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प्रश्न
Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field, and (ii) weak field ligand. (Atomic no.of Fe=26)
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उत्तर
`"Fe"= ["Ar"]3"d"^6 4"s"^2`
`"Fe"^{3+} = ["Ar"]3"d"^6 4"s"^2`
`"Fe"^(+3) = ["AR"] 3"d"^5 4"s"^0`
In octahedral crystal field.
(i) Strong field ligand - Electrons will be paired.
No. of unpaired e- = 1
Paramagnetic
`μ = sqrt(n(n+2) = sqrt(1(3) = sqrt3 B.M`
(ii) Weak field ligand
- Electrons follow Hund’s rule 
No. of unpaired = 5
Paramagnetic
`μ =sqrt(5(5+2))= sqrt(35) B.M.`
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संबंधित प्रश्न
State the superiority of crystal field theory over valence bond theory.
The CFSE for octahedral \[\ce{[CoCl6]^{4-}}\] is 18,000 cm–1. The CFSE for tetrahedral \[\ce{[CoCl4]^{2-}}\] will be ______.
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(i) \[\ce{[MnCl6]^{3-}}\]
(ii) \[\ce{[FeF6]^{3-}}\]
(iii) \[\ce{[CoF6]^{3-}}\]
(iv) \[\ce{[Ni(NH3)6]^{2+}}\]
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\[\ce{[FeF6]^{3-}, [Fe(H2O)6]^{2+}, [Fe(CN)6]^{4-}}\]
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