Question 5

Analyze the energy of orbitals in single-electron and multi-electron systems:

1. For a hydrogen atom (a single-electron system), the energy of an orbital depends only on the principal quantum number $n$. Thus, $E_{2s}(\text{H}) = E_{2p}(\text{H})$. Therefore, option (B) is correct.
2. For multi-electron atoms like lithium, the energy of an orbital depends on both the principal quantum number $n$ and the azimuthal quantum number $l$. Due to the penetration and shielding effects, the $2s$ orbital is more stable (lower in energy) than the $2p$ orbital. Thus, $E_{2s}(\text{Li}) < E_{2p}(\text{Li})$. Therefore, option (A) is correct.
3. The energy of orbitals also depends on the nuclear charge ($Z$). As the atomic number increases, the attraction between the nucleus and electrons increases, lowering the energy of the orbitals (making them more negative). Since Lithium ($Z=3$) has a higher nuclear charge than Hydrogen ($Z=1$), the $n=2$ level in Li is significantly lower in energy than the $n=2$ level in H. Thus, $E_{2s}(\text{H}) > E_{2s}(\text{Li})$ and $E_{2p}(\text{H}) > E_{2p}(\text{Li})$. This makes option (D) correct and option (C) incorrect.