The correct order of the spin-only magnetic m | vedclass

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(B) $(I) [Cr(H_2O)_6]^{2+}$: $Cr^{2+} \Rightarrow [Ar] 3d^4$. $H_2O$ is a weak field ligand. Unpaired $e^- = 4$. Magnetic moment $\mu = \sqrt{4(4+2)}

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$(I) > (IV) > (III) > (II)$

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(B) $(I) [Cr(H_2O)_6]^{2+}$: $Cr^{2+} \Rightarrow [Ar] 3d^4$. $H_2O$ is a weak field ligand. Unpaired $e^- = 4$. Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.89 \ BM$.$(II) [Fe(CN)_6]^{4-}$: $Fe^{2+} \Rightarrow [Ar] 3d^6$. $CN^-$ is a strong field ligand. Unpaired $e^- = 0$. Magnetic moment $\mu = 0 \ BM$.$(III) [Fe(C_2O_4)_3]^{3-}$: $Fe^{3+} \Rightarrow [Ar] 3d^5$. Given $\Delta_0 > P$,it is a low spin complex. Unpaired $e^- = 1$. Magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.$(IV) [CoCl_4]^{2-}$: $Co^{2+} \Rightarrow [Ar] 3d^7$. Tetrahedral complex,$Cl^-$ is a weak field ligand. Unpaired $e^- = 3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$.Comparing the values: $4.89 (I) > 3.87 (IV) > 1.73 (III) > 0 (II)$.Therefore,the correct order is $(I) > (IV) > (III) > (II)$.

The correct order of the spin-only magnetic moments of the following complexes is
$(I) [Cr(H_2O)_6]Br_2$
$(II) Na_4[Fe(CN)_6]$
$(III) Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$
$(IV) (Et_4N)_2[CoCl_4]$

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The correct order of the spin-only magnetic moments of the following complexes is$(I) [Cr(H_2O)_6]Br_2$$(II) Na_4[Fe(CN)_6]$$(III) Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$$(IV) (Et_4N)_2[CoCl_4]$