The 2s and the 2p orbital energies of hydrogen atom are E_ 2s (H)...

For hydrogen atom (a single-electron system), the energy of an orbital depends only on the principal quantum number $n$. Thus, the 2s and 2p orbitals are degenerate.

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

For lithium atom (a multi-electron system), the energy depends on both $n$ and $l$. Due to the penetration effect, the 2s electron penetrates closer to the nucleus and experiences a higher effective nuclear charge than the 2p electron. Thus, the 2s orbital has lower energy than the 2p orbital.

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

The energy of an orbital is given by $E \propto -\dfrac{Z_{\text{eff}}^2}{n^2}$.

For hydrogen, $Z_{\text{eff}} = 1$. For the 2s electron in lithium, the nuclear charge is $Z=3$, and it is shielded by the $1s^2$ electrons. The effective nuclear charge $Z_{\text{eff}}$ for the 2s electron in Li is greater than 1 (approximately 1.3).

Since $Z_{\text{eff}}(\text{Li}, 2s) > Z_{\text{eff}}(\text{H}, 2s)$, the energy of the 2s orbital in Li is more negative than that in H.

$E_{2s}(\text{Li}) < E_{2s}(\text{H}) \Rightarrow E_{2s}(\text{H}) > E_{2s}(\text{Li})$

Since $E_{2p}(\text{H}) = E_{2s}(\text{H})$, we also have $E_{2p}(\text{H}) > E_{2s}(\text{Li})$.

Answer: $E_{2s}(\text{Li}) < E_{2p}(\text{Li})$; $E_{2s}(\text{H}) = E_{2p}(\text{H})$; $E_{2s}(\text{H}) > E_{2s}(\text{Li})$