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Question
The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
1s, 2s, 3s, 2p
Solution
| Orbitals | s | p | d | f |
| l | 0 | 1 | 2 | 3 |
| Orbitals | n | n + l |
| 1s | 1 | 1 |
| 2s | 2 | 2 |
| 3s | 3 | 3 |
| 2p | 2 | 3 |
Thus the increasing order of energy is 1s < 2s < 2p < 3s
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