The effective nuclear charge, \( Z_{\text{eff}} \), is used to estimate the energy of an electron in a given orbital. The energy of an electron in an orbital can be estimated using the formula:
\[
\epsilon_n = -\frac{Z_{\text{eff}}^2 \cdot \text{Ry}}{n^2}
\]
where \( \text{Ry} \) is the Rydberg energy (13.6 eV) and \( n \) is the principal quantum number of the orbital.
For the \( 1s \) orbital, \( n = 1 \) and \( Z_{\text{eff}} = 4.68 \):
\[
\epsilon_{1s} = -\frac{(4.68)^2 \cdot 13.6}{1^2} = -21.9024 \, \text{eV}
\]
For the \( 2s \) orbital, \( n = 2 \) and \( Z_{\text{eff}} = 2.58 \):
\[
\epsilon_{2s} = -\frac{(2.58)^2 \cdot 13.6}{2^2} = -1.6647 \, \text{eV}
\]
For the \( 2p \) orbital, \( n = 2 \) and \( Z_{\text{eff}} = 2.42 \):
\[
\epsilon_{2p} = -\frac{(2.42)^2 \cdot 13.6}{2^2} = -1.4641 \, \text{eV}
\]
The average distance of an electron from the nucleus in terms of the Bohr radius \( a_0 \) can be estimated using:
\[
\bar{r}_{n,\ell} = \frac{n^2}{Z_{\text{eff}}} \cdot a_0
\]
\[
\bar{r}_{1,0} = \frac{1^2}{4.68} \cdot a_0 = 0.2137 \, a_0
\]
\[
\bar{r}_{2,0} = \frac{2^2}{2.58} \cdot a_0 = 1.5504 \, a_0
\]
- Energy of \( 1s \) orbital: \(\boxed{-21.9024 \, \text{eV}}\)
- Energy of \( 2s \) orbital: \(\boxed{-1.6647 \, \text{eV}}\)
- Energy of \( 2p \) orbital: \(\boxed{-1.4641 \, \text{eV}}\)
- Average distance for \( 1s \) orbital: \(\boxed{0.2137 \, a_0}\)
- Average distance for \( 2s \) orbital: \(\boxed{1.5504 \, a_0}\)