Which two of the following orbitals are degenerate in a hydrogen atom?

• A. 2p and 3d
• B. 3s and 3p
• C. 1s and 2p
• D. 2s and 3s

Answer: (B)

Verified step by step guidance

1. Recall that in a hydrogen atom, the energy levels depend only on the principal quantum number \(n\), not on the azimuthal quantum number \(l\). This means orbitals with the same \(n\) value are degenerate (have the same energy).
2. Identify the principal quantum number \(n\) for each orbital: 2p has \(n=2\), 3d has \(n=3\), 3s has \(n=3\), 3p has \(n=3\), 1s has \(n=1\), 2p has \(n=2\), 2s has \(n=2\), and 3s has \(n=3\).
3. Compare orbitals with the same \(n\) value to find degenerate pairs. For example, 3s and 3p both have \(n=3\), so they are degenerate.
4. Check other pairs: 2p and 3d have different \(n\) values (2 and 3), so they are not degenerate; 1s and 2p have \(n=1\) and \(n=2\), so not degenerate; 2s and 3s have \(n=2\) and \(n=3\), so not degenerate.
5. Conclude that the only degenerate pair among the options is 3s and 3p, because they share the same principal quantum number \(n=3\).