Question

which is the electron configuration for nobelium (no)?
○ rn7s²5f¹⁴
○ rn7s²5f⁷
○ ne3s²3p⁷
○ xe6s²5d¹

Explanation:

1. First, determine the atomic number of nobelium (No). Nobelium has an atomic number of 102.
2. Radon (Rn) has an atomic number of 86. So, the electron configuration should build on Rn's configuration.
3. The remaining electrons after Rn (102 - 86 = 16) are filled in the next orbitals. The order of filling for f - block elements (lanthanides and actinides) involves the 5f orbital after the 7s orbital.
4. The 7s orbital can hold 2 electrons, and the 5f orbital can hold up to 14 electrons. For No (atomic number 102), after filling Rn's configuration $[\text{Rn}]$, we fill $7\text{s}^2$ (2 electrons) and then $5\text{f}^{14}$ (14 electrons, since 2 + 14 = 16, which is the number of electrons needed after Rn's 86 electrons to reach 102).
5. The other options are incorrect:

• $[\text{Rn}]7\text{s}^25\text{f}^7$ would correspond to an element with fewer electrons (86 + 2+7 = 95, not 102).
• $[\text{Ne}]3\text{s}^23\text{p}^7$ is invalid because the p - orbital can hold a maximum of 6 electrons, and also Ne has atomic number 10, so this would not be for No.
• $[\text{Xe}]6\text{s}^25\text{d}^1$ corresponds to an element with atomic number 54 (Xe) + 2+1 = 57, not 102.

Answer:

A. $[\text{Rn}]7\text{s}^25\text{f}^{14}$