The complex showing a spin-only magnetic mome | vedclass

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

Question

(B) The spin-only magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $\mu = 2.82 \ B.M.$,$n=2

Vedclass · Accepted Answer

$\left[NiCl_4\right]^{2-}$

Vedclass · Answer

(B) The spin-only magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $\mu = 2.82 \ B.M.$,$n=2$.$(A)$ $Ni(CO)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired $(n=0)$.$(B)$ $\left[NiCl_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. The $3d$ orbitals have $2$ unpaired electrons $(n=2)$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$$(C)$ $Ni\left(PPh_3\right)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $PPh_3$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.$(D)$ $\left[Ni(CN)_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.

The complex showing a spin-only magnetic moment of $2.82 \ B.M.$ is

Similar Questions

The calculated magnetic moments (spin only value) for species $[FeCl_{4}]^{2-}$,$[Co(C_{2}O_{4})_{3}]^{3-}$ and $MnO_{4}^{2-}$ respectively are ..... .

Which of the following is an optically active isomer?

Although $Cr^{3+}$ and $Co^{2+}$ ions have the same number of unpaired electrons,why is the magnetic moment of $Cr^{3+}$ $3.87 \ B.M.$ and that of $Co^{2+}$ $4.87 \ B.M.?$

$[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ are a pair of isomers.

The possible number of stereoisomers for $[Co(en)_2Cl_2]^+$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module