The species with spin only magnetic moment of | vedclass

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

Question

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

Vedclass · Accepted Answer

$[CoF_3(H_2O)_3]$

Vedclass · Answer

(A) The spin only magnetic moment is given by the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu_{eff} = \sqrt{24} \ BM$,we have $n(n+2) = 24$,which gives $n = 4$.In $[CoF_3(H_2O)_3]$,$Co$ is in the $+3$ oxidation state ($3d^6$ configuration). $F^-$ and $H_2O$ are weak field ligands,so the electrons remain unpaired in the $t_{2g}$ and $e_g$ orbitals,resulting in $4$ unpaired electrons.In $[CoCl_4]^{2-}$,$Co$ is in the $+2$ oxidation state ($3d^7$ configuration),resulting in $3$ unpaired electrons.In $[NiCl_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration),resulting in $2$ unpaired electrons.In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration),resulting in $2$ unpaired electrons.Thus,$[CoF_3(H_2O)_3]$ is the correct species.

The species with spin only magnetic moment of $\sqrt{24} \ BM$ is

Similar Questions

Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

If there is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is $10 \text{ cm}$, is

Which of the following radioisotopes is used as anticancerous?

Insectivorous plants live in a soil that is usually deficient in

$Zn$ and $Cd$ metals do not show variable valency because

Vedclass Test Series

Exam Paper Generator

Online Exam Module