Magnetic moment of Gd^{3+} ion (Z = 64) is .. | vedclass

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(C) The electronic configuration of $Gd$ $(Z = 64)$ is $[Xe] 4f^7 5d^1 6s^2$. For the $Gd^{3+}$ ion,three electrons are removed,resulting in the confi

Vedclass · Accepted Answer

$7.9$

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

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

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