Question

General electronic configuration of
actinoids is
$(n–2)f^{1–4}(n-1)d^{0–2}n{s}^2.$ Which
of the following actinoids have one
electron in 6d orbital?

• A. Am
  (Atomic no. 95)
• B. U
  (Atomic no. 92)
• C. Pu
  (Atomic no. 94)
  
• D. Np
  (Atomic no.93)
  

Answer 1

Answer: (B), (D)

Np

(Atomic no.93)

Solution

Analyzing the given options

Option (A)
The atomic number of $U\left(uranium\right)$ is 92.
The electronic configuration of $U\left(uranium\right)$
can be written as [Rn]$5f^{3}6d^{1}7s^2$

Option (B)

• The atomic number of Np (neptunium) is 93. The electronic configuration of Np$\left(Neptunium\right)$ can be written as $\left[Rn\right]5f^{4}6d^{1}7s^2$

• Hence, $U\left(uranium\right)andNp\left(Neptunium\right)$ both have one electron in their 6d orbital

Option(C)

The atomic number of $Pu\left(Plutonium\right)$ is
94.
The electronic configuration of $Pu\left(plutonium\right)$ can be written as
$\left[Rn\right]5f^{6}6d^{0}7s^2$

Option D

• The atomic number of Am (americium) is 95. The electronic configuration of Am$\left(Americium\right)$ can be written as $\left[Rn\right]5f^{7}6d^{0}7s^2$
• Hence, Pu (plutonium) ) and Am |((Americium)\) have zero electron in their 6d orbitals
• Thus, the correct options are (A) and (B).