For a hydrogen atom, which of the following sets of orbitals represent a degenerate group (i.e., all

Question

For a hydrogen atom, which of the following sets of orbitals represent a degenerate group (i.e., all orbitals within the set have the same energy )?

Professor PC Thomas Classes · Answer

The energy of an electron in a hydrogen atom (a one-electron system) depends solely on its principal quantum number, $n$. This means that all orbitals within the same principal energy level (i.e., having the same $\boldsymbol{n}$ value) are degenerate, regardless of their azimuthal quantum number ( $\boldsymbol{l}$ ) or magnetic quantum number ( $\boldsymbol{m}_{\boldsymbol{l}}$ ).

Let's analyze the given options:
A) $2 s, 2 p_x, 3 s$ : The principal quantum numbers are $n=2$ for $2 s$ and $2 p_x$, but $n=3$ for $3 s$. Since the $n$ values are not all the same, these orbitals are not degenerate.
B) $3 p_y, 3 d_{x y}, 3 s$ : The principal quantum numbers for all these orbitals are $n=3$. Since they all have the same principal quantum number, they are degenerate in a hydrogen atom.
C) $1 s, 2 s, 3 s$ : The principal quantum numbers are $n=1$ for $1 s, n=2$ for $2 s$, and $n=3$ for $3 s$. Since the $n$ values are different, these orbitals are not degenerate.
D) $4 s, 4 p_z, 5 d_{z^2}$ : The principal quantum numbers are $n=4$ for $4 s$ and $4 p_{z^{\prime}}$, but $n=5$ for $5 d_{z^2}$. Since the $n$ values are not all the same, these orbitals are not degenerate.
Therefore, the set of orbitals $3 p_y, 3 d_{x y}, 3 s$ represents a degenerate group for a hydrogen atom.
The final answer is □ B.