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प्रश्न
Number of angular nodes for 4d orbital is ______.
पर्याय
4
3
2
1
उत्तर
Number of angular nodes for 4d orbital is 2.
Explanation:
The number of angular nodes is given by n – l
Where n is principal quantum number, l is azimuthal quantum number
For 4d orbital, n = 4 and l = 2
Thus, the number of angular nodes = n – l = 4 – 2 = 2
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संबंधित प्रश्न
Using s, p, d notations, describe the orbital with the following quantum numbers n = 3; l =1.
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Silicon (Z = 14)
If n = 3, what are the quantum number l and m?
Indicate the number of unpaired electron in:
Cr (Z = 24)
The designation of a subshell with n = 6 and l = 2 is ____________.
Which one of the following is NOT possible?
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` |
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 the quantum numbers with the information provided by these.
| Quantum number | Information provided |
| (i) Principal quantum number | (a) orientation of the orbital |
| (ii) Azimuthal quantum number | (b) energy and size of orbital |
| (iii) Magnetic quantum number | (c) spin of electron |
| (iv) Spin quantum number | (d) shape of the orbital |
