Comparison of Orbital Energies in Hydrogen and Lithium Atoms

To determine the correct options regarding orbital energies, let us analyze the behavior of single-electron and multi-electron atomic systems.

1. Hydrogen Atom ($\text{H}$): Single-Electron System

   • For single-electron species like the hydrogen atom, the energy of an orbital depends solely on the principal quantum number $n$ and is independent of the azimuthal quantum number $l$.
   • Using the Bohr model / quantum mechanical model formula for hydrogen: $E_n(\text{H}) = -\frac{13.6}{n^2} \text{ eV}$
   • For $n = 2$: $E_{2\text{s}}(\text{H}) = E_{2\text{p}}(\text{H}) = -\frac{13.6}{2^2} \text{ eV} = -3.4 \text{ eV}$
   • Therefore, Option (B) is correct.

2. Lithium Atom ($\text{Li}$): Multi-Electron System

   • For multi-electron atoms like lithium ($Z = 3$), the energy of an orbital depends on both $n$ and $l$ due to electron-electron repulsion and shielding (screening) effects.
   • The $2\text{s}$ orbital has greater penetrating power than the $2\text{p}$ orbital, which means a $2\text{s}$ electron spends more time closer to the nucleus and experiences a greater effective nuclear charge ($Z_{\text{eff}}$) than a $2\text{p}$ electron.
   • A higher $Z_{\text{eff}}$ results in stronger electrostatic attraction, making the energy more negative (lower energy level): $E_{2\text{s}}(\text{Li}) < E_{2\text{p}}(\text{Li})$
   • Therefore, Option (A) is correct.

3. Comparison between $\text{H}$ and $\text{Li}$ Orbitals

   • In the hydrogen atom ($Z = 1$), the $2\text{s}$ electron experiences a nuclear charge of $Z = 1$.
   • In the lithium atom ($Z = 3$), the $2\text{s}$ electron is shielded by the two $1\text{s}$ electrons, resulting in an effective nuclear charge $Z_{\text{eff}} \approx 1.3$ (Slater's rules: $Z_{\text{eff}} = 3 - 2 \times 0.85 = 1.3$).
   • Since $Z_{\text{eff}}(\text{Li}) > Z_{\text{eff}}(\text{H})$, the $2\text{s}$ orbital in lithium is more strongly bound (has lower energy) than the $2\text{s}$ orbital in hydrogen: $E_{2\text{s}}(\text{Li}) < E_{2\text{s}}(\text{H}) \implies E_{2\text{s}}(\text{H}) > E_{2\text{s}}(\text{Li})$
   • Numerically, $E_{2\text{s}}(\text{H}) = -3.4 \text{ eV}$ while $E_{2\text{s}}(\text{Li}) = -5.39 \text{ eV}$ (equal to the negative of the first ionization energy of $\text{Li}$).
   • Therefore, Option (D) is correct.

4. Evaluation of Option (C)

   • Since $E_{2\text{p}}(\text{H}) = -3.4 \text{ eV}$ and $E_{2\text{s}}(\text{Li}) = -5.39 \text{ eV}$, we have: $E_{2\text{p}}(\text{H}) > E_{2\text{s}}(\text{Li})$
   • Therefore, Option (C) is incorrect.

Thus, the correct options are (A), (B), and (D).