A) Curium
B) Californium
C) Uranium
D) Terbium
Correct Answer: D
Solution :
Terbium is lanthanide as it belongs to 4\[f\] -series having configuration \[[Xe]\,\,4{{f}^{9}},\,\,6{{s}^{2}}\]. However, the remaining members belong to 5\[f\]-series (actinides).
You need to login to perform this action.
You will be redirected in
3 sec
