The magnetic moments of the complexes given b | vedclass

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(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$I. [Ni(CO)_4]$:

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$I < II < III < IV$

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(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$I. [Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. All electrons are paired,so $n = 0$.$II. [Mn(CN)_6]^{4-}$: $Mn$ is in $+2$ oxidation state $(3d^5)$. $CN^-$ is a strong field ligand,causing pairing. $n = 1$.$III. [Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state $(3d^3)$. $n = 3$.$IV. [CoF_6]^{3-}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,no pairing. $n = 4$.Comparing the number of unpaired electrons: $I (0) Thus,the order of magnetic moments is $I < II < III < IV$.

The magnetic moments of the complexes given below are in the order:
$I. [Ni(CO)_4]$
$II. [Mn(CN)_6]^{4-}$
$III. [Cr(NH_3)_6]^{3+}$
$IV. [CoF_6]^{3-}$

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The magnetic moments of the complexes given below are in the order:$I. [Ni(CO)_4]$$II. [Mn(CN)_6]^{4-}$$III. [Cr(NH_3)_6]^{3+}$$IV. [CoF_6]^{3-}$