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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.
5f, 6d, 7s, 7p
उत्तर
| Orbitals | s | p | d | f |
| l | 0 | 1 | 2 | 3 |
| Orbitals | n | n + l |
| 5f | 5 | 8 |
| 6d | 6 | 8 |
| 7s | 7 | 7 |
| 7p | 7 | 8 |
Thus the increasing order of energy is 7s < 5f < 6d < 7p.
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संबंधित प्रश्न
Give the names of quantum numbers.
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Oxygen (Z = 8)
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Nickel atom can lose two electrons to form \[\ce{Ni^{2+}}\] ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.
What is the difference between the terms orbit and orbital?
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 |
Match species given in Column I with the electronic configuration given in Column II.
| Column I | Column II |
| (i) \[\ce{Cr}\] | (a) [Ar]3d84s0 |
| (ii) \[\ce{Fe^{2+}}\] | (b) [Ar]3d104s1 |
| (iii) \[\ce{Ni^{2+}}\] | (c) [Ar]3d64s0 |
| (iv) \[\ce{Cu}\] | (d) [Ar] 3d54s1 |
| (e) [Ar]3d64s2 |
