The spin-only magnetic moments of [Mn(CN)_6]^ | vedclass

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Question

(C) The oxidation state of $Mn$ in $[Mn(CN)_6]^{2-}$ is $+4$. The electronic configuration of $Mn^{4+}$ is $[Ar] 3d^3$.Since $CN^-$ is a strong field

Vedclass · Accepted Answer

$1.73$ and $5.92$

Vedclass · Answer

(C) The oxidation state of $Mn$ in $[Mn(CN)_6]^{2-}$ is $+4$. The electronic configuration of $Mn^{4+}$ is $[Ar] 3d^3$.Since $CN^-$ is a strong field ligand,the $3d$ electrons are arranged as follows: $t_{2g}^3 e_g^0$. The number of unpaired electrons $(n)$ is $1$.The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.Since $Br^-$ is a weak field ligand,no pairing occurs. The number of unpaired electrons $(n)$ is $5$.The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Thus,the values are $1.73 \ BM$ and $5.92 \ BM$.

Complex	Magnetic Moment $(BM)$
$K_4[Mn(CN)_6]$	$2.2$
$[Fe(H_2O)_6]^{2+}$	$5.3$
$K_2[MnCl_4]$	$5.9$

The spin-only magnetic moments of $[Mn(CN)_6]^{2-}$ and $[MnBr_4]^{2-}$ in Bohr magnetons,respectively are

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Complex Magnetic Moment $(BM)$

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