Question

which of the following is not an actinide?
A. Curium ( Z = 96)
B. Californium ( Z = 98)
C. Uranium ( Z = 92)
D. Terbium ( Z = 65)

Answer 1

Hint: The actinides can be identified based on the outermost orbitals. The valence electrons of the actinides enter into the 5f-orbital. The outermost orbitals help us to find the group from which they belong.

Complete answer:
S, p, d and f are four orbitals in which the electron enters. All these orbitals contain electrons in an element but the orbitals that receive the last electrons are known as valence orbitals.
Valence orbital of actinides is 5f-orbital.
The electronic configuration of the curium element is as follows:
$\left[ {{\text{Rn}}} \right]\,{\text{5}}{{\text{f}}^{\text{7}}}\,{\text{6}}{{\text{d}}^1}$
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.
The valence electrons of the curium element enter into the 5f-orbital so the curium is an actinide. So, option (A) is incorrect.
The electronic configuration of the californium element is as follows:
$\left[ {{\text{Rn}}} \right]\,{\text{5}}{{\text{f}}^{10}}\,{\text{7}}{{\text{s}}^{\text{2}}}$
The s-orbitals are completely filled and the f-orbital has ten electrons. 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.
The valence electrons of the californium element enter into 5f-orbital so the californium is an actinide. So, option (B) is incorrect.
The electronic configuration of the uranium element is as follows:
$\left[ {{\text{Rn}}} \right]\,{\text{5}}{{\text{f}}^3}$
The valence electrons of the uranium element enter into 5f-orbital so the uranium is an actinide. So, option (C) is incorrect.
The electronic configuration of the terbium element is as follows:
$\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^9}\,{\text{6}}{{\text{s}}^{\text{2}}}$
The s-orbitals are completely filled and the f-orbital has nine electrons. 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.
The valence electrons of the terbium element enter into 4f-orbital so the terbium is not an actinide. It is a lanthanide. So, option (D) is correct.

Note: F-block elements are divided into two periods. F-block elements from atomic numbers 58 - 71 are known as lanthanides and elements from atomic numbers 90 - 103 are known as actinides. So, terbium is a lanthanide.