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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
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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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4s, 3s, 3p, 4d
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Based upon the above information, solve the questions given below:
Which of the following orbitals has the lowest energy?
5p, 5d, 5f, 6s, 6p
Match the following species with their corresponding ground state electronic configuration.
| Atom / Ion | Electronic configuration |
| (i) \[\ce{Cu}\] | (a) 1s2 2s2 2p6 3s2 3p6 3d10 |
| (ii) \[\ce{Cu^{2+}}\] | (b) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 |
| (iii) \[\ce{Zn^{2+}}\] | (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 |
| (iv) \[\ce{Cr^{3+}}\] | (d) 1s2 2s2 2p6 3s2 3p6 3d9 |
| (e) 1s2 2s2 2p6 3s2 3p6 3d3 |
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