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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.
1s, 2s, 3s, 2p
उत्तर
| 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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संबंधित प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 4; l =3.
State Hund’s rule of maximum multiplicity with a suitable example.
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Write condensed orbital notation of electronic configuration of the following element:
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Write condensed orbital notation of electronic configuration of the following element:
Calcium (Z = 20)
Draw shapes of 2s orbitals.
The electronic configuration of oxygen is written as 1s2 2s2 \[\ce{2p^2_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^1_{{z}}}\] and not as 1s2 2s2 \[\ce{2p^2_{{x}}}\], \[\ce{2p^2_{{y}}}\] \[\ce{2p^0_{{z}}}\], Explain.
Which one of the following orders is CORRECT in case of energy of the given subshells?
P: n = 4; l = 3
Q: n = 5; I = 1
R: n = 5; l = 0
S: n = 4; l = 2
How many electrons can fit in the orbital for which n = 4 and l = 2?
Number of angular nodes for 4d orbital is ______.
