Step-by-step Derivation

1. Analyze the System and Degeneracy: The question asks for the degeneracy of the second excited state of the H⁻ ion, not considering electronic spin. Degeneracy refers to the number of different quantum states that have the same energy level.

2. Contrast H atom with H⁻ ion:

   • H atom: This is a single-electron system. The energy of an electron depends only on the principal quantum number, n. For a given n, all subshells (s, p, d, ...) are degenerate. The question correctly states that for the second excited state (n=3), the degeneracy is 9. This is because the 3s, 3p, and 3d orbitals are all at the same energy level. The total number of orbitals is 1 (from 3s) + 3 (from 3p) + 5 (from 3d) = 9.
   • H⁻ ion: This is a two-electron system. In multi-electron systems, due to electron-electron repulsion and the screening effect, the degeneracy of subshells within the same principal shell is lifted. The energy of an orbital depends on both n and the azimuthal quantum number l. The energy levels follow the Aufbau principle: 1s < 2s < 2p < 3s < 3p < ....

3. Determine the Electronic Configurations of H⁻ ion: We need to identify the electronic configurations for the ground state and the first two excited states to find the second excited state.

   • Ground State: The two electrons of H⁻ will occupy the lowest energy orbital available, which is the 1s orbital. Configuration: 1s²
   • First Excited State: To reach the first excited state, one electron is promoted from the 1s orbital to the next lowest energy orbital, which is 2s. Configuration: 1s¹ 2s¹
   • Second Excited State: To reach the second excited state, one electron is promoted from the 1s orbital to the next available energy level after 2s, which is the 2p subshell. Configuration: 1s¹ 2p¹

4. Calculate the Degeneracy of the Second Excited State: The second excited state has the configuration 1s¹ 2p¹. The energy of this state is determined by the energy of the 2p subshell. The 2p subshell (l=1) consists of three orbitals corresponding to the magnetic quantum numbers $m_l = -1, 0, +1$. These three orbitals ($2p_x$, $2p_y$, $2p_z$) are degenerate, meaning they have the same energy.

   Since the question specifies to ignore electronic spin, the degeneracy is simply the number of these degenerate orbitals. The excited electron can occupy any of the three 2p orbitals, resulting in three states of equal energy.

   Therefore, the degeneracy of the second excited state of H⁻ is 3.

Final Answer

The degeneracy of the second excited state of H⁻ is 3.