Match List-I with List-II. List-I Lis | vedclass

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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$a. [Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $CN

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$a-iv, b-i, c-ii, d-iii$

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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$a. [Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $CN^-$ is a strong field ligand,causing pairing. $n=1$,$\mu = \sqrt{1(1+2)} = 1.73 \ BM$.$b. [Fe(H_{2}O)_{6}]^{3+}$: $Fe^{3+}$ is $3d^{5}$. $H_2O$ is a weak field ligand,no pairing. $n=5$,$\mu = \sqrt{5(5+2)} = 5.92 \ BM$.$c. [Fe(CN)_{6}]^{4-}$: $Fe^{2+}$ is $3d^{6}$. $CN^-$ is a strong field ligand,causing pairing. $n=0$,$\mu = 0 \ BM$.$d. [Fe(H_{2}O)_{6}]^{2+}$: $Fe^{2+}$ is $3d^{6}$. $H_2O$ is a weak field ligand,no pairing. $n=4$,$\mu = \sqrt{4(4+2)} = 4.90 \ BM$.Matching these values: $a-iv, b-i, c-ii, d-iii$.

List-$I$	List-$II$
$a. [Fe(CN)_{6}]^{3-}$	$i. 5.92 \ BM$
$b. [Fe(H_{2}O)_{6}]^{3+}$	$ii. 0 \ BM$
$c. [Fe(CN)_{6}]^{4-}$	$iii. 4.90 \ BM$
$d. [Fe(H_{2}O)_{6}]^{2+}$	$iv. 1.73 \ BM$

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$a. [Fe(CN)_{6}]^{3-}$ $i. 5.92 \ BM$

$b. [Fe(H_{2}O)_{6}]^{3+}$ $ii. 0 \ BM$

$c. [Fe(CN)_{6}]^{4-}$ $iii. 4.90 \ BM$

$d. [Fe(H_{2}O)_{6}]^{2+}$ $iv. 1.73 \ BM$

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Match List-$I$ with List-$II$. List-$I$ List-$II$ $a. [Fe(CN)_{6}]^{3-}$ $i. 5.92 \ BM$ $b. [Fe(H_{2}O)_{6}]^{3+}$ $ii. 0 \ BM$ $c. [Fe(CN)_{6}]^{4-}$ $iii. 4.90 \ BM$ $d. [Fe(H_{2}O)_{6}]^{2+}$ $iv. 1.73 \ BM$ Choose the correct answer from the options given below.