Degeneracy of Second Excited State of Hydride Ion

To determine the degeneracy of the second excited state of the hydride ion ($\text{H}^-$) without considering electronic spin, we analyze the energy level ordering in single-electron versus multi-electron systems.

1. Comparison of Energy Levels

• Single-electron system (e.g., $\text{H}$ atom): The energy depends only on the principal quantum number $n$. Therefore, subshells with the same $n$ are degenerate: $E_{3s} = E_{3p} = E_{3d}$ For $n = 3$, the total number of spatial orbitals (degeneracy) is: $\text{Degeneracy} = 1 \, (3s) + 3 \, (3p) + 5 \, (3d) = 9$

• Multi-electron system (e.g., $\text{H}^-$ ion): The hydride ion ($\text{H}^-$) has $2$ electrons. Due to electron-electron repulsion and shielding effects, the degeneracy of subshells within the same principal energy level is lifted. The energy of an orbital depends on both $n$ and $l$, following the order: $1s < 2s < 2p < 3s < 3p < \dots$

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2. Electronic Configurations of $\text{H}^-$

1. Ground State: Both electrons occupy the lowest energy orbital. $\text{Configuration: } 1s^2$

2. First Excited State: One electron is promoted to the next lowest energy orbital, $2s$. $\text{Configuration: } 1s^1 2s^1$

   • Degeneracy (without spin): $1 \, (1s) \times 1 \, (2s) = 1$

3. Second Excited State: One electron is promoted to the $2p$ orbital. $\text{Configuration: } 1s^1 2p^1$

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3. Calculation of Degeneracy

For the second excited state configuration ($1s^1 2p^1$):

• The $1s$ orbital has $1$ spatial state ($m_l = 0$).
• The $2p$ subshell has $3$ degenerate spatial orbitals ($2p_x, 2p_y, 2p_z$ corresponding to $m_l = -1, 0, +1$).

Without considering electronic spin, the total degeneracy of the second excited state is: $\text{Degeneracy} = 1 \times 3 = 3$