The magnetic moment of Gd^{3+} is ....... B.M | vedclass

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Question

(A) The atomic number of $Gd$ is $64$. The electronic configuration of $Gd$ is $[Xe] 4f^7 5d^1 6s^2$. For $Gd^{3+}$,three electrons are removed,result

Vedclass · Accepted Answer

$7.9$

Vedclass · Answer

(A) The atomic number of $Gd$ is $64$. The electronic configuration of $Gd$ is $[Xe] 4f^7 5d^1 6s^2$. For $Gd^{3+}$,three electrons are removed,resulting in the configuration $[Xe] 4f^7$. The number of unpaired electrons $(n)$ is $7$. The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$. Substituting $n = 7$: $\mu = \sqrt{7(7+2)} = \sqrt{7 	imes 9} = \sqrt{63} \approx 7.94 \ B.M.$ Thus,the closest value is $7.9 \ B.M.$

The magnetic moment of $Gd^{3+}$ is ....... $B.M.$ $(Z = 64)$

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