SATHEE: Coordination Compounds 2 Question 60

Coordination Compounds 2 Question 60

####62. For the octahedral complexes of $F e^{3 +}$ in $S C N^{-}$ (thiocyanato-S) and in $C N^{-}$ ligand environments, the difference between the spin only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is [atomic number of $F e = 26$ ]

(2015 Adv.)

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Solution:

When $S$ is donor atom of $S C N^{-}$ , it produces weak ligand field and forms high spin complex as

${[F e (S C N)_{6}]}^{3 -} : F e^{3 +} (3 d^{5}) =$

Spin only magnetic moment $(μ_{s}) = \sqrt{5 (5 + 2)} B M = \sqrt{35} B M$

In case of $C N^{-}$ ligand, carbon is the donor atom, it produces strong ligand field and forms low spin complex as

${[F e (C N)_{6}]}^{3 -} : F e^{3 +} (3 d^{5})$

Spin only magnetic moment $(μ_{s}) = \sqrt{1 (1 + 2)} B M = \sqrt{3} B M$

Hence, difference in spin only magnetic moment

$= \sqrt{35} - \sqrt{3} \approx 4 B M$

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Coordination Compounds 2 Question 60

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