Explanation:
The actinides can be identified based on the outermost orbitals. The valence electrons of the actinides enter into the 5f-orbital.
Electronic configuration of curium:- [Rn] $${5f}^{7} {6d}^{1}$$
The energy of 6d orbitals is less than the 5f orbital, so the electrons first enter into 6d orbitals, and then the remaining valence electrons enter into 5f orbital. Thus, Curium is an actinoid.
Electronic configuration of californium:- [Rn] $${5f}^{10} {7s}^{2}$$
The energy of 7s orbitals is less than the 5f orbital, so the electrons first enter into 7s orbitals, and then the remaining valence electrons enter into 5f orbital. Thus, Californium is an actinoid.
Electronic configuration of uranium:- [Rn] $${5f}^{3}$$
The valence electrons of the uranium element enter into 5f-orbital so the uranium is an actinoid.
Electronic configuration of terbium:- [Xe] $${4f}^{9} {6s}^{2}$$
The energy of 6s orbitals is less than the 4f orbital, so the electrons first enter into 6s orbitals, and then the remaining valence electrons enter into 4f orbital. Thus, Terbium is lanthanoid and not actinide.
Correct Option: $$D$$