field ligand, so the five electrons occupy \( t_{2g}^{3}e_{g}^{2} \) and all five are unpaired:
\( \displaystyle \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ \mathrm{BM} \)
\( \ce{[Fe(CN)6]^3-} \): \( \ce{CN-} \) is a strong field ligand, so all five electrons go into the \( t_{2g} \) set, \( t_{2g}^{5}e_{g}^{0} \), leaving one unpaired:
\( \displaystyle \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \mathrm{BM} \)
So the moments are about 5.92 BM and 1.74 BM, respectively.