The magnetic moment of a complex ion is 2.83 | vedclass

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

Question

(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$

Vedclass · Accepted Answer

$[V(H_2O)_6]^{3+}$

Vedclass · Answer

(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$,we have $\sqrt{n(n+2)} = 2.83$,which implies $n(n+2) \approx 8$,so $n = 2$.Let us analyze the number of unpaired electrons in each complex:$A$: In $[V(H_2O)_6]^{3+}$,$V^{3+}$ is $3d^2$,so $n = 2$.$B$: In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,so $n = 3$.$C$: In $[Cu(CN)_4]^{2-}$,$Cu^{2+}$ is $3d^9$,so $n = 1$.$D$: In $[MnCl_4]^{2-}$,$Mn^{2+}$ is $3d^5$ (high spin),so $n = 5$.Thus,the complex with $2$ unpaired electrons is $[V(H_2O)_6]^{3+}$.

The magnetic moment of a complex ion is $2.83 \ BM$. The complex ion is

Similar Questions

The correct order of reactivity of hydrogen halides with ethyl alcohol is

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is

$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \, m$. The speed with which he throws the ball is (to the nearest integer) (in $m/s$).

Select incorrect sentence from below.

$A$ particle of charge $q$ and mass $m$ moves in a circular orbit of radius $r$ with angular speed $\omega$. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Vedclass Test Series

Exam Paper Generator

Online Exam Module