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Question 47

The magnetic moment is measured in Bohr Magneton (BM). Spin only magnetic moment of Fe in [Fe(H$$_2$$O)$$_6$$]$$^{3+}$$ and [Fe(CN)$$_6$$]$$^{3-}$$ complexes respectively is:

Solution

Fe³⁺ has d⁵ configuration.

[Fe(H₂O)₆]³⁺: H₂O is weak field. All 5 electrons unpaired (high spin). $$\mu = \sqrt{5 \times 7} = \sqrt{35} = 5.92$$ BM.

[Fe(CN)₆]³⁻: CN⁻ is strong field. t₂g⁵ configuration → 1 unpaired electron. $$\mu = \sqrt{1 \times 3} = \sqrt{3} = 1.732$$ BM.

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