Increasing order of magnetic moments of I. [F | vedclass

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(D) $1)$ $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state. Electronic configuration is $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand,all $6$ el

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$I < II < IV < III$

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(D) $1)$ $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state. Electronic configuration is $[Ar] 3d^6$. Since $CN^-$ is a strong field ligand,all $6$ electrons pair up. Number of unpaired electrons $(n)$ $= 0$. Magnetic moment $= \sqrt{0(0+2)} = 0 \ BM$.$2)$ $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state. Electronic configuration is $[Ar] 3d^5$. $CN^-$ is a strong field ligand,pairing $4$ electrons,leaving $1$ unpaired electron. Magnetic moment $= \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$3)$ $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration is $[Ar] 3d^3$. There are $3$ unpaired electrons. Magnetic moment $= \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$4)$ $[Ni(H_2O)_4]^{2+}$: $Ni$ is in $+2$ oxidation state. Electronic configuration is $[Ar] 3d^8$. $H_2O$ is a weak field ligand,so $2$ electrons remain unpaired. Magnetic moment $= \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.The order of magnetic moments is $0 < 1.73 < 2.83 < 3.87$,which corresponds to $I < II < IV < III$.

Increasing order of magnetic moments of $I. [Fe(CN)_6]^{4-}$,$II. [Fe(CN)_6]^{3-}$,$III. [Cr(NH_3)_6]^{3+}$,$IV. [Ni(H_2O)_4]^{2+}$ is

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