Question

give the noble gas electron configuration for bohrium, element 107. 1 72 53 64

Explanation:

Step1: Identify the nearest noble - gas before bohrium

The nearest noble - gas before bohrium (atomic number 107) is radon (Rn) with an atomic number of 86. So the noble - gas core is [Rn].

Step2: Determine the remaining electrons

Bohrium has 107 electrons. After using the 86 electrons from the radon core, there are 107 - 86=21 remaining electrons.

Step3: Fill the orbitals for the remaining electrons

The electron - filling order is \(7s^25f^{14}6d^5\).

Answer:

[Rn] \(7s^2\) \(5f^{14}\) \(6d^5\)
So the blanks should be filled as follows:

1. Rn
2. \(s^2\)
3. \(f^{14}\)
4. \(d^5\)