Correct increasing order of spin magnetic mom | vedclass

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(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} 	ext{ B.M.}$,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-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} 	ext{ B.M.}$,where $n$ is the number of unpaired electrons.$(I) [FeF_6]^{-3}$: Here,$Fe$ is in $+3$ oxidation state. The electronic configuration of $Fe^{+3}$ is $[Ar] 3d^5$. Since $F^{-}$ is a weak field ligand,pairing does not occur. Number of unpaired electrons,$n = 5$.$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 	ext{ B.M.}$$(II) [V(H_2O)_6]^{+2}$: Here,$V$ is in $+2$ oxidation state. The electronic configuration of $V^{+2}$ is $[Ar] 3d^3$. Number of unpaired electrons,$n = 3$.$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 	ext{ B.M.}$$(III) [Fe(H_2O)_6]^{+2}$: Here,$Fe$ is in $+2$ oxidation state. The electronic configuration of $Fe^{+2}$ is $[Ar] 3d^6$. Since $H_2O$ is a weak field ligand,pairing does not occur. Number of unpaired electrons,$n = 4$.$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 	ext{ B.M.}$Therefore,the increasing order of spin magnetic moment is $II < III < I$.

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}$