Question

The \(2s\) and the \(2p\) orbital energies of hydrogen atom are \(E_{2s}(\mathrm{H})\) and \(E_{2p}(\mathrm{H})\), respectively. The \(2s\) and the \(2p\) orbital energies of lithium atom are \(E_{2s}(\mathrm{Li})\) and \(E_{2p}(\mathrm{Li})\), respectively. The correct option(s) about the orbital energies is(are)

• A. \(E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li})\)
• B. \(E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H})\)
• C. \(E_{2p}(\mathrm{H}) < E_{2s}(\mathrm{Li})\)
• D. \(E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li})\)

Hint:

For hydrogen-like species: \[ \text{Energy depends only on } n \] Hence, \[ 2s = 2p \] For multi-electron atoms: \[ \text{Energy depends on both } n \text{ and } l \] Due to better penetration: \[ s < p < d < f \] in energy for the same principal quantum number.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Orbital energies depend on the nuclear charge (\(Z\)) and the presence of other electrons. - Hydrogen-like species (single electron): The energy of an orbital depends only on the principal quantum number (\(n\)). Orbitals with the same \(n\) but different \(l\) are degenerate. - Multi-electron atoms (like Li): Due to electron-electron repulsion and the screening/shielding effect, the energy depends on both \(n\) and \(l\). s-orbitals are more penetrating and less shielded than p-orbitals of the same shell.
Step 3: Detailed Explanation:
1. Hydrogen atom (B): Since it has only one electron, energy is determined only by \(n\). Thus, \(E_{2s}(H) = E_{2p}(H)\). Statement (B) is True. 2. Lithium atom (A): In multi-electron atoms, the 2s electron is more penetrating and closer to the nucleus than the 2p electron. It experiences a higher effective nuclear charge (\(Z_{eff}\)) and lower shielding. Thus, the 2s orbital is lower in energy. \(E_{2s}(Li)<E_{2p}(Li)\). Statement (A) is True. 3. Comparison between H and Li (D): Energy is proportional to \(-Z_{eff}^2 / n^2\). For H, \(Z=1\). For Li, the valence 2s electron experiences a \(Z_{eff}>1\) (though screened by 1s electrons, it's still higher than 1). Also, the absolute value of the energy of an electron in Li is much more negative (more stable). In chemistry, "higher energy" usually means less stable (closer to zero). \(E = -13.6 \times (Z_{eff}^2/n^2)\) eV. For H: \(E_{2s} = -13.6/4 = -3.4\) eV. For Li: \(E_{2s} \approx -5.39\) eV. Thus, \(-3.4>-5.39\). So \(E_{2s}(H)>E_{2s}(Li)\). Statement (D) is True.
Step 4: Final Answer:
The degeneracy in hydrogen is removed in multi-electron atoms due to shielding. Higher nuclear charge leads to more negative (lower) energy levels.