The spin-only magnetic moments of the complex | vedclass

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(A) In $[Mn(CN)_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $3d^4$. Since $CN^-$ is a strong field liga

Vedclass · Accepted Answer

$2.84 \ BM, 0 \ BM$

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(A) In $[Mn(CN)_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $3d^4$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in $d$-orbitals. Thus,the configuration becomes $t_{2g}^4 e_g^0$,resulting in $n = 2$ unpaired electrons.$\mu_{	ext{spin only}} = \sqrt{n(n+2)} \ BM = \sqrt{2(2+2)} \ BM = \sqrt{8} \ BM \approx 2.84 \ BM$.In $[Co(C_2O_4)_3]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $C_2O_4^{2-}$ is a chelating ligand,in the octahedral field,$Co^{3+}$ $(d^6)$ forms a low-spin complex with $t_{2g}^6 e_g^0$ configuration,resulting in $n = 0$ unpaired electrons.$\mu_{	ext{spin only}} = \sqrt{0(0+2)} \ BM = 0 \ BM$.

The spin-only magnetic moments of the complexes $[Mn(CN)_6]^{3-}$ and $[Co(C_2O_4)_3]^{3-}$ are respectively:

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