The calculated magnetic moments (spin-only va | vedclass

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(B) $(i)$ $[FeCl_{4}]^{2-}$: $Fe$ is in $+2$ oxidation state $(d^{6})$. In a tetrahedral field, the configuration is $e^{3} t_{2}^{3}$, resulting in $

Vedclass · Accepted Answer

$4.90$, $0$ and $1.73 \ BM$

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(B) $(i)$ $[FeCl_{4}]^{2-}$: $Fe$ is in $+2$ oxidation state $(d^{6})$. In a tetrahedral field, the configuration is $e^{3} t_{2}^{3}$, resulting in $n = 4$ unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.(ii) $[Co(C_{2}O_{4})_{3}]^{3-}$: $Co$ is in $+3$ oxidation state $(d^{6})$. In an octahedral field with a strong field ligand $(C_{2}O_{4}^{2-})$, the configuration is $t_{2g}^{6} e_{g}^{0}$, resulting in $n = 0$ unpaired electrons. $\mu = 0 \ BM$.(iii) $MnO_{4}^{2-}$: $Mn$ is in $+6$ oxidation state $(3d^{1})$. There is $n = 1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

The calculated magnetic moments (spin-only value) for species $[FeCl_{4}]^{2-}$, $[Co(C_{2}O_{4})_{3}]^{3-}$ and $MnO_{4}^{2-}$ respectively are ..... .

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