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(A) $[CoF_6]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak-field ligand, so electrons remain unpaired $(n=4)$. $\mu = \sqrt{4(4+2)} =

Vedclass · Accepted Answer

$A-II, B-I, C-IV, D-III$

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(A) $[CoF_6]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak-field ligand, so electrons remain unpaired $(n=4)$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$$[Co(C_2O_4)_3]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $C_2O_4^{2-}$ is a strong-field ligand, so electrons pair up $(n=0)$. $\mu = 0 \ B.M.$$[FeF_6]^{3-}: Fe$ is in $+3$ oxidation state $(3d^5)$. $F^-$ is a weak-field ligand, so electrons remain unpaired $(n=5)$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$$[Mn(CN)_6]^{3-}: Mn$ is in $+3$ oxidation state $(3d^4)$. $CN^-$ is a strong-field ligand, causing pairing $(n=2)$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$Thus, the correct match is $A-II, B-I, C-IV, D-III$.

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

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