Advertisements
Advertisements
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
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 1, l = 0.
Using s, p, d notations, describe the orbital with the following quantum numbers n = 4; l = 2.
State and explain Pauli’s exclusion principle.
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 4, l = 2
Write electronic configurations of \[\ce{Fe, Fe2+, Fe3+}\].
Write condensed orbital notation of electronic configuration of the following element:
Chlorine (Z = 17)
Explain in brief, the significance of the azimuthal quantum number.
Using the concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells, and orbitals.
The number of radial nodes for 3p orbital is ______.
Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals:
| (i) | (a) `n = 3, l = 2, m_l = -2, m_s = - 1/2` |
| (b) `n = 3, l = 2, m_l = -1, m_s = - 1/2` | |
| (ii) | (a) `n = 3, l = 1, m_l = 1, m_s = + 1/2` |
| (b) `n = 3, l = 2, m_l = 1, m_s = + 1/2` | |
| (iii) | (a) `n = 4, l = 1, m_l = 1, m_s = + 1/2` |
| (b) `n = 3, l = 2, m_l = 1, m_s = + 1/2` | |
| (iv) | (a) `n = 3, l = 2, m_l = +2, m_s = - 1/2` |
| (b) `n = 3, l = 2, m_l = +2, m_s = + 1/2` |
In which of the following pairs, the ions are iso-electronic?
(i) \[\ce{Na^{+}, Mg^{2+}}\]
(ii) \[\ce{Al3^{+}, O-}\]
(iii) \[\ce{Na+ , O2-}\]
(iv) \[\ce{N3-, Cl-}\]
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
The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this configuration explained?
What is the difference between the terms orbit and orbital?
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 |
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 |
Choose the INCORRECT statement
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?
