A compound with empirical formula Fe(H₂O)₄(CN)₂ has a magnetic moment corresponding to $2 \frac{2}{3}$ unpaired electrons per iron. The molecular formula of the complex may be

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${[Fe {(H_{2} O)}_{6}]}_{2} [Fe (CN)_{6}]$

$[Fe {(H_{2} O)}_{6}] [Fe (CN)_{6}]$

${[Fe {(H_{2} O)}_{6}]}_{4} {[Fe (CN)_{6}]}_{3}$

$[Fe {(H_{2} O)}_{4} (CN)_{2}]$

answer is D.

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Detailed Solution

Total number of unpaired electrons should be 8 and they should present in 3 iron atoms. So formula must be ${[Fe {(H_{2} O)}_{6}]}_{2} [Fe (CN)_{6}]$ Each [Fe(H₂O)₆]²⁺ contain 4 unpaired electrons whole [Fe(CN)₆]^4– contain zero unpaired electrons. So total unpaired electrons are 8 which give 8/3 or $2 \frac{2}{3}$