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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex. The magnetic moment is given by $\mu = \sqrt{n(n+

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$\left[Co(CN)_6\right]^{3-}$

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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.$(I)$ $\left[Cr(CN)_6\right]^{3-}$: $Cr^{3+}$ is $3d^3$. With $CN^-$ as a strong field ligand,all $3$ electrons remain unpaired in the $t_{2g}$ orbitals. $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM$.$(II)$ $\left[Mn(CN)_6\right]^{3-}$: $Mn^{3+}$ is $3d^4$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 2$ unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.$(III)$ $\left[Fe(CN)_6\right]^{3-}$: $Fe^{3+}$ is $3d^5$. With $CN^-$ as a strong field ligand,electrons pair up,leaving $n = 1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM$.$(IV)$ $\left[Co(CN)_6\right]^{3-}$: $Co^{3+}$ is $3d^6$. With $CN^-$ as a strong field ligand,all electrons pair up in the $t_{2g}$ orbitals $(t_{2g}^6 e_g^0)$. $n = 0$,$\mu = 0 \ BM$.Thus,$\left[Co(CN)_6\right]^{3-}$ has the lowest magnetic moment.

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

Which one of the following coordination complexes exhibits the lowest value of magnetic moment (in $BM$)?

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Complex Magnetic Moment $(BM)$

$K_4[Mn(CN)_6]$ $2.2$

$[Fe(H_2O)_6]^{2+}$ $5.3$

$K_2[MnCl_4]$ $5.9$

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