Which is higher in energy: the \(2 s\) or \(2 p\) orbital in hydrogen? Is this also true for helium? Explain.

Electron configuration refers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. An atomic orbital is an area in which an electron is most likely to be found. Electrons fill orbitals according to their energy levels, starting with the lowest energy orbitals first (Aufbau principle). Step 2: Energy levels of orbitals

For a single-electron atom like hydrogen, the energy level of an orbital can be determined by its principal quantum number (n). Orbitals with the same value of n have the same energy level, which means that for hydrogen, the \(2s\) and \(2p\) orbitals will have the same energy level because they both have n=2. Step 3: Determining the energy levels for multi-electron atoms

For multi-electron atoms, like helium, the energy of an orbital is determined by both the principal quantum number (n) and the azimuthal quantum number (l). In this case, energy increases according to the value of (n+l), and for the same value, energy increases with increasing n. For helium: - The \(2s\) orbital has n=2 and l=0. - The \(2p\) orbital has n=2 and l=1. Considering the rule (n+l), both orbitals have the same value (2). However, for the same value, the energy level increases with n. Since n is the same for both \(2s\) and \(2p\) orbitals in helium, their energy levels are the same. #Conclusion#: For hydrogen, which is a single-electron atom, the energy levels of the \(2s\) and \(2p\) orbitals are the same. For helium, which is a multi-electron atom, the energy levels of the \(2s\) and \(2p\) orbitals are also the same due to the same principal quantum number and the rule of increasing energy with (n+l).