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(C) $(i)$ $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong-field ligand,causing pairing. Unpaired electrons = $1$.$(ii)

Vedclass · Accepted Answer

$iv, i, ii, iii$

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(C) $(i)$ $[Fe(CN)_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ is a strong-field ligand,causing pairing. Unpaired electrons = $1$.$(ii)$ $[MnCl_6]^{3-}$: $Mn$ is in $+3$ oxidation state $(3d^4)$. $Cl^-$ is a weak-field ligand. Unpaired electrons = $4$.$(iii)$ $[FeF_6]^{3-}$: $Fe$ is in $+3$ oxidation state $(3d^5)$. $F^-$ is a weak-field ligand. Unpaired electrons = $5$.$(iv)$ $[Co(NH_3)_6]^{3+}$: $Co$ is in $+3$ oxidation state $(3d^6)$. $NH_3$ is a strong-field ligand,causing pairing. Unpaired electrons = $0$.Increasing order of unpaired electrons: $(iv) < (i) < (ii) < (iii)$.

Arrange the following in order of increasing number of unpaired electrons in them:
$i$. $[Fe(CN)_6]^{3-}$
$ii$. $[MnCl_6]^{3-}$
$iii$. $[FeF_6]^{3-}$
$iv$. $[Co(NH_3)_6]^{3+}$

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Arrange the following in order of increasing number of unpaired electrons in them:$i$. $[Fe(CN)_6]^{3-}$$ii$. $[MnCl_6]^{3-}$$iii$. $[FeF_6]^{3-}$$iv$. $[Co(NH_3)_6]^{3+}$