The magnetic moment of [MnX_4]^{2-} is 5.9 , | vedclass

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

Question

(A) The magnetic moment is given as $5.9 \, BM$. Using the spin-only formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons

Vedclass · Accepted Answer

tetrahedral

Vedclass · Answer

(A) The magnetic moment is given as $5.9 \, BM$. Using the spin-only formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons,we find $n = 5$. In $[MnX_4]^{2-}$,the oxidation state of $Mn$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. Since there are $5$ unpaired electrons,the $d$-orbitals are $d^5$ (high spin). For a $d^5$ configuration with $4$ ligands,the geometry is tetrahedral because the halide ligands are weak field ligands,leading to $sp^3$ hybridization.

The magnetic moment of $[MnX_4]^{2-}$ is $5.9 \, BM$. The geometry of the complex ion is: ($X =$ monodentate halide ion)

Similar Questions

Assertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.

In a square planar complex of $Co^{2+}$ ion,the unpaired electron is present in which orbital?

According to Werner’s theory,the geometry of the complex is determined by:

Which complex compound possesses $sp^3d^2$ hybridisation?

Select the incorrect match.

Vedclass Test Series

Exam Paper Generator

Online Exam Module