Correct increasing order of spin magnetic mom | vedclass

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(B) The spin magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$(I) \ [FeF_6]

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$II < III < I$

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(B) The spin magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$(I) \ [FeF_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,so electrons remain unpaired. $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ B.M$.$(II) \ [V(H_2O)_6]^{2+}$: $V^{2+}$ is $3d^3$. $n = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M$.$(III) \ [Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so electrons are $t_{2g}^4 e_g^2$. $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ B.M$.Comparing the values: $3.87 < 4.90 < 5.92$,which corresponds to $II < III < I$.

Correct increasing order of spin magnetic moment $($in $B.M.)$ for the following complexes is:
$(I) \ [FeF_6]^{3-} \ (II) \ [V(H_2O)_6]^{2+} \ (III) \ [Fe(H_2O)_6]^{2+}$

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Correct increasing order of spin magnetic moment $($in $B.M.)$ for the following complexes is:$(I) \ [FeF_6]^{3-} \ (II) \ [V(H_2O)_6]^{2+} \ (III) \ [Fe(H_2O)_6]^{2+}$