Key principle for hydrogen atom

For a hydrogen atom, the energy of an orbital depends only on the principal quantum number $n$.

$E_n = -\frac{13.6}{n^2}\ \text{eV}$

So, all orbitals having the same $n$ are degenerate (have the same energy), irrespective of $l$ or orientation.

Identify the principal quantum number of each orbital

• $(A)\;4s \Rightarrow n=4$
• $(B)\;3p_x \Rightarrow n=3$
• $(C)\;3d_{x^2-y^2} \Rightarrow n=3$
• $(D)\;3d_{z^2} \Rightarrow n=3$
• $(E)\;4p_z \Rightarrow n=4$

Compare energies

Since lower $n$ means lower energy in hydrogen atom:

$E_3 = -\frac{13.6}{9}, \qquad E_4 = -\frac{13.6}{16}$

Because

$-\frac{13.6}{9} < -\frac{13.6}{16}$

the orbitals with $n=3$ are lower in energy than those with $n=4$.

Find the lowest-energy orbitals among the given choices

The orbitals with lowest energy are:

• $3p_x$
• $3d_{x^2-y^2}$
• $3d_{z^2}$

Thus the correct set is (B), (C) and (D) only.

Match with given options

This corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

Hence, they agree.