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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Which of the following complexes will exhibit the highest number of isomers?

The complexes $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$ can be distinguished by which of the following reagents?

Which of the following square-planar complexes exhibits $cis-trans$ isomerism?

The complex(es),which can exhibit the type of isomerism shown by $[Pt(NH_3)_2 Br_2]$,is(are) [en $= H_2 NCH_2 CH_2 NH_2$ ]
$(A)$ $[Pt(en)(SCN)_2]$
$(B)$ $[Zn(NH_3)_2 Cl_2]$
$(C)$ $[Pt(NH_3)_2 Cl_4]$
$(D)$ $[Cr(en)_2(H_2 O)(SO_4)]^{+}$

Calculate the magnetic moment of the $Fe^{3+}$ ion in the $[Fe(CN)_6]^{3-}$ complex.

Vedclass Test Series

Exam Paper Generator

Online Exam Module