What is the correct order of magnetic moment | vedclass

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(C) $1$. Calculate the number of unpaired electrons $(n)$ for each complex:- $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. $Cl^-$ is a weak field ligand,so $n

Vedclass · Accepted Answer

$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$

Vedclass · Answer

(C) $1$. Calculate the number of unpaired electrons $(n)$ for each complex:- $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. $Cl^-$ is a weak field ligand,so $n = 5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ B.M.$- $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. $Cl^-$ is a weak field ligand,so $n = 3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$- $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,so $n = 0$ (all electrons paired). Magnetic moment $\mu = 0 \ B.M.$$2$. Comparing the values: $5.92 > 3.87 > 0$.$3$. Therefore,the order is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

What is the correct order of magnetic moment (spin-only value in $B.M.$)? (Atomic numbers: $Mn = 25, Fe = 26, Co = 27$)

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