The calculated spin-only magnetic moment valu | vedclass

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

Question

(A) For both complexes,the central metal ion is $Fe^{3+}$.Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.In $[FeCl_4]^-$,$Cl^-$ is a weak field

Vedclass · Accepted Answer

$5.9 \ BM, 1.732 \ BM$

Vedclass · Answer

(A) For both complexes,the central metal ion is $Fe^{3+}$.Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.In $[FeCl_4]^-$,$Cl^-$ is a weak field ligand,so no pairing of electrons occurs. The number of unpaired electrons $(n)$ is $5$.The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \ BM$.In $[Fe(CN)_6]^{3-}$,$CN^-$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^5 e_g^0$,leaving $n = 1$ unpaired electron.The spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.Thus,the values are $5.9 \ BM$ and $1.732 \ BM$.

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

Similar Questions

Which of the following outer octahedral complexes have the same number of unpaired electrons?
$1$. $[MnCl_6]^{3-}$
$2$. $[FeF_6]^{3-}$
$3$. $[CoF_6]^{3-}$
$4$. $[Ni(NH_3)_6]^{2+}$

The metal complex that is diamagnetic is (Atomic number: $Fe = 26, Cu = 29$)

$Fac$ and $Mer$ isomerism is associated with which of the following general formula?

Arrange the following complexes in the increasing order of their spin only magnetic moment (in $B.M$):
$I$. $[Fe(CN)_6]^{4-}$
$II$. $[MnCl_4]^{2-}$
$III$. $[Mn(CN)_6]^{4-}$
$IV$. $[Cr(NH_3)_6]^{3+}$

Magnetic moment of $d^6$ low spin complex is ........ $B.M.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module