The increasing order of magnetic moments of t | vedclass

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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ for each complex:$I. [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $d^6$. $CN^-$

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$I < II < IV < III$

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(D) To find the magnetic moment,we calculate the number of unpaired electrons $(n)$ for each complex:$I. [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $d^6$. $CN^-$ is a strong field ligand,so electrons pair up. $n = 0$. Magnetic moment $\mu = \sqrt{0(0+2)} = 0 \ BM$.$II. [Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong field ligand,so $n = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$III. [Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $d^3$. $n = 3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$IV. [Ni(H_2O)_4]^{2+}$: $Ni^{2+}$ is $d^8$. $H_2O$ is a weak field ligand,tetrahedral geometry,$n = 2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.Comparing the values: $0 < 1.73 < 2.83 < 3.87$,which corresponds to $I < II < IV < III$.

The increasing order of magnetic moments of the following complexes is:-
$I. [Fe(CN)_6]^{4-}$
$II. [Fe(CN)_6]^{3-}$
$III. [Cr(NH_3)_6]^{3+}$
$IV. [Ni(H_2O)_4]^{2+}$

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The increasing order of magnetic moments of the following complexes is:-$I. [Fe(CN)_6]^{4-}$$II. [Fe(CN)_6]^{3-}$$III. [Cr(NH_3)_6]^{3+}$$IV. [Ni(H_2O)_4]^{2+}$