Match List-I with List-II. List-I (Compl | vedclass

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(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$(A) [Cr(

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

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(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$(A) [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. Number of unpaired electrons $n=3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M. \ (II)$.$(B) [NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. In a tetrahedral field,$n=2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ B.M. \ (IV)$.$(C) [CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so $n=4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M. \ (I)$.$(D) [Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing,so $n=0$. $\mu = 0 \ B.M. \ (III)$.Thus,the correct match is $A-II, B-IV, C-I, D-III$.

List-$I$ (Complex)	List-$II$ (Spin only magnetic moment in $B$.$M$.)
$A$. $[Cr(NH_3)_6]^{3+}$	$I$. $4.90$
$B$. $[NiCl_4]^{2-}$	$II$. $3.87$
$C$. $[CoF_6]^{3-}$	$III$. $0.0$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $2.83$

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Match List-$I$ with List-$II$. List-$I$ (Complex) List-$II$ (Spin only magnetic moment in $B$.$M$.) $A$. $[Cr(NH_3)_6]^{3+}$ $I$. $4.90$ $B$. $[NiCl_4]^{2-}$ $II$. $3.87$ $C$. $[CoF_6]^{3-}$ $III$. $0.0$ $D$. $[Ni(CN)_4]^{2-}$ $IV$. $2.83$ Choose the correct answer from the options given below: