The 2s and the 2p orbital energies of hydrogen atom are E_2s(H) and E_2p(H), respectively.
The 2s and the 2p orbital energies of lithium atom are E_2s(Li) and E_2p(Li), respectively. The correct option(s) about the orbital energies is(are)

The correct options are:
E_2s(Li) < E_2p(Li)
E_2s(H) = E_2p(H)
E_2s(H) > E_2s(Li)

To understand the relative energies of the 2s and 2p orbitals in hydrogen (H) and lithium (Li) atoms, we need to analyze single-electron systems versus multi-electron systems.

1. Orbital Energies in Hydrogen (Single-electron system):

Hydrogen (H) is a single-electron species. For single-electron species (like H, He⁺, Li²⁺), the energy of an orbital depends strictly on the principal quantum number ($n$) and is independent of the azimuthal quantum number ($l$).

Therefore, for the hydrogen atom where $n=2$:

$E_{2s}\left(\text{H}\right)=E_{2p}\left(\text{H}\right)$

This confirms that E_2s(H) = E_2p(H) is a correct statement.

2. Orbital Energies in Lithium (Multi-electron system):

Lithium (Li) has an atomic number $Z=3$ and contains 3 electrons. In multi-electron systems, electron-electron repulsions and shielding/penetration effects split orbitals of the same principal energy level ($n$) based on their azimuthal quantum number ($l$).

The 2s orbital penetrates closer to the nucleus than the 2p orbital. Because the 2s electron experiences greater penetration, it is less effectively shielded by the inner 1s electrons and feels a higher effective nuclear charge ($Z_{\text{eff}}$). Consequently, the 2s orbital is held more tightly by the nucleus and has lower energy (is more stable) than the 2p orbital.

Therefore, for the lithium atom:

$E_{2s}\left(\text{Li}\right)<E_{2p}\left(\text{Li}\right)$

This confirms that E_2s(Li) < E_2p(Li) is a correct statement.

3. Comparing 2s Orbital Energies of H and Li:

The energy of an electron in an orbital increases (becomes more negative / lower energy) with an increase in the nuclear charge $Z$. The formula for the energy of a hydrogen-like orbital is given by:

$E\propto -\frac{Z^2}{n^2}$

Even after taking shielding into account for multi-electron atoms, the effective nuclear charge experienced by a 2s electron in lithium ($Z_{\text{eff}}>1$) is significantly higher than that in hydrogen ($Z=1$). A higher effective nuclear charge stabilizes the orbital, making its energy more negative (lower in value).

Since the 2s orbital in Li is more strongly bound (more negative in energy) than the 2s orbital in H:

$E_{2s}\left(\text{H}\right)>E_{2s}\left(\text{Li}\right)$

This confirms that E_2s(H) > E_2s(Li) is also a correct statement.