Amongst FeCl_{3} cdot 3H_{2}O,K_{3}[Fe(CN)_{6 | vedclass

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(C) The inner-orbital complexes are $K_{3}[Fe(CN)_{6}]$ and $[Co(NH_{3})_{6}]Cl_{3}$.$K_{3}[Fe(CN)_{6}]$ contains $Fe^{3+}$ $(3d^{5})$,which forms a $

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$2$

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(C) The inner-orbital complexes are $K_{3}[Fe(CN)_{6}]$ and $[Co(NH_{3})_{6}]Cl_{3}$.$K_{3}[Fe(CN)_{6}]$ contains $Fe^{3+}$ $(3d^{5})$,which forms a $d^{2}sp^{3}$ complex with $1$ unpaired electron. Its spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ B.M.$$[Co(NH_{3})_{6}]Cl_{3}$ contains $Co^{3+}$ $(3d^{6})$,which forms a $d^{2}sp^{3}$ complex with $0$ unpaired electrons. Its spin-only magnetic moment is $\mu = 0 \ B.M.$$CN^{-}$ is a stronger field ligand than $NH_{3}$,so $K_{3}[Fe(CN)_{6}]$ has a larger crystal field splitting energy $(\Delta_{0})$ than $[Co(NH_{3})_{6}]Cl_{3}$.Since $\Delta_{0} = \frac{hc}{\lambda}$,a larger $\Delta_{0}$ corresponds to a shorter wavelength $(\lambda)$ of absorbed light.Therefore,$K_{3}[Fe(CN)_{6}]$ is the complex that absorbs light at the shortest wavelength.Its spin-only magnetic moment is $1.732 \ B.M.$,which rounds to $2$ to the nearest integer.

Amongst $FeCl_{3} \cdot 3H_{2}O$,$K_{3}[Fe(CN)_{6}]$ and $[Co(NH_{3})_{6}]Cl_{3}$,the spin-only magnetic moment value of the inner-orbital complex that absorbs light at the shortest wavelength is $B.M.$ [nearest integer].

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