To determine the correct electron configuration for nobelium (No), we need to consider its atomic number and the arrangement of electrons in its orbitals.
Nobelium has an atomic number of 102, meaning it has 102 electrons. These electrons populate the available atomic orbitals according to the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
Here are the options provided:
1. [Rn] 7s² 5f¹⁴
2. [Rn] 7s² 5f⁷
3. [Ne] 3s² 3p⁷
4. [Xe] 6s² 5d¹
Let's analyze each option in detail:
1. [Rn] 7s² 5f¹⁴:
- `[Rn]` refers to the electron configuration of radon (Ra) which has 86 electrons.
- Adding 2 electrons to the 7s orbital: `[Rn] 7s²`
- Adding 14 electrons to the 5f orbital: `[Rn] 7s² 5f¹⁴`
- This configuration correctly accounts for the 102 electrons of nobelium.
2. [Rn] 7s² 5f⁷:
- This configuration suggests that only 7 electrons are in the 5f orbital.
- This does not account for all 102 electrons properly.
3. [Ne] 3s² 3p⁷:
- `[Ne]` represents the electron configuration of neon, which has only 10 electrons.
- Adding electrons would still result in a configuration far fewer than the 102 electrons required by nobelium. Moreover, the 3p orbital cannot hold 7 electrons (as it exceeds its maximum capacity of 6).
4. [Xe] 6s² 5d¹:
- `[Xe]` stands for the electron configuration of xenon, which has 54 electrons.
- Adding 2 electrons to the 6s orbital and 1 electron to the 5d orbital results in 57 electrons, which falls short of the 102 electrons required.
After analyzing all four options, it is clear that the correct electron configuration for nobelium (No) is:
[tex]\[ \text{[Rn] 7s}^2 \text{5f}^{14} \][/tex]
Thus, the correct choice is option 1.
