Question

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

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

Answer 1

The correct option is A $\left[Xe\right]4f^{7}6s^2,\left[Xe\right]4f^{7}5d^{1}6s^2\text{and}\left[Xe\right]4f^{9}6s^2$

${}_{63}Eu\to \left[Xe\right]4f^{7}6s^2$
${}_{64}Gd\to \left[Xe\right]4f^{7}5d^{1}6s^2$
${}_{65}Tb\to \left[Xe\right]4f^{9}5d^{0}6s^2$
$Gd$ possesses a single electron in the 5d subshell. In the case of other lanthanoids, the 5d orbital is empty.
Option (a) is correct