The electronic configurations of Cr^{2+}, Mn^ | vedclass

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

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$[Co(H_2O)_6]^{2+}$

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(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.All the given complexes are octahedral complexes with $H_2O$ as a weak field ligand,so the number of unpaired electrons remains the same as in the free metal ions.For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $d^4$,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 	ext{ BM}$.For $[Mn(H_2O)_6]^{2+}$,$Mn^{2+}$ is $d^5$,$n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 	ext{ BM}$.For $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $d^6$,$n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 	ext{ BM}$.For $[Co(H_2O)_6]^{2+}$,$Co^{2+}$ is $d^7$,$n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 	ext{ BM}$.Comparing the values,$[Co(H_2O)_6]^{2+}$ has the lowest magnetic moment.

The electronic configurations of $Cr^{2+}, Mn^{2+}, Fe^{2+},$ and $Co^{2+}$ are $d^4, d^5, d^6,$ and $d^7$ respectively. Which of the following has the lowest magnetic moment? $[Cr = 24, Mn = 25, Fe = 26, Co = 27]$

Similar Questions

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$X$ is the number of geometrical isomers exhibited by $[Pt(NH_3)(H_2O)BrCl]$.
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Magnetic moment of $d^6$ low spin complex is ........ $B.M.$

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