Question

The maximum number of possible oxidation states of actinides are shown by:
A.) Berkelium(Bk) and Californium(Cf)
B.) Nobelium(No) and Lawrencium(Lr)
C.) Actinium(Ac) and Thorium(Th)
D.) Neptunium(Np) and Plutonium(Pu)

Answer 1

Hint: You know that the actinide series contains elements with atomic numbers 89 to 103 and is the third group in the periodic table. The series is the row below the lanthanide series, which is located underneath the main body of the periodic table. Now try to recall the elements to try to find the oxidation number of elements.

Complete step by step answer:

Actinides: The elements in which the extra electron enters 5f- orbitals of (n-2)th main shall are known as 5f-block elements, actinides, or actinides.

Electronic configuration of actinides: General electronic configuration of actinides is 2,8,18, 32, $5s^{ 2 }p^{ 6 }d^{ 10 }f^{ 0-14 },\quad 6s^{ 2 }p^{ 6 }d^{ 0-2 },\quad 7s^{ 2 }$

Oxidation states of actinide elements: The oxidation state of actinide element is given below:

Ac

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lw

+2

+3

-

-

+3

+3

+3

+3

+3

+3

+3

+3

+3

+3

+3

+3

+4

+4

+4

+4

+4

+4

+4

+4

+5

+5

+5

+5

+5

+5

+6

+6

+6

+6

+7

+7

Here we can see that the maximum oxidation state is +7. It is shown only by Np and Pu.

Therefore, the correct answer to this question is option D.



Note: According to the definition of actinides, only thirteen elements from $Th_{ 90 }(5f^{ 0 }6d^{ 2 }7s^{ 2 })$ to $No_{ 102 }(5f^{ 14 }6d^{ 0 }7s^{ 2 })$ should be the members of the actinide series. However, you should know that all the fifteen elements from $Ac_{ 89 }(5f^{ 0 }6d^{ 1 }7s^{ 2 })$ to $Lw_{ 103 }(5f^{ 14 }6d^{ 1 }7s^{ 2 })$ are considered as the members of the actinide series since all these fifteen elements have same physical and chemical properties. In fact, actinium is the prototype of actinides as lanthanum is the prototype of lanthanides.