Advertisements
Advertisements
प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 4; l = 2.
उत्तर
For n = 4 and l = 2
The orbital is 4d.
APPEARS IN
संबंधित प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 1, l = 0.
State the order of filling atomic orbitals following Aufbau principle.
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 3, l = 2
If n = 3, what are the quantum number l and m?
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.
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.
Which mineral among the following contains vanadium in it?
The designation of a subshell with n = 6 and l = 2 is ____________.
How many electrons can fit in the orbital for which n = 4 and l = 2?
The number of radial nodes for 3p orbital is ______.
Which of the following sets of quantum numbers are correct?
| `n` | `l` | `m_l` | |
| (i) | 1 | 1 | +2 |
| (ii) | 2 | 1 | +1 |
| (iii) | 3 | 2 | –2 |
| (iv) | 3 | 4 | –2 |
Which of the following orbitals are degenerate?
3dxy, 4dxy 3dz2, 3dyz, 4dyz, 4dz2
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
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, solve the questions given below:
Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
The electronic configuration of valence shell of Cu is 3d104s1 and not 3d94s2. How is this configuration explained?
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 |
