Question

rank the following orbitals (4p, 2s, 3d, and 3s) in terms of increasing electron energy
4p 2s 3d 3s
○ (lowest) 4p < 3d < 3s < 2s (highest)
○ (lowest) 3s < 2s < 4p < 3d (highest)
○ (lowest) 2s < 3d < 4p < 3s (highest)
○ (lowest) 2s < 3s < 3d < 4p (highest)

Explanation:

To rank the orbitals (4p, 2s, 3d, 3s) by increasing electron energy, we use the Aufbau principle and energy level order. The general order of increasing energy for orbitals is based on the principal quantum number (n) and the angular momentum quantum number (l). Orbitals with lower n (and appropriate l) have lower energy. For s - orbitals, as n increases, energy increases: 2s < 3s. For d - orbitals, 3d has a higher energy than 3s. For p - orbitals, 4p has a higher energy than 3d (since the energy order follows the sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p...). So the order from lowest to highest energy is 2s < 3s < 3d < 4p.

Answer:

(lowest) 2s < 3s < 3d < 4p (highest)