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 $\sqrt{15} \ B.M.$

$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

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

Similar Questions

An aqueous solution of $[Ni(H_2O)_6]^{2+}$ contains $Ni^{2+}$ ions and has a magnetic moment of $2.83 \ BM$. What is the effect on the magnetic moment of the solution when ammonia is added?

Which one of the following does not exhibit geometrical isomerism?

Which of the following molecules/ions can exhibit isomerism:
$A$. Tetrahedral $NiCl_2Br_2^{2-}$ $B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$ $D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Here,$en = 1,2-\text{diaminoethane}$.

Among $TiF_6^{2-}$,$CoF_6^{3-}$,$Cu_2Cl_2$ and $NiCl_4^{2-}$,the colourless species are:

Draw all the isomers (geometrical and optical) of:
$(i)$ $[CoCl_2(en)_2]^+$
$(ii)$ $[Co(NH_3)Cl(en)_2]^{2+}$
$(iii)$ $[Co(NH_3)_2Cl_2(en)]^+$

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