Question

The electronic configuration of $Eu$ (Atomic No. $63$), $Gd$ (Atomic No. $64$) and $Tb$ (Atomic No. $65$) are :

• A. $\left[Xe\right]4f^{7}6s^2,\left[Xe\right]4f^{8}6s^2$ and $\left[Xe\right]4f^{8}5d^{1}6s^2$
• B. $\left[Xe\right]4f^{7}5d^{1}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2$ and $\left[Xe\right]4f^{9}6s^2$
• C. $\left[Xe\right]4f^{6}5d^{1}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2$ and $\left[Xe\right]4f^{8}5d^{1}6s^2$
• D. $\left[Xe\right]4f^{7}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2$ and $\left[Xe\right]4f^{9}6s^2$

Answer 1

Answer: (D)

$\left[Xe\right]4f^{7}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2$ and $\left[Xe\right]4f^{9}6s^2$

The electronic configuration of $Eu$ (Atomic No. 63), $Gd$ (Atomic No. 64) and $Tb$ (Atomic No. 65) are :

$\left[Xe\right]4f^{7}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2$ and $\left[Xe\right]4f^{9}6s^2$ respectively.

$Gd$ possesses a single electron in the $5d$ subshell. In the case of other lanthanoids, the $5d$ orbital is empty.

Hence, the correct option is D.