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$The complex k_{3} [Fe (CN)_{6}] should have a spin only magnetic moment of$

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$\sqrt[2]{5} B . M$

$\sqrt{35} B . M$

6 B.M

$\sqrt{48} B . M$

answer is C.

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Detailed Solution

$\begin{array}{l} In K_{3} [Fe (CN)_{6}] \Rightarrow F e^{+ 3} = d^{5} \Rightarrow 5 unpaired electron \\ ∴ Magnetic moment (μ) = \sqrt{n (n + 2)} B.M \\ μ = \sqrt{5 (5 + 2)} B . M \\ μ = \sqrt{35} B . M \end{array}$

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