The calculated spin-only magnetic moments of | vedclass

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(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).

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