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प्रश्न
Explain the anomalous behaviour of chromium.
उत्तर
- Chromium (Cr) has atomic number 24.
- Its expected electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d4.
- The 3d orbital is less stable as it is not half-filled.
- Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes 1s2 2s2 2p6 3s2 3p6 4s1 3d5.
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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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Explain the anomalous behaviour of copper.
Write orbital notations for the electron in orbitals with the following quantum numbers.
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Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 4, l = 2
Write orbital notations for the electron in orbitals with the following quantum numbers.
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Write condensed orbital notation of electronic configuration of the following element:
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Explain in brief, the significance of the azimuthal quantum number.
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Indicate the number of unpaired electrons in \[\ce{Si}\] (Z = 14).
Indicate the number of unpaired electron in:
Cr (Z = 24)
How many electrons in 19K have n = 3, l = 1?
The three electrons have the following set of quantum numbers:
X = 6, 1, −1, `+1/2`
Y = 6, 0, 0, `+1/2`
Z = 5, 1, 0, `+1/2`
Identify the CORRECT statement.
The number of radial nodes for 3p orbital is ______.
Total number of orbitals associated with third shell will be ______.
Orbital angular momentum depends on ______.
The pair of ions having same electronic configuration is ______.
Nickel atom can lose two electrons to form \[\ce{Ni^{2+}}\] ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.
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.
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Based upon the above information, arrange the following orbitals in the increasing order of energy.
5f, 6d, 7s, 7p
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Based upon the above information, solve the questions given below:
Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
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 |
In the case of R, S configuration the group having the highest priority is ______.
