The calculated magnetic moments spin only value for species - FeCl4 -2- -Co C2 O4 3 -3- and MnO2-4 respectively are -

The spin only magnetic moment formula μ=√(n(n+2))B.M n = no. of unpaired electrons. [ FeCl_4 ]^2 Oxidation state of Fe in complex is +2, Fe^2+ Cl is weak lig ...

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Question

The calculated magnetic moments (spin only value) for species ${[F e C l_{4}]}_{4}^{2 -}$ , $[C o (C_{2} O_{4})_{3}]^{3 -}$ and $M n O_{4}^{2 -}$ respectively are :

5.92, 4.90 a n d 0 B M

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4.90, 0 a n d 1.73 B M

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4.90, 0 a n d 2.83 B M

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5.82, 0 a n d 0 B M

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Solution

The correct option is B $4.90, 0 a n d 1.73 B M$
The spin only magnetic moment formula-
$μ = \sqrt{n (n + 2)} B . M$
n = no. of unpaired electrons.

${[F e C l_{4}]}_{4}^{2 -}$
Oxidation state of Fe in complex is +2, $F e^{2 +}$
Cl is weak ligand, its configuration is $(e)^{3} (t_{2})^{3}$
Thus, it has 4 unpaired electrons,
$μ = 4.9 B M$

${[C o {(C_{2} O_{4})}_{3}]}_{3}^{3 -} \to C o^{3 +}$ and strong field.
Thus, its configuration is $(t_{2 g})^{6}$
No unpaired electrons, n=0
$μ = 0 B M$

$M n O_{4}^{2 -} \to M n^{+ 6}$
Its configuration is $[A r] 3 d^{1}$
Thus, it has one unpaired electron, $n = 1$
$μ = 1.73 B M$

(n is number of unpaired electron and $μ$ is spin only magnetic moment)

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The calculated magnetic moments (spin only value) for species [FeCl4]2−, [Co(C2O4)3]3− and MnO2−4 respectively are :

The calculated magnetic moments (spin only value) for species ${[F e C l_{4}]}_{4}^{2 -}$ , $[C o (C_{2} O_{4})_{3}]^{3 -}$ and $M n O_{4}^{2 -}$ respectively are :