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 Ion	Label
$[Mn(CN)_6]^{3-}$	$A$
$[Fe(CN)_6]^{3-}$	$B$
$[CoF_6]^{3-}$	$C$
$[Co(C_2O_4)_3]^{3-}$	$D$

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

Similar Questions

Observe the following complex ions:
Complex Ion Label
$[Mn(CN)_6]^{3-}$ $A$
$[Fe(CN)_6]^{3-}$ $B$
$[CoF_6]^{3-}$ $C$
$[Co(C_2O_4)_3]^{3-}$ $D$

Identify the option in which the unpaired electrons in the complex ions are in correct increasing order.

The complex that can show optical activity is:

The complexes $[Co(NH_3)_6][Cr(C_2O_4)_3]$ and $[Cr(NH_3)_6][Co(C_2O_4)_3]$ exhibit which type of isomerism?

Which of the following complexes can show coordination isomerism?

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