The species with spin only magnetic moment of | vedclass

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(A) The spin only magnetic moment is given by the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu_{eff

Vedclass · Accepted Answer

$[CoF_3(H_2O)_3]$

Vedclass · Answer

(A) The spin only magnetic moment is given by the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu_{eff} = \sqrt{24} \ BM$,we have $n(n+2) = 24$,which gives $n = 4$.In $[CoF_3(H_2O)_3]$,$Co$ is in the $+3$ oxidation state ($3d^6$ configuration). $F^-$ and $H_2O$ are weak field ligands,so the electrons remain unpaired in the $t_{2g}$ and $e_g$ orbitals,resulting in $4$ unpaired electrons.In $[CoCl_4]^{2-}$,$Co$ is in the $+2$ oxidation state ($3d^7$ configuration),resulting in $3$ unpaired electrons.In $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration),resulting in $2$ unpaired electrons.In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration),resulting in $2$ unpaired electrons.Thus,$[CoF_3(H_2O)_3]$ is the correct species.

Column-$I$	Column-$II$
$a$. Superior Vena Cava	$p$. Carries deoxygenated blood to the lungs
$b$. Inferior Vena Cava	$q$. Carries oxygenated blood to the heart
$c$. Pulmonary Artery	$r$. Brings deoxygenated blood from lower body parts to the right atrium
$d$. Pulmonary Vein	$s$. Brings deoxygenated blood from upper body parts to the right atrium

The species with spin only magnetic moment of $\sqrt{24} \ BM$ is

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