Which of the following compounds has a magnet | vedclass

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(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15

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$[Cr(CN)_6]^{-3}$

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(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15} \ B.M.$,we have $n(n+2) = 15$,which implies $n^2 + 2n - 15 = 0$. Solving this quadratic equation gives $(n+5)(n-3) = 0$,so $n = 3$.Now,let us analyze the complexes:$A$: In $[Fe(H_2O)_6]^{+2}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand,so $Fe^{+2}$ has $4$ unpaired electrons $(n=4)$.$B$: In $[Cr(CN)_6]^{-3}$,$Cr$ is in $+3$ oxidation state $(3d^3)$. $CN^-$ is a strong field ligand,so $Cr^{+3}$ has $3$ unpaired electrons $(n=3)$.$C$: In $[Ni(NH_3)_6]^{+2}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $NH_3$ is a strong field ligand,so $Ni^{+2}$ has $2$ unpaired electrons $(n=2)$.$D$: In $[PtCl_4]^{-2}$,$Pt$ is in $+2$ oxidation state $(5d^8)$. $Pt$ is a $5d$ series metal,so it forms square planar complexes with $0$ unpaired electrons $(n=0)$.Thus,$[Cr(CN)_6]^{-3}$ has $3$ unpaired electrons,corresponding to $\sqrt{15} \ B.M.$

Which of the following compounds has a magnetic moment of $\sqrt{15} \ B.M.$?

Similar Questions

The compounds $[Cr(H_2O)_6]Cl_3$,$[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ and $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$ exhibit:

Among $(a) - (d)$,the complexes that can display geometrical isomerism are:
$(a) [Pt(NH_3)_3 Cl]^+$
$(b) [Pt(NH_3) Cl_5]^-$
$(c) [Pt(NH_3)_2 Cl(NO_2)]$
$(d) [Pt(NH_3)_4 ClBr]^{2+}$

What is the spin-only magnetic moment (in Bohr Magneton units) of $Ni^{2+}$ in an aqueous solution? (Atomic number of $Ni = 28$)

The calculated spin-only magnetic moment values in $BM$ for $[FeCl_4]^-$ and $[Fe(CN)_6]^{3-}$ are

The coordination number,oxidation number,number of electrons in $d$-orbital,and number of unpaired electrons respectively in the complex $[Co(NH_3)_4CO_3]ClO_4$ are . . . . . . .

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