The calculated spin-only magnetic moments of | vedclass

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

Question

(A) For $[Cr(NH_3)_6]^{3+}$:$Cr$ $(Z=24)$ has configuration $[Ar] 3d^5 4s^1$. Thus,$Cr^{3+}$ is $[Ar] 3d^3$.Number of unpaired electrons $(n)$ = $3$.$

Vedclass · Accepted Answer

$3.87$ and $2.84$

Vedclass · Answer

(A) For $[Cr(NH_3)_6]^{3+}$:$Cr$ $(Z=24)$ has configuration $[Ar] 3d^5 4s^1$. Thus,$Cr^{3+}$ is $[Ar] 3d^3$.Number of unpaired electrons $(n)$ = $3$.$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 	ext{ B.M.}$For $[CuF_6]^{3-}$:$Cu$ $(Z=29)$ has configuration $[Ar] 3d^{10} 4s^1$. Thus,$Cu^{3+}$ is $[Ar] 3d^8$.In an octahedral field,$3d^8$ configuration has $2$ unpaired electrons in the $e_g$ orbitals.Number of unpaired electrons $(n)$ = $2$.$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 	ext{ B.M.}$Therefore,the values are $3.87 	ext{ B.M.}$ and $2.84 	ext{ B.M.}$ respectively.

The calculated spin-only magnetic moments of $[Cr(NH_3)_6]^{3+}$ and $[CuF_6]^{3-}$ in $BM$,respectively,are (Atomic numbers of $Cr$ and $Cu$ are $24$ and $29$,respectively).

Similar Questions

The electronic configurations of $Cr^{2+}, Mn^{2+}, Fe^{2+}$ and $Ni^{2+}$ ions are $3d^4, 3d^5, 3d^6$ and $3d^8$ respectively. Which of the following aqua complexes exhibits the minimum paramagnetic behavior?

Which of the following complex ions will exhibit paramagnetic behavior?

The complex compounds $[Co(NH_3)_5 SO_4] Br$ and $[Co(NH_3)_5 Br] SO_4$ are

The spin-only magnetic moment of $[Mn(H_2O)_6]^{2+}$ complex is $........ \ B.M.$ (Nearest integer) (Given: Atomic number of $Mn$ is $25$)

Explain paramagnetism and diamagnetism.

Vedclass Test Series

Exam Paper Generator

Online Exam Module