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(B) To find the magnetic moment $(\mu)$,we use the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$I. [FeF_6]^{3-}$:

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$B-A-E-D-C$

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(B) To find the magnetic moment $(\mu)$,we use the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$I. [FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so $n=5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.93 \ BM$ $(B)$.$II. [Ti(H_2O)_6]^{3+}$: $Ti^{3+}$ is $3d^1$. $n=1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$ $(A)$.$III. [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. $n=3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.88 \ BM$ $(E)$.$IV. [Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $n=2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$ $(D)$.$V. [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing,so $n=0$. $\mu = 0.00 \ BM$ $(C)$.Thus,the correct matching is $I-B, II-A, III-E, IV-D, V-C$.

List-$I$	List-$II$
$I. [FeF_6]^{3-}$	$A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$	$B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$	$C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$	$D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$	$E. 3.88 \ BM$

Match List-$I$ with List-$II$ and select the correct answer using the codes given below:
List-$I$ List-$II$
$I. [FeF_6]^{3-}$ $A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$ $B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$ $C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$ $D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$ $E. 3.88 \ BM$

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below:List-$I$List-$II$$I. [FeF_6]^{3-}$$A. 1.73 \ BM$$II. [Ti(H_2O)_6]^{3+}$$B. 5.93 \ BM$$III. [Cr(NH_3)_6]^{3+}$$C. 0.00 \ BM$$IV. [Ni(H_2O)_6]^{2+}$$D. 2.83 \ BM$$V. [Fe(CN)_6]^{4-}$$E. 3.88 \ BM$