To determine the electron configuration of nobelium (No), we need to know its atomic number, which is 102. The electron configuration describes the distribution of electrons in an atom's orbitals.
Let's build up the electron configuration step-by-step:
1. Core electrons (inner shells):
- The electron configuration for argon (Ar) is: [tex]\([Ne] 3s^2 3p^6\)[/tex].
- Moving forward from argon: [tex]\([Ar] 4s^2 3d^{10} 4p^6\)[/tex] which is krypton [tex]\([Kr]\)[/tex].
- Continuing: [tex]\([Kr] 5s^2 4d^{10} 5p^6\)[/tex] which is xenon [tex]\([Xe]\)[/tex].
- Next: [tex]\([Xe] 6s^2 4f^{14} 5d^{10} 6p^6\)[/tex] which is radon [tex]\([Rn]\)[/tex].
2. Outermost electrons:
- Following [tex]\([Rn]\)[/tex], we move into the period 7 orbitals:
- The 7s orbital is filled first: [tex]\([Rn] 7s^2\)[/tex].
- Then the electrons fill the 5f orbital. Since nobelium is part of the f-block, the electrons fill the 5f sublevel.
For nobelium:
- After [tex]\([Rn] 7s^2\)[/tex], the next 14 electrons will fill up the 5f orbital completely.
So, the electron configuration for nobelium (No) is:
[tex]\[
[Rn] 7s^2 5f^{14}
\][/tex]
Hence, the correct answer is:
[tex]\[ [Rn] 7s^2 5f^{14} \][/tex]
This matches the first option provided in the question.
