The most paramagnetic complex is

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(D) To determine the most paramagnetic complex,we calculate the number of unpaired electrons in each:$1$. $[Sc(CN)_6]^{-3}$: $Sc^{+3}$ is $[Ar] 3d^0$,

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$[Cr(CN)_6]^{-3}$

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(D) To determine the most paramagnetic complex,we calculate the number of unpaired electrons in each:$1$. $[Sc(CN)_6]^{-3}$: $Sc^{+3}$ is $[Ar] 3d^0$,so $n = 0$.$2$. $[Co(CN)_6]^{-3}$: $Co^{+3}$ is $[Ar] 3d^6$. In the presence of a strong field ligand $(CN^-)$,electrons pair up,resulting in $n = 0$.$3$. $[Fe(CN)_6]^{-4}$: $Fe^{+2}$ is $[Ar] 3d^6$. In the presence of a strong field ligand $(CN^-)$,electrons pair up,resulting in $n = 0$.$4$. $[Cr(CN)_6]^{-3}$: $Cr^{+3}$ is $[Ar] 3d^3$. There are $3$ unpaired electrons $(n = 3)$.Since $[Cr(CN)_6]^{-3}$ has the highest number of unpaired electrons,it is the most paramagnetic.

List-$I$ (Complex)	List-$II$ (Spin only Magnetic moment)
$A$. $[CoF_6]^{3-}$	$I$. $0$
$B$. $[Co(C_2O_4)_3]^{3-}$	$II$. $\sqrt{24}$
$C$. $[FeF_6]^{3-}$	$III$. $\sqrt{8}$
$D$. $[Mn(CN)_6]^{3-}$	$IV$. $\sqrt{35}$
	$V$. $\sqrt{15}$

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Match the following:

List-$I$ (Complex) List-$II$ (Spin only Magnetic moment)

$A$. $[CoF_6]^{3-}$ $I$. $0$

$B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$

$C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$

$D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$

$V$. $\sqrt{15}$

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