Final answer:
The total number of electrons in the s and p orbitals for Helium, Boron, Nitrogen, and Sodium are 2 (2s, 0p), 3 (2s, 1p), 5 (2s, 3p), and 8 (2s, 6p) respectively, according to their electronic configurations.
Step-by-step explanation:
To determine the total number of electrons in the s and p orbitals for helium (He), boron (B), nitrogen (N), and sodium (Na), we look at their electronic configurations. These configurations tell us how electrons are distributed among the orbitals.
- Helium (He): The electron configuration is 1s². Helium has 2 electrons in its s orbital and no electrons in p orbitals because it has only one energy level.
- Boron (B): The electron configuration of boron is [He]2s²2p¹. Boron has 2 electrons in its s orbital and 1 electron in its p orbital.
- Nitrogen (N): The electron configuration of nitrogen is [He]2s²2p³. Nitrogen has 2 electrons in its s orbital and 3 electrons in its p orbitals.
- Sodium (Na): The electron configuration of sodium is [Ne]3s¹. In the context of s and p orbitals, we ignore the 3s electron, and sodium has 2 electrons in its s orbital and 6 electrons in its p orbitals (from the second energy level).