Animated Solution for Chemistry - Atomic Structure: Not considering the electronic spin the degeneracy of the second excited state ($n=3$) of H-atom is 9, where the degeneracy of the second excited state of ${\text{H}}^{-}$ is

Imagine you are looking at the simplest atom in the universe: Hydrogen. It has just one proton and one electron. Because there is only one electron, there is absolutely no electron-electron repulsion.

In this single-electron utopia, the energy of an electron depends solely on its principal quantum number, $n$. It doesn't care about the shape of the orbital, whether it's $s$, $p$, or $d$. If they share the same shell, they share the same energy.

But what happens when we add an extra electron to create the ${\text{H}}^{-}$ ion? Everything changes.

Suddenly, we have two electrons repelling each other. This electron-electron repulsion shatters the perfect symmetry. The energy now depends on both the principal quantum number $n$ and the azimuthal quantum number $l$. This is the multi-electron reality, governed by the $(n+l)$ rule.

A Single-Electron Utopia

Let's first map out the energy levels of the neutral Hydrogen atom.

The ground state is simply $n=1$, which contains the $1s$ orbital.

When the electron gets excited, it moves to the first excited state, which is the entire $n=2$ shell. Because energy only depends on $n$, the $2s$ and $2p$ subshells have the exact same energy.

Moving up again, the second excited state is the $n=3$ shell. This shell contains the $3s$, $3p$, and $3d$ subshells, all sitting at the exact same energy level.

Degeneracy is simply the number of orbitals that share the same energy. For the second excited state ($n=3$) of Hydrogen, we have one $3s$ orbital, three $3p$ orbitals, and five $3d$ orbitals.

Adding them up: $1+3+5=9$. This perfectly matches the degeneracy of 9 given in the question!

The Multi-Electron Reality

Now, let's shift our focus to the ${\text{H}}^{-}$ ion. Because it is a multi-electron system, the subshells within the same shell split in energy.

Following the $(n+l)$ rule (Aufbau principle), the energy order becomes:

The ground state is the lowest energy level, which is the $1s$ subshell.

The first excited state is the next available energy level, which is strictly the $2s$ subshell.

And the second excited state? It is the next level up, which is the $2p$ subshell. Notice how different this is from the neutral Hydrogen atom!

We have successfully identified that the second excited state of the ${\text{H}}^{-}$ ion is the $2p$ subshell.

The question asks for the degeneracy of this state, specifically instructing us not to consider electronic spin.

The $2p$ subshell consists of exactly three orbitals: $p_x$, $p_y$, and $p_z$.

Since these three orbitals share the exact same energy, the degeneracy is simply 3.

Our final answer is 3.

But let's think like a true JEE advanced student. What if the question hadn't explicitly said "not considering the electronic spin"?

If we had to consider spin, each of those three $2p$ orbitals could hold two electrons—one with spin up ($+1/2$) and one with spin down ($-1/2$).

In that scenario, the degeneracy would double, becoming $3\times 2=6$.

This is a classic trap! Always read the question carefully to see if spin is included or excluded. Great job navigating through this conceptual nuance!