A magnetic moment of 1.73 BM will be shown b | vedclass

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(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 1.73 \ BM$,we have

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$[Cu(NH_3)_4]^{2+}$

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(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 1.73 \ BM$,we have $\sqrt{n(n+2)} = 1.73$,which implies $n = 1$.$1$. $TiCl_4$: $Ti$ is in $+4$ oxidation state $(d^0)$,so $n = 0$.$2$. $[CoCl_6]^{4-}$: $Co$ is in $+2$ oxidation state $(d^7)$. $Cl^-$ is a weak field ligand,so electrons remain unpaired,$n = 3$.$3$. $[Cu(NH_3)_4]^{2+}$: $Cu$ is in $+2$ oxidation state $(d^9)$. In the square planar complex,there is $1$ unpaired electron in the $3d$ orbital,$n = 1$.$4$. $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(d^8)$. $CN^-$ is a strong field ligand,causing pairing,$n = 0$.Thus,$[Cu(NH_3)_4]^{2+}$ has $1$ unpaired electron and a magnetic moment of $1.73 \ BM$.

$A$ magnetic moment of $1.73 \ BM$ will be shown by which of the following?

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