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प्रश्न
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.
5p, 4d, 5d, 4f, 6s
उत्तर
| Orbitals | s | p | d | f |
| l | 0 | 1 | 2 | 3 |
| Orbitals | n | n + l |
| 5p | 5 | 6 |
| 4d | 4 | 6 |
| 5d | 5 | 7 |
| 4f | 4 | 7 |
| 6s | 6 | 6 |
Thus the increasing order of energy is 4d < 5p < 6s < 4f < 5d.
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संबंधित प्रश्न
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X = 6, 1, −1, `+1/2`
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(i) \[\ce{Na^{+}, Mg^{2+}}\]
(ii) \[\ce{Al3^{+}, O-}\]
(iii) \[\ce{Na+ , O2-}\]
(iv) \[\ce{N3-, Cl-}\]
Match the following
| (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 |
Which of the following is the correct plot for the probability density ψ2 (r) as a function of distance 'r' of the electron from the nucleus for 2s orbitals?
