The spin-only magnetic moment of [MnBr_4]^{x- | vedclass

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

Question

(D) The spin-only magnetic moment is given by the formula $\mu_{s} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu_{s}

Vedclass · Accepted Answer

tetrahedral,$2$

Vedclass · Answer

(D) The spin-only magnetic moment is given by the formula $\mu_{s} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu_{s} = 5.9 \ BM$,we have $\sqrt{n(n+2)} = 5.9$.Squaring both sides,$n(n+2) \approx 34.81$,which gives $n = 5$.Since $Mn$ $(Z=25)$ has the electronic configuration $[Ar] 3d^5 4s^2$,to have $5$ unpaired electrons,$Mn$ must be in the $+2$ oxidation state $(Mn^{2+}: [Ar] 3d^5)$.$Br^-$ is a weak-field ligand,so it does not cause pairing of electrons,resulting in $sp^3$ hybridization.For the complex $[MnBr_4]^{x-}$,the oxidation state of $Mn$ is $x-4 = -2$,so $x = 2$.The geometry of $sp^3$ hybridized complexes is tetrahedral.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

The spin-only magnetic moment of $[MnBr_4]^{x-}$ is $5.9 \ BM$. The geometry of the complex and $x$ respectively are

Similar Questions

Which of the following statements is incorrect?

The hybridization of the central atom in $NH_3, [PtCl_4]^{2-}, PCl_5$,and $BCl_3$ respectively is:

Which statement is correct?

Identify the number of unpaired electrons present and the geometry of the $[Co(NH_3)_6]^{3+}$ complex,respectively.

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module