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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.
4s, 3s, 3p, 4d
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Solution
| Orbitals | s | p | d | f |
| l | 0 | 1 | 2 | 3 |
| Orbitals | n | n + l |
| 4s | 4 | 4 |
| 3s | 3 | 3 |
| 3p | 3 | 4 |
| 4d | 4 | 6 |
Thus the increasing order of energy is 3s < 3p < 4s < 4d.
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| (i) Photon | (a) Value is 4 for N shell |
| (ii) Electron | (b) Probability density |
| (iii) ψ2 | (c) Always positive value |
| (iv) Principal quantum number n | (d) Exhibits both momentum and wavelength |
Choose the INCORRECT statement
