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  • Amongst <img alt="\mathrm{FeCl}_{3} .3 \mathrm{H}_{2} \mathrm{O}, \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BFeCl

Amongst \mathrm{FeCl}_{3} .3 \mathrm{H}_{2} \mathrm{O}, \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] and \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is ____________B.M [nearest integer]

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\\\left[\mathrm{Fe}(\mathrm{H}_{2} \mathrm{O})_{3} \mathrm{Cl}_{3}\right] \rightarrow \text{outer orbital complex}\\ \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightarrow \text{inner orbital complex}\\ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\mathrm{Cl}_{3} \rightarrow \text{inner orbital complex}

but \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] will have more spliting than \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] , because \mathrm{CN^{\ominus }} more stronger ligand.

\mathrm{more\ spliting \rightarrow more\ \Delta_{0} \rightarrow lesser \ wavelength\ absorbed.}

So, spin -only magnetic moment value of \left [ \mathrm{Fe(CN)_{6}} \right ]^{3-}

\begin{aligned} &\mathrm{\mu=\sqrt{n(n+2) B M} }\quad \left [ \because \text{1 unpaired }e^{\circleddash } \right ]\\ &\mathrm{\mu=\sqrt{3}\ B M} \\ &\mathrm{\mu=1.732 \; \mathrm{BM} \; \approx \; 2 \mathrm{\ BM}} \end{aligned}

Hence, the correct answer is 2

Posted by

himanshu.meshram

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