MCQ
 ........ $B.M.$ is the magnetic moment of ${K_3}[Fe{F_6}]$
  • $5.91$
  • B
    $4.89$
  • C
    $3.87$
  • D
    $6.92$

Answer

Correct option: A.
$5.91$
a
(a)${K_3}[Fe{F_6}]$

$F{e^{3 + }} = [Ar]3{d^5}4{s^0}$

Number of unpaired electrons $= 5$

Magnetic moment $ = \sqrt {n(n + 2)} = \sqrt {5(5 + 2)} $

$ = \sqrt {35} $ $= 5.91\, BM.$ 

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