For the complex ion [Ni(Cl)_4]^{2-}, what is | vedclass

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(B) $1$. In the complex ion $[Ni(Cl)_4]^{2-}$, the oxidation state of $Ni$ is $+2$. $2$. The electronic configuration of $Ni^{2+}$ $(Z=28)$ is $[Ar] 3

Vedclass · Accepted Answer

$2.83$

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(B) $1$. In the complex ion $[Ni(Cl)_4]^{2-}$, the oxidation state of $Ni$ is $+2$. $2$. The electronic configuration of $Ni^{2+}$ $(Z=28)$ is $[Ar] 3d^8$. $3$. Since $Cl^-$ is a weak field ligand, it does not cause pairing of electrons in the $3d$ orbitals. $4$. The $3d^8$ configuration has two unpaired electrons $(n=2)$. $5$. The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$. $6$. Substituting $n=2$, we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.

List-$I$	List-$II$
$(a).$ $[Co(NH_3)_6][Cr(CN)_6]$	$(i).$ Linkage isomerism
$(b).$ $[Co(NH_3)_3(NO_2)_3]$	$(ii).$ Solvate isomerism
$(c).$ $[Cr(H_2O)_6]Cl_3$	$(iii).$ Coordination isomerism
$(d).$ $cis-[CrCl_2(ox)_2]^{3-}$	$(iv).$ Optical isomerism

For the complex ion $[Ni(Cl)_4]^{2-}$, what is the theoretical value of the magnetic moment (in $BM$)?

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