The calculated magnetic moments (spin only va | vedclass

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(B) $(i)$ $[FeCl_{4}]^{2-}$: $Fe^{2+}$ is $d^{6}$ in a tetrahedral field. Configuration is $e^{3} t_{2}^{3}$,so $n = 4$ unpaired electrons. $\mu = \sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) $(i)$ $[FeCl_{4}]^{2-}$: $Fe^{2+}$ is $d^{6}$ in a tetrahedral field. Configuration is $e^{3} t_{2}^{3}$,so $n = 4$ unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$(ii)$ $[Co(C_{2}O_{4})_{3}]^{3-}$: $Co^{3+}$ is $d^{6}$ in an octahedral field with a strong field ligand ($C_{2}O_{4}^{2-}$ is often considered strong enough to cause pairing in $Co^{3+}$). Configuration is $t_{2g}^{6} e_{g}^{0}$,so $n = 0$ unpaired electrons. $\mu = 0 \ BM$.$(iii)$ $MnO_{4}^{2-}$: $Mn$ is in $+6$ oxidation state. $Mn^{6+}$ is $3d^{1}$. $n = 1$ 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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